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Vectors, EXT1 EQ-Bank 40

Determine the component of  \(\textbf{a} = 2\textbf{i}-\textbf{j} + 3\textbf{k}\)  that is perpendicular to  \(\textbf{b} = \textbf{i} + \textbf{j}-\textbf{k}.\)   (3 marks)

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\(\textbf{a}-\operatorname{proj}_{\textbf{b}}\textbf{a}=\left(\begin{array}{c}2 \\ -1 \\ 3\end{array}\right)+\dfrac{2}{3}\left(\begin{array}{c}1 \\ 1 \\ -1\end{array}\right)=\left(\begin{array}{c}2 \frac{2}{3} \\ -\frac{1}{3} \\ 2 \frac{1}{3}\end{array}\right)\)

Show Worked Solution

\(\textbf{a}=\left(\begin{array}{c}2 \\ -1 \\ 3\end{array}\right), \ \textbf{b}=\left(\begin{array}{c}1 \\ 1 \\ -1\end{array}\right)\)

\(\textbf{a} \cdot \textbf{b}=2-1-3=-2\)

\(\abs{\textbf{b}}^2=1^2+1^2+(-1)^2=3\)
 

\(\text{Projection of} \ \textbf{a} \ \text{in the direction of} \  \textbf{b}\):

\(\operatorname{proj}_{\textbf{b}} \textbf{a}=\left(\dfrac{\textbf{a} \cdot \textbf{b}}{\abs{\textbf{b}}^2}\right) \textbf{b}=-\dfrac{2}{3}\left(\begin{array}{c}1 \\ 1 \\ -1\end{array}\right)\)
 

\(\text {Component of} \ \textbf{a} \ \text {that is perpendicular to} \ \textbf{b}\):

\(\textbf{a}-\operatorname{proj}_{\textbf{b}}\textbf{a}=\left(\begin{array}{c}2 \\ -1 \\ 3\end{array}\right)+\dfrac{2}{3}\left(\begin{array}{c}1 \\ 1 \\ -1\end{array}\right)=\left(\begin{array}{c}2 \frac{2}{3} \\ -\frac{1}{3} \\ 2 \frac{1}{3}\end{array}\right)\)

Vectors, EXT1 EQ-Bank 8 MC

If  \(\underset{\sim}{u}=2 \underset{\sim}{i}-2 j+\underset{\sim}{k}\)  and  \(\underset{\sim}{v}=3 \underset{\sim}{i}-6 j+2 \underset{\sim}{k}\), the projection of \(\underset{\sim}{v}\) onto \(\underset{\sim}{u}\) is

  1. \(\dfrac{20}{49}(3 \underset{\sim}{i}-6 \underset{\sim}{j}+2 \underset{\sim}{k})\)
  2. \(\dfrac{20}{3}(2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k})\)
  3. \(\dfrac{20}{7}(3 \underset{\sim}{i}-6 \underset{\sim}{j}+2 \underset{\sim}{k})\)
  4. \(\dfrac{20}{9}(2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k})\)
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\(D\)

Show Worked Solution

\(\underset{\sim}{u}=\left(\begin{array}{c}2 \\ -2 \\ 1\end{array}\right), \ \ \underset{\sim}{v}=\left(\begin{array}{c}3 \\ -6 \\ 2\end{array}\right)\)

\(\underset{\sim}{u} \cdot \underset{\sim}{v}=6+12+2=20\)

\(\abs{\underset{\sim}{u}}^2=2^2+(-2)^2+1^2=9\)

\(\operatorname{proj}_{\underset{\sim}{u}} \underset{\sim}{v}=\left(\dfrac{\underset{\sim}{u} \cdot \underset{\sim}{v}}{\abs{\underset{\sim}{u}}^2}\right) \underset{\sim}{u}=\dfrac{20}{9}\left(\begin{array}{c}2 \\ -2 \\ 1\end{array}\right)\)

\(\Rightarrow D\)

Vectors, EXT1 EQ-Bank 3 MC

Given that  \(\overrightarrow{OP}=\left(\begin{array}{c}-3 \\ 1 \\ -1\end{array}\right)\)  and  \(\overrightarrow{O Q}=\left(\begin{array}{c}2 \\ 5 \\ -3\end{array}\right)\), what is \(\overrightarrow{P Q}\) ?

  1. \(\left(\begin{array}{c}1 \\ -6 \\ 4\end{array}\right)\)
  2. \(\left(\begin{array}{c}-1 \\ 6 \\ -4\end{array}\right)\)
  3. \(\left(\begin{array}{c}5 \\ 4 \\ -2\end{array}\right)\)
  4. \(\left(\begin{array}{c}-5 \\ -4 \\ 2\end{array}\right)\)
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\(C\)

Show Worked Solution

\(\overrightarrow{PQ}=\overrightarrow{O Q}-\overrightarrow{O P}=\left(\begin{array}{c}2 \\ 5 \\ -3\end{array}\right)-\left(\begin{array}{c}-3 \\ 1 \\ -1\end{array}\right)=\left(\begin{array}{c}5 \\ 4 \\ -2\end{array}\right)\)

\(\Rightarrow C\)

Vectors, EXT1 EQ-Bank 21

Given the vectors  \(\textbf{a} = \textbf{i}+3\textbf{j}\)  and  \(\textbf{b} =4\textbf{i} +2\textbf{j}\), find the projection of \(\textbf{a}\) onto \(\textbf{b}\).   (2 marks)

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\(\operatorname{proj}_{\textbf{b}}\textbf{a}=\left(\dfrac{\textbf{a} \cdot \textbf{b}}{\abs{\textbf{b}}^2}\right) \textbf{b}=\left(\dfrac{10}{20}\right) \textbf{b}=\dfrac{1}{2} \displaystyle \binom{4}{2}=\binom{2}{1}\)

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\(\displaystyle \textbf{a}=\binom{1}{3}, \ \ \textbf{b}=\binom{4}{2}\)

\(\textbf{a}\cdot \textbf{b}=1 \times 4+3 \times 2=10\)

\(\abs{\textbf{b}}^2=4^2+2^2=20\)

\(\operatorname{proj}_{\textbf{b}}\textbf{a}=\left(\dfrac{\textbf{a} \cdot \textbf{b}}{\abs{\textbf{b}}^2}\right) \textbf{b}=\left(\dfrac{10}{20}\right) \textbf{b}=\dfrac{1}{2} \displaystyle \binom{4}{2}=\binom{2}{1}\)

Vectors, EXT1 EQ-Bank 38

Let \(\underset{\sim}{a}=2 \underset{\sim}{i}-3 j+\underset{\sim}{k}\) and \(\underset{\sim}{b}=\underset{\sim}{i}+m j-\underset{\sim}{k}\), where \(m\) is an integer.

The vector resolute of \(\underset{\sim}{a}\) in the direction of \(\underset{\sim}{b}\) is \(-\dfrac{11}{18}(\underset{\sim}{i}+m\underset{\sim}{j}-\underset{\sim}{k})\).

  1. Find the value of  \(m\).   (3 marks)

  2. Find the component of \(\underset{\sim}{a}\) that is perpendicular to \(\underset{\sim}{b}\).   (1 mark)

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a.    \(m=4\)

b.    \(\left(\begin{array}{c}2 \frac{11}{18} \\ -\frac{5}{9} \\ \frac{7}{18}\end{array}\right)\)

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a.    \(\underset{\sim}{a}=\left(\begin{array}{c}2 \\ -3 \\ 1\end{array}\right), \quad \underset{\sim}{b}=\left(\begin{array}{c}1 \\ m \\ -1\end{array}\right)\)

\(\underset{\sim}{b} \cdot \underset{\sim}{a}=2-3 m-1=1-3 m\)

\(\abs{\underset{\sim}{b}}=\sqrt{1^2+m^2+(-1)^2}=\sqrt{2+m^2}\)

\(\operatorname{proj}_{\underset{\sim}{b}}\underset{\sim}{a}=\left(\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\abs{b}^2}\right) \underset{\sim}{b}=\dfrac{1-3 m}{2+m^2}\, \underset{\sim}{b}\)
 

\(\text{Equating projection vectors:}\)

\(\dfrac{1-3 m}{m^2+2}\) \(=-\dfrac{11}{18}\)  
\(18-54 m\) \(=-11 m^2-22\)  
\(11 m^2-54 m+40\) \(=0\)  
\((11 m-10)(m-4)\) \(=0\)  

 
\(\therefore m=4\ \left(m \neq \frac{10}{11}, m \in Z\right)\)
 

b.    \(\text{Component of \(\underset{\sim}{a}\) perpendicular to \(\underset{\sim}{b}\):}\)

\(\underset{\sim}{a}-\operatorname{proj}_{\underset{\sim}{b}} \underset{\sim}{a}=\left(\begin{array}{c}2 \\ -3 \\ 1\end{array}\right)+\dfrac{11}{18}\left(\begin{array}{c}1 \\ 4 \\ -1\end{array}\right)=\left(\begin{array}{c}2 \frac{11}{18} \\ -\frac{5}{9} \\ \frac{7}{18}\end{array}\right)\)

Vectors, EXT1 EQ-Bank 33

Let  `underset ~a = 3 underset ~i-2 underset ~j + m underset ~k`  and  `underset ~b = 2 underset ~i-underset ~j + 3 underset ~k`, where  `m in R`.

Find the value(s) of `m` such that the projection of `underset ~a` onto `underset ~b` has magnitude `sqrt 14`.   (3 marks)

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`m = -22/3, 2`

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\(\underset{\sim}{a}=\left(\begin{array}{c}3 \\ -2 \\ m\end{array}\right), \quad \underset{\sim}{b}=\left(\begin{array}{c}2 \\ -1 \\ 3\end{array}\right)\)

\(\underset{\sim}{a} \cdot \underset{\sim}{b}=6+2+3 m=8+3 m\)

\(\abs{\underset{\sim}{b}}=\sqrt{2^2+(-1)^2+3^2}=\sqrt{14}\)
 

\(\text{Since magnitude of projection}=\sqrt{14}\):

\(\abs{\operatorname{proj}_{\underset{\sim}{b}} \underset{\sim}{a}}=\dfrac{\abs{\underset{\sim}{a} \cdot \underset{\sim}{b}}}{\abs{\underset{\sim}{b}}}=\dfrac{\abs{8+3 m}}{\sqrt{14}}=\sqrt{14}\)

\(\abs{8+3 m}\) \(=14\)
\(8+3 m\) \(= \pm 14\)
\(3 m\) \(=-8 \pm 14\)
\(m\) \(=-\dfrac{22}{3}, 2\)

Vectors, EXT1 2017 NHT 10

Consider the vectors  `underset ~a =-underset ~i-2 underset ~j + 3 underset ~k`  and  `underset ~b = 2 underset ~i + c underset ~j + underset ~k`.

Find the value of `c` if the angle between `underset ~a` and `underset ~b` is `pi/3`.   (4 marks)

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`c = -3`

Show Worked Solution
`underset ~a ⋅ underset ~b` `= -1 xx 2 + (-2) xx c + 3 xx 1`
  `= -2-2c + 3`
  `= 1-2c`

 

`1-2c` `= sqrt((-1)^2 + (-2)^2 + 3^3) *sqrt(2^2 + c^2 + 1^2) xx cos (pi/3)`
`1-2c` `= 1/2(sqrt 14 ⋅ sqrt(5 + c^2))`
`2-4c` `= sqrt(14(5 + c^2))`
`(2-4c)^2` `= 14(5 + c^2)`
`4-16c + 16c^2` `= 70 + 14c^2`
`2c^2-16c-66` `= 0`
`c^2-8c-33` `= 0`
`(c-11)(c + 3)` `= 0`

 
`c = 11 or c = -3`

`text(S)text(ince)\ \ 2-4c = sqrt(15(5 + c^2))`

`2-4c > 0\ \ =>\ \ c<2`

`:. c = -3`

Vectors, EXT1 2024 SPEC1 4

Consider the vectors  \(\underset{\sim}{ a }=3 \underset{\sim}{ j }+3 \underset{\sim}{ k }\)  and  \(\underset{\sim}{ b }=2 \underset{\sim}{ i }-\underset{\sim}{ j }-2 \underset{\sim}{ k }\).

Find the angle between \(\underset{\sim}{ a }\) and \(\underset{\sim}{ b }\).   (2 marks)

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\(\theta=\dfrac{3 \pi}{4}\left(\text{or} \ 135^{\circ}\right)\)

Show Worked Solution

\(\underset{\sim}{a}=\left(\begin{array}{l}0 \\ 3 \\ 3\end{array}\right) \Rightarrow \abs{\underset{\sim}{a}}=\sqrt{18}=3 \sqrt{2}\)

     \(\underset{\sim}{b}=\left(\begin{array}{c}2 \\ -1 \\ -2\end{array}\right) \Rightarrow\abs{\underset{\sim}{b}}=\sqrt{9}=3\)

     \(\cos \theta=\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\abs{\underset{\sim}{a}} \cdot \abs{\underset{\sim}{b}}}=\dfrac{-3-6}{3 \sqrt{2} \times 3}=-\dfrac{1}{\sqrt{2}}\)

    \(\therefore \theta=\cos ^{-1}\left(-\dfrac{1}{\sqrt{2}}\right)=\dfrac{3 \pi}{4}\left(\text{or} \ 135^{\circ}\right)\)

Vectors, EXT1 2014 SPEC1 1

Consider the vector  `underset ~a = sqrt 3 underset ~i-underset ~j-sqrt 2 underset ~k`, where `underset ~i, underset ~j` and `underset ~k` are unit vectors in the positive directions of the `x, y` and `z` axes respectively.

  1. Find the unit vector in the direction of  `underset ~a`.   (1 mark)

  2. Find the acute angle that `underset ~a` makes with the positive direction of the `x`-axis.   (2 marks)

  3. The vector  `underset ~b = 2 sqrt 3 underset ~i + m underset ~j-5 underset ~k`.
  4. Given that `underset ~b` is perpendicular to `underset ~a,` find the value of `m`.  (2 marks)

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a.    `1/sqrt 6 (sqrt 3 underset ~i-underset ~j-sqrt 2 underset ~k)`

b.    `theta = 45^@`

c.    `m = 6 + 5 sqrt 2`

Show Worked Solution

a.    `|underset ~a|= sqrt((sqrt 3)^2 + (-1)^2 + (-sqrt 2)^2)= sqrt 6`

`hat underset ~a= underset ~a/|underset ~a|= 1/sqrt 6 (sqrt 3 underset ~i-underset ~j-sqrt 2 underset ~k)`
 

b.    `x text{-axis vectors include}\ (1,0,0).`

`underset ~a ⋅ underset ~i = ((\sqrt3),(-1),(-\sqrt2))((1),(0),(0))=\sqrt3`

  `underset ~a ⋅ underset ~i` `= |underset ~a||underset ~i| cos theta= sqrt 6 cos theta`
  `sqrt 3` `= sqrt 6 cos theta`
  `cos theta` `=1/sqrt 2`
  `:. theta` `= 45^@`

 
c.   `underset ~a ⋅ underset ~b = sqrt 3 (2 sqrt 3) + (-1)(m) + (-sqrt 2)(-5) = 0`

`6-m + 5 sqrt 2` `=0`  
`:. m` `=6 + 5 sqrt 2`  

Vectors, EXT1* V1 2025 HSC 11d

  1. Force \({\underset{\sim}{F}}_1\) has magnitude 12 newtons in the direction of vector  \(2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}\).   
  2. Show that  \({\underset{\sim}{F}}_1=8 \underset{\sim}{i}-8 \underset{\sim}{j}+4 \underset{\sim}{k}\).   (1 mark)

  3. Force \({\underset{\sim}{F}}_1\) from part (i) and a second force,  \({\underset{\sim}{F}}_2=-6 \underset{\sim}{i}+12 \underset{\sim}{j}+4 \underset{\sim}{k}\), both act upon a particle.
  4. Show that the resultant force acting on the particle is given by:
  5.      \({\underset{\sim}{F}}_3=2 \underset{\sim}{i}+4 \underset{\sim}{j}+8 \underset{\sim}{k}.\)   (1 mark)

  6. Calculate  \({\underset{\sim}{F}}_3 \cdot \underset{\sim}{d}\), where \({\underset{\sim}{F}}_3\) is the resultant force from part (ii) and  \(\underset{\sim}{d}=\underset{\sim}{i}+\underset{\sim}{j}+2 \underset{\sim}{k}\).   (1 mark)

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i.    \(\text{Unit vector of the direction vector:}\)

\(\dfrac{2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}}{\abs{2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}}} = \dfrac{2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}}{\sqrt{2^2+(-2)^2 + 1^2}} = \dfrac{1}{3} \left( 2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k} \right)\)
 

\(\text{Since \({\underset{\sim}{F}}_1\) has magnitude 12:}\)

\({\underset{\sim}{F}}_1=12 \times \dfrac{1}{3}\left(\begin{array}{c}2 \\ -2 \\ 1\end{array}\right)=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)\)

\({\underset{\sim}{F}}_1=8\underset{\sim}{i}-8 \underset{\sim}{j}+4 \underset{\sim}{k}\)
    

ii.    \({\underset{\sim}{F}}_3={\underset{\sim}{F}}_1+{\underset{\sim}{F}}_2=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)+\left(\begin{array}{c}-6 \\ 12 \\ 4\end{array}\right)=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)\)

\({\underset{\sim}{F}}_3=2 \underset{\sim}{i}+4 \underset{\sim}{j}+8 \underset{\sim}{k}\)
 

iii.  \({\underset{\sim}{F}}_3 \cdot d=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=2+4+16=22\)

Show Worked Solution

i.    \(\text{Unit vector of the direction vector:}\)

\(\dfrac{2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}}{\abs{2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}}} = \dfrac{2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}}{\sqrt{2^2+(-2)^2 + 1^2}} = \dfrac{1}{3} \left( 2 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k} \right)\)
 

\(\text{Since \({\underset{\sim}{F}}_1\) has magnitude 12:}\)

\({\underset{\sim}{F}}_1=12 \times \dfrac{1}{3}\left(\begin{array}{c}2 \\ -2 \\ 1\end{array}\right)=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)\)

\({\underset{\sim}{F}}_1=8\underset{\sim}{i}-8 \underset{\sim}{j}+4 \underset{\sim}{k}\)
 

ii.    \({\underset{\sim}{F}}_3={\underset{\sim}{F}}_1+{\underset{\sim}{F}}_2=\left(\begin{array}{c}8 \\ -8 \\ 4\end{array}\right)+\left(\begin{array}{c}-6 \\ 12 \\ 4\end{array}\right)=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)\)

\({\underset{\sim}{F}}_3=2 \underset{\sim}{i}+4 \underset{\sim}{j}+8 \underset{\sim}{k}\)
 

iii.  \({\underset{\sim}{F}}_3 \cdot d=\left(\begin{array}{l}2 \\ 4 \\ 8\end{array}\right)\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=2+4+16=22\)

Vectors, EXT1* V1 2024 HSC 11c

Find the angle between the two vectors  \(\underset{\sim}{u}=\left(\begin{array}{c}1 \\ 2 \\ -2\end{array}\right)\) and  \(\underset{\sim}{v}=\left(\begin{array}{c}4 \\ -4 \\ 7\end{array}\right)\), giving your answer in radians, correct to 1 decimal place.   (2 marks)

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\(\theta=2.3^c \ \ \text{(1 d.p.)}\)

Show Worked Solution

\(\underset{\sim}{u}=\left(\begin{array}{c}1 \\ 2 \\ -2\end{array}\right),\abs{\underset{\sim}{u}}=\sqrt{1+4+4}=3\)

\(\underset{\sim}{v}=\left(\begin{array}{c}4 \\ -4 \\ 7\end{array}\right),\abs{\underset{\sim}{v}}=\sqrt{16+16+49}=9\)

\(\cos \theta=\dfrac{\underset{\sim}{u} \cdot \underset{\sim}{v}}{|\underset{\sim}{u}||\underset{\sim}{v}|}=\dfrac{1 \times 4-2 \times 4-2 \times 7}{3 \times 9}=-\dfrac{2}{3}\)

\(\theta=\cos ^{-1}\left(-\dfrac{2}{3}\right)=2.30 \ldots=2.3^c \ \ \text{(1 d.p.)}\)

Vectors, EXT1* V1 2022 HSC 11d

A triangle is formed in three-dimensional space with vertices `A(1,-1,2)`, `B(0,2,-1)`  and `C(2,1,1)`.

Find the size of `/_ABC`, giving your answer to the nearest degree.   (3 marks)

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`33°`

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`vec(BA)=((1),(-1),(2))-((0),(2),(-1))=((1),(-3),(3))`

`abs(vec(BA))=sqrt(1^2+3^2+3^2)=sqrt19`
 

`vec(BC)=((2),(1),(1))-((0),(2),(-1))=((2),(-1),(2))`

`abs(vec(BC))=sqrt(2^2+1^2+2^2)=sqrt9=3`
 

`vec(BA)*vec(BC)=1xx2+ -3xx-1+3xx2=11`

`cos/_ABC=(vec(BA)*vec(BC))/(abs{vec(BA)}abs{vec(BC)})=11/(3sqrt19)`

`:./_ABC=cos^(-1)(11/(3sqrt19))=32.733…=33°\ \ text{(nearest degree)}`

Vectors, EXT1* V1 2020 HSC 11d

Consider the two vectors  `underset~u = 2 underset~i-underset~j + 3 underset~k`  and  `underset~v = p underset~i +  underset~j + 2 underset~k`.
 
For what values of `p` are  `underset~u-underset~v`  and  `underset~u + underset~v`  perpendicular?   (3 marks)

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`p= ± 3`

Show Worked Solution

`underset~u-underset~v = ((-2),(-1),(3))-((p),(1),(2)) = ((-2-p),(-2),(1))`
 

`underset~u + underset~v = ((-2),(-1),(3)) + ((p),(1),(2)) = ((p-2),(0),(5))`
 

`⊥ \ text{when} \ \ (underset~u-underset~v) · (underset~u + underset~v ) = 0 :`
 

`((-2-p),(-2),(1)) · ((p-2),(0),(5)) = 0`
 

`-(p + 2)(p-2) + 5` `= 0`
`-(p^2-4) + 5` `= 0`
`-p^2 + 9` `= 0`
`p^2` `= 9`
`p` `= ± 3`

Vectors, EXT1* V1 2024 HSC 12a

The vector \(\underset{\sim}{a}\) is \(\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)\) and the vector \(\underset{\sim}{b}\) is \(\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)\).

  1. Find \(\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}\).   (1 mark)

  2. Show that  \(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}\)  is perpendicular to \(\underset{\sim}{b}\).   (2 marks)

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i.     \(\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)\)

ii.    \(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)-\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\)

\( \left(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\underset{\sim}{b}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right) = -2+0+2=0\)

\(\therefore\ \text {Vectors are perpendicular.}\)

Show Worked Solution

i.    \(\underset{\sim}{a}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right), \quad \underset{\sim}{b}=\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)\)
 

\(\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}=\dfrac{2+0-12}{4+0+16}\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)=-\dfrac{1}{2}\left(\begin{array}{c}2 \\ 0 \\ -4\end{array}\right)=\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)\)

 
ii.    \(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\, \underset{\sim}{b}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)-\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\)
 

\( \left(\underset{\sim}{a}-\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\underset{\sim}{b} \cdot \underset{\sim}{b}}\,\underset{\sim}{b}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right)\left(\begin{array}{c}-1 \\ 0 \\ 2\end{array}\right) = -2+0+2=0\)

 
\(\therefore\ \text{Vectors are perpendicular.}\)

Vectors, EXT1* V1 2023 HSC 11b

Find the angle between the vectors

\(\underset{\sim}{a}=\underset{\sim}{i}+2 \underset{\sim}{j}-3 \underset{\sim}{k}\)

\(\underset{\sim}{b}=-\underset{\sim}{i}+4 \underset{\sim}{j}+2 \underset{\sim}{k}\),

giving your answer to the nearest degree.   (3 marks)

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\(87^{\circ} \)

Show Worked Solution

\[\underset{\sim}{a}=\left(\begin{array}{c} 1 \\ 2 \\ -3 \end{array}\right),\ \  \underset{\sim}{b}=\left(\begin{array}{c} -1 \\ 4 \\ 2 \end{array}\right) \]

\(\Big{|} \underset{\sim}{a} \Big{|} = \sqrt{1+4+9} = \sqrt{14} \)

\(\Big{|} \underset{\sim}{b} \Big{|} = \sqrt{1+16+4} = \sqrt{21} \)

\( \underset{\sim}{a} \cdot \underset{\sim}{b} = -1 + 8-6=1 \)

\(\cos\ \theta \) \(=\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\Big{|}\underset{\sim}{a}\Big{|} \cdot \Big{|}\underset{\sim}{b}\Big{|}} \)  
  \(=\dfrac{1}{\sqrt{294}} \)  
\( \theta\) \(=\cos ^{-1} \Big{(}\dfrac{1}{\sqrt{294}}\Big{)} \)  
  \(=86.65…\)  
  \(=87^{\circ} \)  

Vectors, EXT1* V1 2021 HSC 11c

Find the angle between the vectors  `underset~a = ((2),(0),(4))`  and  `underset~b = ((-3),(1),(2))`, giving the angle in degrees correct to 1 decimal place.   (3 marks)

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`83.1^@`

Show Worked Solution

`underset~a = ((2),(0),(4)) \ , \ |underset~a| \ = sqrt{2^2 + 4^2} = sqrt20`

`underset~b = ((-3),(1),(2)) \ , \ |underset~b| \ = sqrt{(-3)^2 + 1^2 + 2^2} = sqrt14`

`underset~a * underset~b` `= ((2),(0),(4)) ((-3),(1),(2)) = – 6 + 0 + 8 = 2`
`underset~a * underset~b` `= |underset~a| |underset~b| \ cos theta`
`2` `= sqrt20 sqrt14 \ cos theta`
`cos theta` `= 2/sqrt280`
`theta` `= cos^(-1) (1/sqrt70)`
  `= 83.1^@ \ text{(1 d.p,)}`

Vectors, EXT1 EQ-Bank 20

Two vectors are given by  `underset ~a = 4 underset ~i + m underset ~j - 3 underset ~k`  and  `underset ~b = −2 underset ~i + n underset ~j - underset ~k`, where `m`, `n in R^+`.

If  `|\ underset ~a\ | = 10`  and `underset ~a` is perpendicular to `underset ~b`, determine the exact values of `m` and `n`.   (3 marks)

Show Answers Only

`m=5sqrt3, \ n=\sqrt{3}/3`

Show Worked Solution

`text(Using)\ \ |\ underset ~a\ | = 10:`

`10` `= sqrt(4^2 + m^2 + (-3)^2)`
`100` `=m^2+75`
`m^2` `= 25`
`m` `=5sqrt3\ \ (m in R^+)`

 

`text(S)text(ince)\ \ underset ~a _|_ underset ~b\ \ =>\ \ underset ~a xx underset ~b=0`

`0` `=4 xx (−2) + mn + (−3) xx (−1)`
`0` `=n xx 5sqrt3-5`
`n` `=5/(5\sqrt{3})`
`n` `=1/\sqrt{3}=\sqrt{3}/3`

Vectors, EXT1 2012 SPEC2 15 MC

The vectors  `underset~a = 2underset~i + m underset~j-3underset~k`  and  `underset~b = m^2underset~i-underset~j + underset~k`  are perpendicular for

  1. `m = −2/3`  and  `m = 1`
  2. `m = −3/2`  and  `m = 1`
  3. `m = 2/3`  and  `m = −1`
  4. `m = 3/2`  and  `m = −1`
Show Answers Only

`D`

Show Worked Solution

`underset ~a ⊥ underset ~b\ \ =>\ \ underset ~a ⋅ underset ~b=0`

`underset ~a ⋅ underset ~b` `= 2m^2 + m(-1) + (-3)(1)`
`0` `= 2m^2-m-3`
`0` `= (2m-3)(m + 1)`

 
`:. m = 3/2, quad m = -1`

`=> D`

Vectors, EXT1 2019 SPEC2 11 MC

Let point `M` have coordinates `(a, 1,-2)` and let point `N` have coordinates `(-3, b,-1)`.

If the coordinates of the midpoint of `vec(MN)` are `(-5, 3/2, c)` and `a, b` and `c` are real constants, the the values of `a, b` and `c` are respectively

  1. `−13, 2 and −1/2`
  2. `−7, −2 and −3/2`
  3. `−2, −1/2 and −3`
  4. `−7, 2 and −3/2`
Show Answers Only

`D`

Show Worked Solution

`M = 1/2 ([(a),(1),(−2)] + [(−3),(b),(−1)]) = 1/2 [(a-3),(1 + b),(−3)]`

`1/2(a-3)` `= −5`
`a-3` `= −10`
`a` `= −7`
`1/2(1 + b)` `= 3/2`
`1 + b` `= 3`
`b` `= 2`
`c` `= −3/2`

 
`=>D`

Vectors, EXT1 2011 SPEC2 12 MC

The angle between the vectors  `3underset~i + 6underset~j-2underset~k`  and  `2underset~i-2underset~j + underset~k`, correct to the nearest tenth of a degree, is

  1. 2.0°
  2. 91.0°
  3. 112.4°
  4. 121.3°
Show Answers Only

`C`

Show Worked Solution

`|3underset~i + 6underset~j-2underset~k| = sqrt(9 + 36 + 4) = sqrt49 = 7`

`|2underset~i-2underset~j + underset~k| = sqrt(4 + 4 + 1) = sqrt9 = 3`

`(3underset~i + 6underset~j-2underset~k) * (2underset~i-2underset~j + underset~k)`

`= 3 xx 2 + 6 xx (−2) + (−2) xx 1`

`= 6-12-2`

`= -8`  

`costheta` `= ((3tildei + 6tildej-2tildek).(2tildei-2tildej + tildek))/(|\ 3tildei + 6tildej-2tildek\ ||\ 2tildei-2tildej + tildek\ |)= -8/21`
`:. theta `= cos^(−1)(−8/12)~~ 112.4^@`

 
`=> C`

Vectors, EXT1 V1 2025 HSC 9 MC

The vectors \(\underset{\sim}{a}, \underset{\sim}{b}\) and \(\underset{\sim}{c}\) have magnitudes 3, 5 and 7 respectively.
 

Given that \(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c}=\underset{\sim}{0}\), what is the size of angle \(\theta\) between \(\underset{\sim}{a}\) and \(\underset{\sim}{b}\) ?

  1. \(\dfrac{\pi}{6}\)
  2. \(\dfrac{\pi}{3}\)
  3. \(\dfrac{2 \pi}{3}\)
  4. \(\dfrac{5 \pi}{6}\)
Show Answers Only

\(B\)

Show Worked Solution

\(\text{Find}\ \theta\ \text{using:}\ \ \cos\,\theta = \dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\abs{\underset{\sim}{a}} \, \abs{\underset{\sim}{b}}}\)

\(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c}=0 \ \Rightarrow\  \underset{\sim}{a}+\underset{\sim}{b}=-\underset{\sim}{c}\)

\(\abs{\underset{\sim}{a}+\underset{\sim}{b}}=\abs{-\underset{\sim}{c}}=7\)

♦ Mean mark 45%.
\(\abs{\underset{\sim}{a}+\underset{\sim}{b}}^2\) \(=(\underset{\sim}{a}+\underset{\sim}{b})(\underset{\sim}{a}+\underset{\sim}{b})=49\)
\(49\) \(=\underset{\sim}{a} \cdot \underset{\sim}{a}+2 a \cdot \underset{\sim}{b}+\underset{\sim}{b} \cdot \underset{\sim}{b}\)
\(49\) \(=\abs{\underset{\sim}{a}}^2+2 a \cdot b+\abs{\underset{\sim}{b}}^2\)
\(49\) \(=9+2 \underset{\sim}{a} \cdot \underset{\sim}{b}+25\)
\(2 \underset{\sim}{a} \cdot \underset{\sim}{b}\) \(=15\)
\(\underset{\sim}{a} \cdot \underset{\sim}{b}\) \(=\dfrac{15}{2}\)

\(\cos \theta\) \(=\dfrac{\frac{15}{2}}{3 \times 5}=\dfrac{1}{2}\)
\(\therefore \theta\) \(=\dfrac{\pi}{3}\)

 
\(\Rightarrow B\)

Vectors, EXT1 V1 2025 HSC 14c

The hands of an analogue clock are \(OA\) and \(OB\),

where \(A\) is \(\left(\sin \left(\dfrac{\pi t}{360}\right), \cos \left(\dfrac{\pi t}{360}\right)\right), B\) is \(\left(2 \sin \left(\dfrac{\pi t}{30}\right), 2 \cos \left(\dfrac{\pi t}{30}\right)\right)\),

\(O\) is the origin, and  \(t \geq 0\)  is the number of minutes past midnight.

Find the values of \(t\) when the hands are perpendicular for the first and second time after midnight. Give your answers to 3 decimal places.   (3 marks)

Show Answers Only

\(t=16.364, 49.091 \ \text{mins}\).

Show Worked Solution

\(\text{Express \(OA\) and \(OB\) as vectors:}\)

\(\overrightarrow{O A}=\displaystyle \binom{\sin \left(\frac{\pi t}{360}\right)}{\cos \left(\frac{\pi t}{360}\right)}, \quad \overrightarrow{O B}=\displaystyle \binom{2\, \sin \left(\frac{\pi t}{30}\right)}{2\, \cos \left(\frac{\pi t}{30}\right)}\)
 

\(\text{When hands are perpendicular,} \ \ \overrightarrow{OA} \cdot \overrightarrow{OB}=0:\)

\(\sin \left(\dfrac{\pi t}{360}\right) \times 2\, \sin \left(\dfrac{\pi t}{30}\right)+\cos \left(\dfrac{\pi t}{360}\right) \times 2\, \cos \left(\dfrac{\pi t}{30}\right)=0\)

\(\cos \left(\dfrac{\pi t}{30}-\dfrac{\pi t}{360}\right)\) \(=0\)
\(\cos \left(\dfrac{11 \pi t}{360}\right)\) \(=0\)

 

\(\dfrac{11 \pi t}{360}=\dfrac{\pi}{2}, \dfrac{3 \pi}{2}\)

\(t=\dfrac{\pi}{2} \times \dfrac{360}{11 \pi}=16.364 \ \text{mins}\)

\(t=\dfrac{3 \pi}{2} \times \dfrac{360}{11 \pi}=49.091 \ \text{mins}\).

Vectors, EXT1 V1 2025 HSC 11e

For what value of \(m\) is the vector \(\displaystyle \binom{1}{m}\) parallel to the vector \(\displaystyle \binom{2}{6}\)?   (1 mark)

Show Answers Only

\(m=3\)

Show Worked Solution

\(\text{If vectors are parallel:}\)

\(\displaystyle \binom{2}{6}=k\binom{1}{m} \ \Rightarrow \ k=2\)

\(2m\) \(=6\)
\(m\) \(=3\)

Vectors, EXT1 V1 2025 HSC 2 MC

The projection of \(\underset{\sim}{u}\) onto \(\underset{\sim}{v}\) is given by  \(\left(\dfrac{\underset{\sim}{u} \cdot \underset{\sim}{v}}{|\underset{\sim}{v}|^2}\right) \underset{\sim}{v}\).

What is the projection of  \(\underset{\sim}{u}=\underset{\sim}{i}+2 \underset{\sim}{j}\)  onto  \(\underset{\sim}{v}=2 \underset{\sim}{i}-3 \underset{\sim}{j}\) ?

  1. \(-\dfrac{4}{5}(\underset{\sim}{i}+2 \underset{\sim}{j})\)
  2. \(-\dfrac{4}{13}(2 \underset{\sim}{i}-3 \underset{\sim}{j})\)
  3. \(-\dfrac{4}{\sqrt{5}}(\underset{\sim}{i}+2 \underset{\sim}{j})\)
  4. \(-\dfrac{4}{\sqrt{13}}(2 \underset{\sim}{i}-3 \underset{\sim}{j})\)
Show Answers Only

\(B\)

Show Worked Solution

\(\underset{\sim}{u}=\displaystyle\binom{1}{2},|\underset{\sim}{u}|=\sqrt{1^2+2^2}=\sqrt{5}\)

\(\underset{\sim}{v}=\displaystyle \binom{2}{-3},|\underset{\sim}{v}|=\sqrt{2^2+(-3)^2}=\sqrt{13}\)

\(\operatorname{proj}_{\underset{\sim}{v}}{\underset{\sim}{u}}\) \(=\dfrac{\underset{\sim}{u} \cdot \underset{\sim}{v}}{|\underset{\sim}{v}|^2} \times \underset{\sim}{v}\)
  \(=\dfrac{2-6}{13}(\underset{\sim}{2i}-3\underset{\sim}{j})\)
  \(=-\dfrac{4}{13}(\underset{\sim}{2i}-3\underset{\sim}{j})\)

 
\(\Rightarrow B\)

Vectors, EXT1 V1 EQ-Bank 6 MC

Find the direction of  \(\overrightarrow{BA}\)  given,

\(\overrightarrow{OA}=\dbinom{-1}{2}\)  and  \(\overrightarrow{OB}=\dbinom{1}{5}\)

  1. 56°
  2. 143°
  3. 217°
  4. 236°
Show Answers Only

\(\Rightarrow D\)

Show Worked Solution

\(\overrightarrow{BA}=\overrightarrow{OA}-\overrightarrow{OB}=\dbinom{-1}{2}-\dbinom{1}{5}=\dbinom{-2}{-3}\)
 

\(\text{Angle is in 3rd quadrant.}\)

\(\text{Reference angle } (\theta)=\tan ^{-1}\left(\dfrac{3}{2}\right)=56^{\circ}\)

\(\text{Direction}=180+56=236^{\circ}\)

\(\Rightarrow D\)

Vectors, EXT1 V1 EQ-Bank 12

Given  \(\overrightarrow{OP}=2\underset{\sim}{i}-3\underset{\sim}{j}, \ \overrightarrow{P Q}=-\underset{\sim}{i}+2 \underset{\sim}{j}\), find the expression for \(\overrightarrow{O Q}.\)    (2 marks)

Show Answers Only

\(\overrightarrow{O Q}=\underset{\sim}{i}-\underset{\sim}{j}\)

Show Worked Solution

\(\overrightarrow{PQ}\) \(=\overrightarrow{OQ}-\overrightarrow{OP}\)  
\(\displaystyle\binom{-1}{2}\) \(=\displaystyle\binom{x}{y}-\displaystyle\binom{2}{-3}\)  
\(\displaystyle\binom{x}{y}\) \(=\displaystyle\binom{-1}{2}+\displaystyle\binom{2}{-3}=\binom{1}{-1}\)  

 
\(\therefore \overrightarrow{O Q}=\underset{\sim}{i}-\underset{\sim}{j}\)

Vectors, EXT1 V1 EQ-Bank 7 MC

Which of the following vectors is perpendicular to \(\displaystyle \binom{3}{-2}\) and has a magnitude of 3?

  1. \(\displaystyle 3\binom{-3}{2}\)
  2. \(\displaystyle \frac{3}{\sqrt{13}}\binom{-3}{2}\)
  3. \(\displaystyle \frac{\sqrt{10}}{\sqrt{13}}\binom{2}{3}\)
  4. \(\displaystyle \frac{3}{\sqrt{13}}\left(\frac{2}{3}\right)\)
Show Answers Only

\(\Rightarrow D\)

Show Worked Solution

\(\text{If} \ \perp \ \Rightarrow \text {dot product}=0:\)

\(\displaystyle \binom{3}{-2}\binom{-3}{2}=-9-4=-13 \neq 0  \quad \text{(Eliminate A and B)}\)
  

\(\text{Consider Option D:}\)

\(\displaystyle \frac{3}{\sqrt{13}}\left(\frac{2}{3}\right)=\binom{\frac{6}{\sqrt{13}}}{\frac{9}{\sqrt{13}}} \)

\(\text{Magnitude }=\sqrt{\left(\frac{6}{\sqrt{13}}\right)^2+\left(\frac{9}{\sqrt{13}}\right)^2}=\sqrt{\dfrac{36+81}{13}}=3\)

\(\Rightarrow D\)

Vectors, EXT1 V1 EQ-Bank 4 MC

Evaluate  \(\abs{\underset{\sim}{u}+\underset{\sim}{w}}\underset{\sim}{v}\)  given  \(\underset{\sim}{u}=\displaystyle\binom{2}{1}, \underset{\sim}{v}=\binom{1}{3}\)  and  \(\underset{\sim}{w}=\displaystyle\binom{-4}{3}\)

  1. \(10\)
  2. \(0\)
  3. \(\displaystyle \binom{\sqrt{20}}{3 \sqrt{20}}\)
  4. \(\displaystyle \binom{-20}{15}\)
Show Answers Only

\(\Rightarrow C\)

Show Worked Solution

\(\displaystyle \underset{\sim}{u}+\underset{\sim}{w}=\binom{2}{1}+\binom{-4}{3}=\binom{-2}{4} \Rightarrow \abs{\underset{\sim}{u}+\underset{\sim}{w}}=\sqrt{4+16}=\sqrt{20}\)

\(\displaystyle \abs{\underset{\sim}{u}+\underset{\sim}{w}} \underset{\sim}{v}=\sqrt{20}\binom{1}{3}=\binom{\sqrt{20}}{3 \sqrt{20}}\)

\(\Rightarrow C\)

Vectors, EXT1 V1 2024 HSC 13c

The vector \(\underset{\sim}{a}\) is \(\displaystyle \binom{1}{3}\) and the vector \(\underset{\sim}{b}\) is \(\displaystyle\binom{2}{-1}\).

The projection of a vector \(\underset{\sim}{x}\) onto the vector \(\underset{\sim}{a}\) is \(k \underset{\sim}{a}\), where \(k\) is a real number.

The projection of the vector \(\underset{\sim}{x}\) onto the vector \(\underset{\sim}{b}\) is \(p \underset{\sim}{b}\), where \(p\) is a real number.

Find the vector \(\underset{\sim}{x}\) in terms of \(k\) and \(p\).   (4 marks)

Show Answers Only

\(\underset{\sim}{x}=\dfrac{5}{7} \displaystyle \binom{2 k+3 p}{4 k-p}\)

Show Worked Solution

\(\underset{\sim}{a}=\displaystyle \binom{1}{3}, \ \abs{\underset{\sim}{a}}=\sqrt{1^2+3^2}=\sqrt{10}\)

\(\underset{\sim}{b}=\displaystyle \binom{2}{-1}, \ \abs{\underset{\sim}{b}}=\sqrt{2^2+(-1)^2}=\sqrt{5}\)

\(\text{Let } \underset{\sim}{x}=\displaystyle \binom{x_1}{x_2}\)

\(\operatorname{proj}_{\underset{\sim}{a}} \underset{\sim}{x}=\dfrac{\underset{\sim}{x} \cdot \underset{\sim}{a}}{|\underset{\sim}{a}|^2} \underset{\sim}{a}=\dfrac{x_1+3 x_2}{10} \cdot \underset{\sim}{a}\)

\(k=\dfrac{x_1+3 x_2}{10} \ \Rightarrow \ x_1+3 x_2=10 k\ \ldots\ (1)\)

♦ Mean mark 47%.

\(\operatorname{proj}_{\underset{\sim}{b}} \underset{\sim}{x}=\dfrac{\underset{\sim}{b} \cdot \underset{\sim}{x}}{|\underset{\sim}{b}|^2} b=\dfrac{2 x_1-x_2}{5} \cdot \underset{\sim}{b}\)

\(p=\dfrac{2 x_1-x_2}{5} \ \Rightarrow \ 2 x_1-x_2=5 p\ \ldots\\ (2)\)
 

  \(\text {Multiply } (2) \times 3\)

\(6 x_1-3 x_2=15 p\ \ldots\ (3)\)

  \((1)+(3)\)

\(7 x_1\) \(=10 k+15 p\)  
\(x_1\) \(=\dfrac{1}{7}(10 k+15)\)  

 
\(\text {Multiply } (1) \times 2\)

\(2 x_1+6 x_2=20 k\ \ldots\ (4)\)

  \(\text {Subtract} (4)-(2)\)

\(7x_2\) \(=20 k-5 p\)  
\(x_2\) \(=\dfrac{1}{7}(20 k-5 p)\)  

 
\(\therefore \underset{\sim}{x}=\displaystyle \frac{1}{7}\binom{10 k+15 p}{20 k-5 p}=\frac{5}{7}\binom{2 k+3 p}{4 k-p}\)

Vectors, EXT1 V1 2024 HSC 12a

The vectors \(\displaystyle \binom{a^2}{2}\) and \(\displaystyle  \binom{a+5}{a-4}\) are perpendicular.

Find the possible values of \(a\).   (3 marks)

Show Answers Only

\(x=1,-4 \text { or }-2\)

Show Worked Solution

\(\text{If vectors are }\perp:\)

\(\displaystyle\binom{a^2}{2} \cdot\binom{a+5}{a-4}=0\)

\(a^3+5 a^2+2 a-8=0\)
 

\(\text{Test for roots:}\)

\(1^3+5 \times 1^2+2\times 1-8=0 \, \checkmark\)

\((a-1) \text{ is a factor.}\)

\(\text{By polynomial long division:}\)

\((a-1)\left(a^2+6 a+8\right)=0\)

\((a-1)(a+4)(a+2)=0\)

\(\therefore x=1,-4 \text { or }-2\)

Vectors, EXT1 V1 2024 HSC 11a

Consider the vectors  \(\underset{\sim}{a}=3 \underset{\sim}{i}+2 \underset{\sim}{j}\)  and  \(\underset{\sim}{b}=-\underset{\sim}{i}+4 \underset{\sim}{j}\).

  1. Find  \(2 \underset{\sim}{a}-\underset{\sim}{b}\).   (1 mark)

  2. Find  \(\underset{\sim}{a} \cdot \underset{\sim}{b}\).   (1 mark)

Show Answers Only

i.    \(\displaystyle \binom{7}{0}\)

ii.   \(5\)

Show Worked Solution

i.     \(\underset{\sim}{a}=3 \underset{\sim}{i}+2 \underset{\sim}{j}, \ \underset{\sim}{b}=-\underset{\sim}{i}+4 \underset{\sim}{j}\)

\(2 \underset{\sim}{a}-\underset{\sim}{b}=2 \displaystyle \binom{3}{2}-\binom{-1}{4}=\binom{6}{4}-\binom{-1}{4}=\binom{7}{0}\)
 

ii.    \(\underset{\sim}{a} \cdot \underset{\sim}{b}=\displaystyle\binom{3}{2}\binom{-1}{4}=3 \times(-1)+2 \times 4=5\).

Vectors, EXT1* V1 2024 HSC 1 MC

Which of the following vectors is perpendicular to  \(3 \underset{\sim}{i}+2 \underset{\sim}{j}-5 \underset{\sim}{k}\) ?

  1. \(-\underset{\sim}{i}-\underset{\sim}{j}+\underset{\sim}{k}\)
  2. \(\underset{\sim}{i}+\underset{\sim}{j}-\underset{\sim}{k}\)
  3. \(-2 \underset{\sim}{i}+3 \underset{\sim}{j}+\underset{\sim}{k}\)
  4. \( 3 \underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{k}\)
Show Answers Only

\(D\)

Show Worked Solution

\(\text{Consider option}\ D:\)

\[\underset{\sim}{a} \cdot \underset{\sim}{b}=\left(\begin{array}{c} 3 \\ 2 \\ -5 \end{array}\right) \left(\begin{array}{c} 3 \\ -2 \\ 1 \end{array}\right) = 9-4-5=0 \]

\(\Rightarrow D\)

Vectors, EXT1 V1 2023 HSC 6 MC

Given the two non-zero vectors \(\underset{\sim}{a}\) and \(\underset{\sim}{b}\), let \(\underset{\sim}{c}\) be the projection of \(\underset{\sim}{a}\) onto \(\underset{\sim}{b}\).

What is the projection of \(10 \underset{\sim}{a}\) onto \(2 \underset{\sim}{b}\) ?

  1. \(2 \underset{\sim}{c}\)
  2. \(5 \underset{\sim}{c}\)
  3. \(10 \underset{\sim}{c}\)
  4. \(20 \underset{\sim}{c}\)
Show Answers Only

\(C\)

Show Worked Solution

\(\underset{\sim}c=\text{proj}_{\underset{\sim}b}\underset{\sim}a =\dfrac{\underset{\sim}a \cdot \underset{\sim}b}{|b|^2} \underset{\sim}b \)

♦ Mean mark 49%.
\(\text{proj}_{2\underset{\sim}b} 10\underset{\sim}a \) \(=\dfrac{10\underset{\sim}a \cdot 2\underset{\sim}b}{\big{|}2\underset{\sim}b\big{|}^2} 2\underset{\sim}b \)  
  \(=\dfrac{20 \times 2}{2^2} \Bigg{(}\dfrac{\underset{\sim}a \cdot \underset{\sim}b}{|\underset{\sim}b|^2} \underset{\sim}b \Bigg{)} \)  
  \(=10 \underset{\sim}c \)  

 
\(\Rightarrow C\)

Vectors, EXT1 V1 2022 HSC 14b

The vectors `\vec{u}` and `\vec{v}` are not parallel. The vector `\vec{p}` is the projection of `\vec{u}` onto the vector `\vec{v}`.

The vector `\vec{p}` is parallel to `\vec{v}` so it can be written `\lambda_0 \vec{v}` for some real number `\lambda_0`. (Do NOT prove this.)

Prove that  `|\vec{u}-\lambda \vec{v}|`  is smallest when `\lambda=\lambda_0` by showing that, for all real numbers `\lambda,\|\vec{u}-\lambda_0 \vec{v}\| \leq|\vec{u}-\lambda \vec{v}|`.  (3 marks)

Show Answers Only

`text{Proof (See Worked Solutions)}`

Show Worked Solution
`overset(->)p` `=text{proj}_(overset(->)v)overset(->)u`  
`lambda_0 overset(->)v` `=(overset(->)u*overset(->)v)/(|overset(->)v|^2) overset(->)v`  
`lambda_0` `=(overset(->)u*overset(->)v)/(|overset(->)v|^2 )\ \ \ …\ (1)`  

 
`text{Show}\ \ |\vec{u}-\lambda_0 \vec{v}\| \leq|\vec{u}-\lambda \vec{v}| :`

`|\vec{u}-\lambda \vec{v}\|^2 -|\vec{u}-\lambda_0 \vec{v}|^2`

`=(vec{u}-\lambda \vec{v})*(vec{u}-\lambda \vec{v})-(vec{u}-\lambda_0 \vec{v})*(vec{u}-\lambda_0 \vec{v})`

`=vec{u}*vec{u}-2lambda vec{u}*vec{v}+lambda^2vec{v}*vec{v}-(vec{u}*vec{u}-2lambda_0vec{u}*vec{v}+lambda_0^2vec{v}*vec{v})`

`=-2lambdavec{u}*vec{v}+lambda^2|vec{v}|^2+2lambda_0vec{u}*vec{v}-lambda_0^2|vec{v}|^2`

`=|vec{v}|^2(lambda^2-lambda_0^2)-2vec{u}*vec{v}(lambda-lambda_0)`

`=|vec{v}|^2(lambda-lambda_0)[lambda+lambda_0-2(vec{u}*vec{v})/|vec{v}|^2]`

`=|vec{v}|^2(lambda-lambda_0)[lambda+lambda_0-2lambda_0]\ \ \ text{(see (1))}`

`=|vec{v}|^2(lambda-lambda_0)^2>=0`
 

`text{S}text{ince}\ \ |\vec{u}-\lambda \vec{v}\|^2 -|\vec{u}-\lambda_0 \vec{v}|^2>=0`

`=>\ |\vec{u}-\lambda_0 \vec{v}|^2<=|\vec{u}-\lambda \vec{v}\|^2 `

`=>\ |\vec{u}-\lambda_0 \vec{v}|<=|\vec{u}-\lambda \vec{v}\| \ \ text{… as required}`

`:. |\vec{u}-\lambda \vec{v}|\ \ text{is smallest when}\ \ lambda=\lambda_0`


♦♦♦ Mean mark 22%.

Vectors, EXT1 V1 2022 HSC 11d

The vectors  `underset~u=([a],[2])`  and  `underset~v=([a-7],[4a-1])`  are perpendicular.

What are the possible values of `a`?  (2 marks)

Show Answers Only

`a=1, -2`

Show Worked Solution

`text{If}\ \ underset~u ⊥ underset~v:`

`([a],[2])*([a-7],[4a-1])` `=0`  
`a(a-7)+2(4a-1)` `=0`  
`a^2-7a+8a-2` `=0`  
`a^2+a-2` `=0`  
`(a+2)(a-1)` `=0`  

 
`:.a=1\ \ text{or}\ \ -2`

Vectors, EXT1 V1 2022 HSC 11a

For the vectors  `underset~u= underset~i- underset~j`  and  `underset~v=2 underset~i+ underset~j`, evaluate each of the following. 

  1. `underset~u+3 underset~v`   (1 mark)

  2. `underset~u * underset~v`   (1 mark)

Show Answers Only
  1. `((7),(2))`
  2. `1`
Show Worked Solution

i.  `underset~u= ((1),(-1)),\ \ underset~v= ((2),(1))`

`underset~u+3 underset~v` `=((1),(-1))+3((2),(1))`  
  `=((1+3xx2),(-1+3xx1))`  
  `=((7),(2))`  

 

ii.    `underset~u * underset~v` `=((1),(-1))*((2),(1))`
    `=1xx2+(-1)xx1`
    `=1`

Vectors, EXT1 V1 2022 HSC 8 MC

The angle between two unit vectors `underset~a` and `underset~b` is `theta` and  `|underset~a+ underset~b| < 1`.

Which of the following best describes the possible range of values of  `theta` ?

  1. `0 <= theta < (pi)/(3)`
  2. `0 <= theta < (2pi)/(3)`
  3. `(pi)/(3) < theta <= pi`
  4. `(2pi)/(3) < theta <= pi`
Show Answers Only

`D`

Show Worked Solution

`text{By Elimination:}`

`text{Consider}\ \ underset~a=((1),(0)) and underset~b=((-1),(0))`

`|underset~a+ underset~b| =0 < 1\ \ and\ \ theta=pi`

`text{→ Eliminate A and B}`
 

`text{Consider}\ \ theta=(2pi)/3 and underset~a=((1),(0)),\ \ underset~b=((costheta),(sintheta))`

`underset~a+underset~b=((1+cos((2pi)/3)),(sin((2pi)/3)))=((1/2),(sqrt3/2))`

`|underset~a+ underset~b|^2 = (1/2)^2+(sqrt3/2)^2=1`

`:. theta !=(2pi)/3`

`text{→ Eliminate C}`

`=>D`


♦♦ Mean mark 34%.

Vectors, EXT1 V1 2022 HSC 6 MC

The following diagram shows the vector `underset∼u` and the vectors `underset∼i+underset∼j,-underset∼i+ underset∼j,-underset∼i- underset∼j` and `underset∼i-underset∼j`. 
 


 

Which statement regarding this diagram could be true?

  1. The projection of `underset∼u` onto  `underset∼i+ underset∼j`  is the vector  `1.1 underset∼i+ 1.8 underset∼j`.
  2. The projection of `underset∼u` onto  `-underset∼i+ underset∼j`  is the vector  `-0.4 underset∼i+0.4 underset∼j`.
  3. The projection of `underset∼u` onto  `- underset∼i- underset∼j`  is the vector  `3.2 underset∼i+3.2 underset∼j`. 
  4. The projection of `underset∼u` onto  `underset∼i- underset∼j`  is the vector  `0.5 underset∼i-0.5 underset∼j`. 
Show Answers Only

`B`

Show Worked Solution

`text{Consider each option by tracing projections on the graph:}`
 

`overset(->)(OM)= text(proj)_((-underset~i+underset~j)) underset~u`

`text{Option B’s projection is only possible correct option.}`

`=>B`


♦♦ Mean mark 38%.

 

Vectors, EXT1 V1 2021 SPEC2 12 MC

Consider the vectors  `underset~a = x underset~i + underset~j, \ underset~b = underset~i - underset~j`  and  `underset~c = underset~i + x underset~j`.

Given that  `theta`  is the angle between  `underset~a`  and  `underset~b`,  and  `phi`  is the angle between  `underset~b`  and  `underset~c, cos(theta) cos (phi)`  is

  1. `(2(1 + x^2))/(1 - x^2)`
  2. `(sqrt2(1 - x^2))/(1 + x^2)`
  3. `-((x + 1)^2)/(2(1 + x^2))`
  4. `-((x - 1)^2)/(2(1 + x^2))`
Show Answers Only

`D`

Show Worked Solution

`underset~a = x underset~i – underset~j, \ underset~b = underset~i – underset~j, \ underset~c = underset~i + x underset~j`

`underset~a · underset~b = |underset~a||underset~b|costheta`

`costheta = (x – 1)/(sqrt(x^2 + 1)sqrt2)`

`cos phi = (underset~b · underset~c)/(|underset~b||underset~c|) = (1 – x)/(sqrt2 sqrt(1 + x^2))`

`costheta · cos phi` `= ((x – 1)(1 – x))/(2(1 + x^2))`
  `= -((x – 1)^2)/(2(1 + x^2))`

 
`=>\ D`

Vectors, EXT1 V1 2021 HSC 5 MC

For the two vectors  `overset->(OA)`  and  `overset->(OB)`  it is know that

`overset->(OA) · overset->(OB) < 0`

Which of the following statements MUST be true?

  1. Either, `overset->(OA)`  is negative and  `overset->(OB)`  is positive, or  `overset->(OA)`  is positive and  `overset->(OB)`  is negative.
  2. The angle between  `overset->(OA)`  and  `overset->(OB)`  is obtuse.
  3. The product  `|overset->(OA)||overset->(OB)|`  is negative.
  4. The points `O`, `A` and `B` are collinear.
Show Answers Only

`B`

Show Worked Solution

`overset->(OA) · overset->(OB) < 0`

`|overset->(OA)||overset->(OB)| cos theta` `< 0`
`cos theta` `< 0`

 
`text(If)\ \ cos theta < 0, theta\ \ text{is in 2nd quadrant (obtuse).}`

`=> B`

Vectors, EXT1 V1 2021 HSC 1 MC

Given that  `overset->(OP) = ((-3),(1))`  and  `overset->(OQ) = ((2),(5))`, what is `overset->(PQ)`?

  1. `((1),(-6))`
  2. `((-1),(6))`
  3. `((5),(4))`
  4. `((-5),(-4))`
Show Answers Only

`C`

Show Worked Solution
`overset->(PQ)` `= overset->(OQ) – overset->(OP)`
  `= ((2),(5))-((-3),(1))`
  `= ((5),(4))`

 
`=>\ C`

Vectors, EXT1 V1 2020 HSC 11b

For what values(s) of  `a`  are the vectors  `((a),(−1))`  and  `((2a - 3),(2))`  perpendicular?  (3 marks)

Show Answers Only

`a = −1/2\ text(or)\ 2`

Show Worked Solution
`((a),(−1)) · ((2a – 3),(2))` `= 0`
`a(2a – 3) + (−1) xx 2` `= 0`
`2a^2 – 3a – 2` `= 0`
`(2a + 1)(a – 2)` `= 0`

 
`:. a = −1/2\ \ text(or)\ \ 2`

Vectors, EXT1 V1 2020 HSC 9 MC

The projection of the vector  `((6),(7))`  onto the line  `y = 2x`  is  `((4),(8))`.

The point  `(6, 7)`  is reflected in the line  `y = 2x`  to a point `A`.

What is the position vector of the point `A`?

  1. `((6),(12))`
  2. `((2),(9))`
  3. `((−6),(7))`
  4. `((−2),(1))`
Show Answers Only

`B`

Show Worked Solution

`text(Graph the projection and reflection:)`

 

`=>B`

Vectors, EXT1 V1 2020 HSC 4 MC

Maria starts at the origin and walks along all of the vector  `2underset~i + 3underset~j`, then walks along all of the vector  `3underset~i - 2underset~j`  and finally along all of the vector  `4underset~i - 3underset~j`.

How far from the origin is she?

  1. `sqrt77`
  2. `sqrt85`
  3. `2sqrt13 + sqrt5`
  4. `sqrt5 + sqrt7 + sqrt13`
Show Answers Only

`B`

Show Worked Solution
`underset~v` `= ((2),(3)) + ((3),(−2)) + ((4),(−3))`
  `= ((9),(−2))`
`|underset~v|` `= sqrt(9^2 + (−2)^2)`
  `= sqrt85`

 
`=>B`

Vectors, EXT1 V1 EQ-Bank 15

The vectors  `underset~a = 6underset~i + 2underset~j, \ underset~b = underset~i - 5underset~j`  and  `underset~c = 4underset~i + 4underset~j`

Find the values of  `m`  and  `n`  such that  `m underset~a + n underset~b = underset~c`.  (2 marks)

Show Answers Only

`n= −1/2`

`m = 3/4`

Show Worked Solution

`m underset~a + n underset~b= underset~c`

`m((6),(2)) + n((1),(−5))` `= ((4),(4))`

 
`6m + n = 4\ \ …\ (1)`

`2m – 5n = 4\ \ …\ (2)`
 

`text(Multiply)\ (2) xx 3`

`6m – 15n = 12\ \ …\ (3)`
 

`text(Subtract)\ \ (1) – (3)`

`16n = –8 \ => \ n= −1/2`

`text(Substitute)\ \ n = –1/2\ \ text{into (2):}`

`2m + 5/2` `= 4`
`m` `= 3/4`

 
`:. m=3/4, \ n= −1/2`

Vectors, EXT1 V1 EQ-Bank 27

Using vectors, calculate the acute angle between the line that passes through  `A(1, 3)`  and  `B(2,–6)`  and the line that passes through  `C(1, 5)`  and  `D(3,–2)`.

Give your answer correct to one decimal place.  (2 marks)

Show Answers Only

`9.6°`

Show Worked Solution

`underset~a = ((1),(3)), \ underset~b = ((2),(−6)), \ underset~c = ((1),(5)), \ underset~d = ((3),(−2))`

`overset(->)(AB)` `= underset~b – underset~a = ((2),(−6)) – ((1),(3)) = ((1),(−9))`
`overset(->)(CD)` `= underset~d – underset~c = ((3),(−2)) – ((1),(5)) = ((2),(−7))`

 

`costheta` `= (overset(->)(AB) · overset(->)(CD))/(|overset(->)(AB)| · |overset(->)(CD)|)`
  `= (2 + 63)/(sqrt82 · sqrt53)`
  `= 0.985…`

 

`:. theta` `=cos^(-1) 0.985…`
  `= 9.605…`
  `= 9.6°\ \ (text(to 1 d.p.))`

Vectors, EXT1 V1 EQ-Bank 13

Let the vectors  `underset~a=4 underset~i - underset~j, \ underset~ b = 3underset~i+2 underset~j`  and  `underset~c=-2 underset~i +5underset~j`.

  1. Calculate  `underset~a*(underset~b+underset~c)`   (1 mark)
  2. Verify  `underset~a*(underset~b+underset~c) = underset~a * underset~b + underset~a * underset~c`   (1 mark)

Show Answers Only

`text{Proof (See Worked Solution)}`

Show Worked Solution

a.    `underset~a=((4),(-1)),\ \ underset~b=((3),(2)),\ \ underset~c=((-2),(5))`

`(underset~b+underset~c) = ((3),(2)) + ((-2),(5)) = ((1),(7))`

`underset~a*(underset~b+underset~c)` `=((4),(-1)) *((1),(7))`   
  `=(4 xx 1) -(1 xx 7)`  
  `=-3`  

 

b.     `underset~a * underset~b + underset~a * underset~c` `=((4),(-1)) *((3),(2)) + ((4),(-1))*((-2),(5))`  
    `=(4 xx 3) -(1 xx 2) + (4xx-2) -(1 xx 5)`
    `=-3`
    `=underset~a*(underset~b+underset~c)`

 

Vectors, EXT1 V1 EQ-Bank 5 MC

What is the angle between the vectors  `((2),(1))`  and  `((-4),(2))`?

A.   `cos^(-1)(0.06)`

B.   `cos^(-1)(–0.06)`

C.   `cos^(-1)(0.6)`

D.   `cos^(-1)(–0.6)`

Show Answers Only

`D`

Show Worked Solution
`cos theta` `=(underset~a * underset~b)/(|underset~a||underset~b|)`  
  `=(-8+2)/(sqrt(2^2+1^2) xx sqrt((-4)^2+2^2)`  
  `=(-6)/(sqrt5 sqrt20)`  
  `=-0.6`  
`:. theta` `= cos^(-1) (-0.6)`  

  
`=> D`

Vectors, EXT1 V1 EQ-Bank 29

If  `theta`  is the angle between  `underset~a = underset~i + 3j`  and  `underset~b = 3underset~i + underset~j`, then find the exact value of  `cos 2theta`.  (2 marks)

Show Answers Only

`−7/25`

Show Worked Solution

`underset~a = [(1),(3)],\ \ underset~b = [(3),(1)]`

`|underset~a| = sqrt(1^2 + 3^2) = sqrt10`

`|underset~b| = sqrt(3^2 + 1^2) = sqrt10`

`cos theta= (1 xx 3 + 3 xx 1)/(sqrt10 sqrt10)= 3/5`

`cos2theta` `= 2cos^2theta – 1`
  `= 2(3/5)^2 – 1`
  `= −7/25`

Vectors, EXT1 V1 EQ-Bank 28

Two vectors are given by `underset~a = 2underset~i + m underset~j`  and  `underset~b = −5underset~i + n underset~j`  where  `m, n > 0`.

If  `|underset~a| = 3`  and  `underset~a`  is perpendicular to  `underset~b`, find the values of  `m`  and  `n`.  (2 marks)

Show Answers Only

`2sqrt5`

Show Worked Solution

`underset~a = [(2),(m)],\ \ underset~b = [(−5),(n)]`

 
`text(Using)\ |underset~a| = 3:`

`3` `= sqrt(2^2 + m^2)`
`m^2` `= 5`
`:.m` `= sqrt5,\ \ \ (m > 0)`

 
`text(S)text(ince)\ underset~a ⊥ underset~b:`

`a · b` `= 0`
`2xx −5 + mn` `= 0`
`sqrt5 n` `= 10`
`n` `= 10/sqrt5`
  `= 2sqrt5`

Vectors, EXT1 V1 EQ-Bank 31

Relative to a fixed origin, the points `A`, `B` and `C` are defined respectively by the position vectors  `underset~a = −underset~i - underset~j, \ underset~b = 3underset~i + 2underset~j`  and  `underset~c = −aunderset~i + 2underset~j`, where  `a`  is a real constant.

If the magnitude of angle `ABC`  is  `pi/3`, find `a`.  (3 marks)

Show Answers Only

`−3`

Show Worked Solution

`text(Angle between)\ overset(->)(BA)\ text(and)\ overset(->)(BC) = pi/3`

`overset(->)(BA)` `= overset(->)(OA) – overset(->)(OB)`
  `= [(−1),(−1)] – [(3),(2)] = [(−4),(−3)]`

 

`overset(->)(BC)` `= overset(->)(OC) – overset(->)(OB)`
  `= [(−a),(2)] – [(3),(2)] = [(−a−3),(0 )]`

 

`overset(->)(BA) · overset(->)(BC)` `= [(−4),(−3)] · [(−a −3),(0 )]`
  `= 4a + 12`

 
`overset(->)(BA) · overset(->)(BC) = |overset(->)(BA)| · |overset(->)(BC)|costheta`

`4a + 12` `= sqrt((−4)^2 + (−3)^2) · sqrt((-a-3)^2) · cos\ pi/3`
`4a + 12` `= 5(-a-3) · 1/2`
`4a + 12` `= -(5a)/2-15/2`
`(13a)/2` `= -39/2`
`:.a` `= -3`

Vectors, EXT1 V1 EQ-Bank 18

Consider the vector  `underset~a = underset~i + sqrt3underset~j`, where  `underset~i`  and  `underset~j`  are unit vectors in the positive direction of the `x` and `y` axes respectively.

  1. Find the unit vector in the direction of  `underset~a`.    (1 mark)

  2. Find the acute angle that  `underset~a`  makes with the positive direction of the `x`-axis.   (1 mark)

  3. The vector  `underset~b = m underset~i - 2underset~j`.

     

    Given that  `underset~b`  is perpendicular to  `underset~a`, find the value of  `underset~m`.   (1 mark)

Show Answers Only

a.    `1/2(underset~i + sqrt3underset~j)`

b.    `60°`

c.    `2sqrt3`

Show Worked Solution

a.    `underset~a = underset~i + sqrt3underset~j`

`|underset~a| = sqrt(1 + (sqrt(3))^2) = 2`

`overset^a = (underset~a)/(|underset~a|) = 1/2(underset~i + sqrt3underset~j)`

 

b.    `text(Solution 1)`

`underset~a\ =>\ text(Position vector from)\ \ O\ \ text{to}\ \ (1, sqrt3)`

`tan theta` `=sqrt3`  
`:. theta` `=60°`  
     

`text(Solution 2)`

`text(Angle with)\ xtext(-axis = angle with)\ \ underset~b = underset~i`

`underset~a · underset~i = 1 xx 1 = 1`

`underset~a · underset~i` `= |underset~a||underset~i|costheta`
`1` `= 2 xx 1 xx costheta`
`costheta` `= 1/2`
`:. theta` `= 60°`

 

c.     `underset~b = m underset~i – 2underset~j`

`underset~a · underset~b = [(1),(sqrt3)] · [(m),(−2)] = m – 2sqrt3`

`text(S)text(ince)\ underset~a ⊥ underset~b:`

`m – 2sqrt3` `= 0`
`m` `= 2sqrt3`

Vectors, EXT1 V1 EQ-Bank 19

Consider the following vectors

`overset(->)(OA) = 2underset~i + 2underset~j,\ \  overset(->)(OB) = 3underset~i - underset~j,\ \ overset(->)(OC) = 5underset~i + 3underset~j`

  1. Find  `overset(->)(AB)`.  (1 mark)

  2. The points `A`, `B` and `C` are vertices of a triangle. Prove that the triangle has a right angle at `A`.  (2 marks)

  3. Find the length of the hypotenuse of the triangle.  (1 mark)

Show Answers Only

a.    `underset~i – 3underset~j`

b.    `text(See Worked Solutions)`

c.    `2sqrt5`

Show Worked Solution

a.    `text(Find)\ overset(->)(AB):`

COMMENT: Many teachers recommend column vector notation to simplify calculations and minimise errors – we agree!

`overset(->)(OA) = [(2),(2)],\ \ overset(->)(OB)[(3),(−1)]`

`overset(->)(AB)` `= overset(->)(OB) – overset(->)(OA)`
  `= [(3),(−1)] – [(2),(2)]`
  `= [(1),(−3)]`
  `= underset~i – 3underset~j`

 

b.      `overset(->)(AC)` `= overset(->)(OC) – overset(->)(OA)`
    `= [(5),(3)] – [(2),(2)]`
    `= [(3),(1)]`
    `= 3underset~i + underset~j`

 

`overset(->)(AB) · overset(->)(AC)` `= 1 xx 3 + −3 xx 1=0`

`=> AB ⊥ AC`

`:. DeltaABC\ text(has a right angle at)\ A.`

 

c.    `overset(->)(BC)\ text(is the hypotenuse)`

`overset(->)(BC)` `= overset(->)(OC) – overset(->)(OB)`
  `= [(5),(3)] – [(3),(−1)]`
  `= [(2),(4)]`
`|overset(->)(BC)|` `=\ text(length of hypotenuse)`
  `= sqrt(2^2 + 4^2)`
  `= sqrt(20)`
  `= 2sqrt5`

Vectors, EXT1 V1 EQ-Bank 1 MC

The vectors  `underset~a = 2underset~i + m underset~j`  and  `underset~b = m^2underset~i-underset~j`  are perpendicular for

  1. `m = -2`  and  `m = 0`
  2. `m = 2`  and  `m = 0`
  3. `m = -1/2`  and  `m = 0`
  4. `m = 1/2`  and  `m = 0`
Show Answers Only

`D`

Show Worked Solution

`underset ~a ⊥ underset ~b\ \ =>\ \ underset ~a ⋅ underset ~b=0`

`underset ~a ⋅ underset ~b` `= 2m^2 + m(-1)`
`0` `= 2m^2-m`
`0` `= m(2m-1)`

 
`:. m = 0, quad m = 1/2`

`=> D`

Vectors, EXT1 V1 EQ-Bank 17

Consider the vectors

`underset~a = 6underset~i + 2underset~j,\ \ underset~b = 2underset~i - m underset~j`

  1. Calculate  `2underset~a - 3underset~b`.  (1 mark)

  2. Find the values of  `m`  for which  `|underset~b| = 3sqrt2`.  (2 marks)

  3. Find the value of  `m`  such that  `underset~a`  is perpendicular to  `underset~b`.  (1 mark)

Show Answers Only

a.    `[(6),(4 + 3m)]`

b.    `±sqrt14`

c.    `6`

Show Worked Solution
a.     `2underset~a – 3underset~b` `= 2[(6),(2)] – 3[(2),(−m)]`
    `= [(12),(4)] – [(6),(−3m)]`
    `= [(6),(4 + 3m)]`

 

b.    `underset~a = [(6),(2)], \ \ underset~b = [(2),(−m)]`

`|underset~b|` `= sqrt(4 + m^2)`
`3sqrt2` `= sqrt(4 + m^2)`
`18` `= 4 + m^2`
`m^2` `= 14`
`m` `= ±sqrt14`

 

c.    `text(If)\ \ underset~a ⊥ underset~b \ => \ underset~a · underset~b = 0`

`6 xx 2 + 2 xx – m` `= 0`
`2m` `= 12`
`:. m` `= 6`

Vectors, EXT1 V1 EQ-Bank 26

Consider the vectors,  `underset~a = overset(->)(OA)`  where  `|OA| = 5`  and  `underset~b = overset(->)(OB)`  where  `|OB| = 7`.

If  `angleAOB = 30°`, find  `text(proj)_(underset~b)underset~a`  as a multiple of  `underset~b`.   (2 marks)

Show Answers Only

`(5sqrt3)/14 · underset~b`

Show Worked Solution

`underset~overset^b = (underset~b)/(|OB|) = (underset~b)/7`

`text(proj)_underset~bunderset~a` `= (|underset~a|\ cos30°) · underset~overset^b`
  `= 5 xx sqrt3/2 xx (underset~b)/7`
  `= (5sqrt3)/14 · underset~b`