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

Show Worked Solution

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`

Show Worked Solution

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

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

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

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

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

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

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°`

Show Worked Solution

`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)

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`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 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

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

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

\(\Rightarrow D\)