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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 2 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 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

i.   `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`  

 

ii.   `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 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)

Show Answers Only
  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 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
  1. `underset~i – 3underset~j`
  2. `text(See Worked Solutions)`
  3. `2sqrt5`
Show Worked Solution

i.  `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`

 

ii.    `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.`

 

iii.   `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
  1. `[(6),(4 + 3m)]`
  2. `±sqrt14`
  3. `6`
Show Worked Solution
i.    `2underset~a – 3underset~b` `= 2[(6),(2)] – 3[(2),(−m)]`
    `= [(12),(4)] – [(6),(−3m)]`
    `= [(6),(4 + 3m)]`

 

ii.   `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`

 

iii.   `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`

Vectors, EXT1 V1 EQ-Bank 24

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)

Show Answers Only

`(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 EQ-Bank 16

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

Show Answers Only

`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`

Vectors, EXT1 V1 EQ-Bank 23

Points  `A`  and  `B`  have position vectors  `underset~a = 2underset~i - 2underset~j`  and  `underset~b =  4i + sqrt2 underset~j` respectively.

Find the angle between  `underset~a`  and  `underset~b`  to the nearest minute.  (2 marks)

Show Answers Only

`64°28′`

Show Worked Solution

`underset~a = 2underset~i – 2underset~j \ => \ |underset~a| = sqrt(2^2 + 2^2) = sqrt8`

`underset~b = 4 i + sqrt2 j \ => \ |underset~b| = sqrt(4^2 + (sqrt2)^2) = sqrt18`

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

`costheta` `= (underset~a · underset~b)/(|underset~a||underset~b|)`
  `= (8 – 2sqrt2)/(sqrt8sqrt18)`
  `= (8 – 2sqrt2)/12`
`:. theta` `=64.471…`
  `=64°28′\ \ \ text{(nearest minute)}`

Vectors, EXT1 V1 EQ-Bank 32

Consider the vectors given by  `underset ~a = m underset ~i + underset ~j`  and  `underset ~b = underset ~i + m underset ~j`, where  `m in R`.

Find the value(s) of  `m`  if the acute angle between  `underset ~a`  and  `underset ~b`  is 30°.   (2 marks)

Show Answers Only

`sqrt 3, 1/sqrt 3`

Show Worked Solution
`underset ~a *underset ~b` `= m xx 1 + 1 xxm`
  `=2m`
   
`underset ~a *underset ~b` `= |underset ~a||underset ~b| cos 30^@`
  `= sqrt(m^2 + 1) *sqrt(1 + m^2) *cos 30^@`
  `= {(m^2 + 1) sqrt 3}/2`

 

`{(m^2 + 1) sqrt 3}/2` `=2m`  
`m^2 sqrt 3 + sqrt 3` `=4m`  
`m^2 sqrt 3-4m + sqrt 3` `=0`  
`(sqrt 3 m)^2-4(sqrt 3 m) + 3` `=0`  
`(sqrt 3 m)^2-4(sqrt 3 m) + 2^2-1` `=0`  
`(sqrt 3 m-2)^2-1` `=0`  
`sqrt 3 m-2` `= +-1`  
`sqrt 3 m` `= 2 +- 1`  

 
`:. m = (2 +- 1)/sqrt 3 = 3/sqrt 3 or 1/sqrt 3`

`= sqrt 3, 1/sqrt 3`

Vectors, EXT1 V1 2015 SPEC2 15 MC

The projection of the force  `underset~F = a underset~i + b underset~j`, where `a` and `b` are non-zero real constants, in the direction of the vector  `underset~w = underset~i + underset~j`, is

  1. `((a + b)/2)underset~w`
  2. `underset~F/(a + b)`
  3. `((a + b)/(a^2 + b^2))underset~F`
  4. `((a + b)/sqrt2)underset~w` 
Show Answers Only

`A`

Show Worked Solution

`hatw= underset~w/sqrt(1+1)= (underset~i + underset~j)/sqrt2`

`underset~F*hat w = (a + b)/sqrt2`

♦ Mean mark 49%.

`(underset~F*hat w)hatw= ((a + b)/sqrt2) underset~w/sqrt2= ((a + b)/2) underset~w`

`=> A`

Vectors, EXT1 V1 2018 SPEC2 12 MC

If  `|underset ~a + underset ~b| = |underset ~a| + |underset ~b|`  and  `underset ~a, underset ~b != underset ~0`, which one of the following is necessarily true?

A.   `underset ~a\ text(is parallel to)\ underset ~b`

B.   `|underset ~a| = |underset ~b|`

C.   `underset ~a = underset ~b`

D.   `underset ~a\ text(is perpendicular to)\ underset ~b` 

Show Answers Only

`A`

Show Worked Solution
`|underset ~a + underset ~b|^2` `= (|underset ~a| + |underset ~b|)^2\ \ \ text{(given)}`
  `= |underset ~a|^2 + 2|underset ~a||underset ~b|+|underset ~b|^2`
`underset ~a ⋅ underset ~b` `= |underset ~a||underset ~b| cos theta`

 
`=> 2|underset ~a||underset ~b| = (2 underset ~a ⋅ underset ~b)/(cos theta)`

♦♦♦ Mean mark 36%.

`=>|underset ~a + underset ~b|^2 = |underset ~a|^2 + (2 underset ~a ⋅ underset ~b)/(cos theta) + |b|^2`

`(underset ~a + underset ~b) * (underset ~a + underset ~b) = underset ~a ⋅ underset ~a + (2 underset ~a ⋅ underset ~b)/(cos theta) + underset ~b ⋅ underset ~b`

`underset ~a ⋅ underset ~a + 2underset ~a ⋅ underset ~b + underset ~b ⋅ underset ~b = underset ~a ⋅ underset ~a + (2 underset ~a ⋅ underset ~b)/(cos theta) + underset ~b ⋅ underset ~b`

`2 underset ~a ⋅ underset ~b = (2 underset ~a ⋅ underset ~b)/(cos theta)`

`:. cos theta = 1\ \ =>\ \  theta = 0`

`=>  A`