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

Show Worked Solution

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

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

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

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

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

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 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 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 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 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 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)

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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 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)

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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 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)

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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`

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`