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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. \(-\dfrac{4}{\sqrt{13}}(2 \underset{\sim}{i}-3 \underset{\sim}{j})\)
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\(B\)

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 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)

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

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 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. \(20 \underset{\sim}{c}\)
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\(C\)

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)

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`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 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 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 SM-Bank 20

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)

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  1. `1/2(underset~i + sqrt3underset~j)`
  2. `60°`
  3. `2sqrt3`
Show Worked Solution

i.   `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)`

 

ii.   `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°`

 

iii.   `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 SM-Bank 14

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)

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`(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`

Vectors, EXT1 V1 SM-Bank 13

Given  `underset~a = 4underset~i - 3underset~j`  and  `underset~b = 7underset~i - underset~j`, what is the magnitude of the projection of  `underset~a`  onto  `underset~b`. Give your answer in simplest form.  (3 marks)

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`(31sqrt2)/10`

Show Worked Solution

`underset~a = [(4),(−3)],\ \ underset~b = [(7),(−1)]`

`text(proj)_(underset~b) underset~a` `= (28 + 3)/(49 + 1)(7underset~i – underset~j)`
  `= 31/50(7underset~i – underset~j)`
  `= 217/50 underset~i – 31/50 underset~j`

 

`|\ text(proj)_(underset~b) underset~a\ |` `= sqrt((217/50)^2 + (31/50)^2)`
  `= sqrt((217^2 + 31^2))/50`
  `= sqrt(48\ 050)/50`
  `= (155sqrt2)/50`
  `= (31sqrt2)/10`

Vectors, EXT1 V1 SM-Bank 12

Find the projection of  `underset~a`  onto  `underset~b`  given  `underset~a = 2underset~i + underset~j`  and  `b = 3underset~i - 2underset~j`.  (2 marks)

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`12/13underset~i – 8/13underset~j`

Show Worked Solution

`underset~a = [(2),(1)],\ \ underset~b = [(3),(−2)]`

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

`text(proj)_(underset~b) underset~a` `= (underset~a · underset~b)/(underset~b · underset~b) xx underset~b`
  `= (6 – 2)/(9 + 4)(3underset~i – 2underset~j)`
  `= 4/13(3underset~i – 2underset~j)`
  `= 12/13underset~i – 8/13underset~j`