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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* 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 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 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 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. \(\displaystyle \binom{-20}{15}\)
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\(\Rightarrow C\)

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 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)

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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 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 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 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 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)

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`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 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)

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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 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`