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Vectors, EXT2 EQ-Bank 12

A curve has a vector equation

   \(r=(2 \sin t-1)\mathbf{i}+(2 \cos t+3) \mathbf{j} \)

  1. Write the Cartesian equation of this curve.   (2 marks)

  2. Sketch the curve on the Cartesian plane below.   (1 mark)

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a.    \((x+1)^2+(y-3)^2=4\)

b.    \((x+1)^2+(y-3)^2=2^2\)

\(\Rightarrow \ \text{Circle with centre} \ (-1,3), \ \text{radius}=2\)
 

Show Worked Solution

a.    \(r=(2 \sin t-1) \mathbf{i}+(2 \cos t+3) \mathbf{j}\)

\(x=2 \sin t-1 \ \Rightarrow \ \sin t=\dfrac{x+1}{2}\)

\(y=2 \cos t+3 \ \Rightarrow \ \cos t=\dfrac{y-3}{2}\)

\(\text{Using} \ \ \sin ^2 t+\cos ^2 t=1:\)

\(\dfrac{(x+1)^2}{4}+\dfrac{(y-3)^2}{4}\) \(=1\)  
\((x+1)^2+(y-3)^2\) \(=4\)  

  
b.    \((x+1)^2+(y-3)^2=2^2\)

\(\Rightarrow \ \text{Circle with centre} \ (-1,3), \ \text{radius}=2\)
 

 

Vectors, EXT2 EQ-Bank 26

Let \(S\) be a sphere with equation

\begin{align*}
\left|r-\left(\begin{array}{c} 2 \\ -3 \\ 3
\end{array}\right)\right|=15
\end{align*}

Show the point  \(P(1,-1,2)\) lies outside the sphere \(S\).   (2 marks)

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\(\text{Centre of circle} \ (C)=(2,-3,3)\)

\(\text{Radius}=15 \ \text{(given)}\)

\(\text{Find distance from} \ C \text { to } P(1,-1,2):\)

\(\operatorname{dist}=\sqrt{(1-2)^2+(-1+3)^2+(2-3)^2}=\sqrt{6}\)

\(\text{Since \(\ \sqrt{6}<15, P\) lies inside sphere.}\)

Show Worked Solution

\(\text{Centre of circle} \ (C)=(2,-3,3)\)

\(\text{Radius}=15 \ \text{(given)}\)

\(\text{Find distance from} \ C \text { to } P(1,-1,2):\)

\(\operatorname{dist}=\sqrt{(1-2)^2+(-1+3)^2+(2-3)^2}=\sqrt{6}\)

\(\text{Since \(\ \sqrt{6}<15, P\) lies inside sphere.}\)

Vectors, EXT2 EQ-Bank 34

A sphere of radius  \(r=3\)  is centred at \(C(6,-3,2)\).

A line passes through \(A(3,-1,6)\) and \(B(5,-1,-5)\).

Show that this line is a tangent to the sphere.    (4 marks)

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\(\text{See Worked Solutions}\)

Show Worked Solution

\(\text{The sphere centred at} \ \ C(6,-3,2) \ \ \text {with radius} \ \ r=3\)

\((x-6)^2+(y+3)^2+(z-2)^2=3^2=9\)
 

\(\text{The line}\ A B\ \text{has direction}\)

\(\overrightarrow{O B}-\overrightarrow{O A}=\left[\begin{array}{c}5-3 \\ -1-(-1) \\ -5-6\end{array}\right]=\left[\begin{array}{c}2 \\ 0 \\ -11\end{array}\right]\)
 

\(\text{Line \(A B\) has parametric equation:}\)

\(\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\overrightarrow{O A}+\lambda \overrightarrow{A B}=\left[\begin{array}{c}3 \\ -1 \\ 6\end{array}\right]+\lambda\left[\begin{array}{c}2 \\ 0 \\ -11\end{array}\right]=\left[\begin{array}{c}3+2 \lambda \\ -1 \\ 6-11 \lambda\end{array}\right]\)
 

\(\text{Substitute into the equation for the sphere:}\)

\((3+2 \lambda-6)^2+(-1+3)^2+(6-11 \lambda-2)^2\) \(=9\)
\((2 \lambda-3)^2+4+(4-11 \lambda)^2\) \(=9\)
\(9-12 \lambda+4 \lambda^2+4+16-88 \lambda+121 \lambda^2\) \(=9\)
\(125 \lambda^2-100 \lambda+20\) \(=0\)
\(5\left(25 \lambda^2-20 \lambda+4\right)\) \(=0\)
\(5(5 \lambda-2)^2\) \(=0\)

 

\(\text{There is only one}\ \lambda\ \text{that solves this equation.}\)

\(\text{i.e. one common point on the line and the sphere.}\)

\(\therefore\ \text{The line is a tangent to the sphere.}\)

Vectors, EXT2 EQ-Bank 27

Line 1 is given by the equations  \(x=-1+2 s, \ y=1-2 s\)  and  \(z=1+2 s\), where \(s\) is a parameter.

Line 2 is given by the equations  \(x=1+2 t, \ y=-1-t\)  and  \(z=4+3 t\), where \(t\) is a parameter.

Show that Line 1 and Line 2 are skew.   (3 marks)

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Show Worked Solution

\(\text{Skew lines do not intersect and are not parallel.}\)

\(\text{Solving for \(s\) and \(t\) in \(x\) and \(y\):}\)

\(\text{From} \ x: \quad\) \(-1+2 s\) \(=1+2 t\ \ldots\ (1)\)
\(\text{From} \ y: \quad\) \(1-2 s\) \(=-1-t\ \ldots\ (2)\)
\((1)+(2)\) \(0\) \(=t\)

 

\(\text{Substitute}\ \ t=0\ \ \text{into (1):}\)

\(-1+2 s=1\ \ \Rightarrow\ \ s=1\)
 

\(\text{Substitute}\ \ s=1\ \ \text{and}\ \ t=0\ \ \text{into}\ z:\)

\(z=1+2=3\ \text{(line 1)}, \ z=4\ \text{(line 2)}\)

\(\Rightarrow\ \text{Lines 1 and 2 do not intersect.}\)
 

\(\text{Direction vectors:}\)

\(\text{Line 1 = }\left( \begin{array}{r}2 \\ -2 \\ 2\end{array}\right),\ \ \text{Line 2 = } \left(\begin{array}{r}2 \\ -1 \\ 3\end{array}\right).\)

\(\left(\begin{array}{r}2 \\ -2 \\ 2\end{array}\right) \neq k\left(\begin{array}{r}2 \\ -1 \\ 3\end{array}\right) \ \text{for any}\ \ k \in R\)

\(\Rightarrow\ \text{Lines 1 and 2 are not parallel.}\)

\(\therefore\ \text{Lines 1 and 2 are skew as they do not intersect and are not parallel.}\)

Vectors, EXT2 V1 2025 SPEC2 18 MC

The lines given by  \({\underset{\sim}{r}}_1(\lambda)=2 \underset{\sim}{i}+r \underset{\sim}{j}-3 \underset{\sim}{k}+\lambda(\underset{\sim}{i}-\underset{\sim}{j}+4 \underset{\sim}{k})\) and  \({\underset{\sim}{r}}_2(\mu)=\underset{\sim}{i}+s \underset{\sim}{k}+\mu(\underset{\sim}{i}+\underset{\sim}{j}-\underset{\sim}{k})\) intersect at the point \((4,3, t)\), where \(\lambda, \mu \in R\)  and \(r, s\) and \(t\) are real constants.

The values of \(r, s\) and \(t\) respectively are

  1. 2, 3 and 5
  2. 5, 3 and 5
  3. 5, 5 and 8
  4. 5, 8 and 5
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\(D\)

Show Worked Solution
\({\underset{\sim}{r}}_1(\lambda)=2\underset{\sim}{i}+r\underset{\sim}{j}-3\underset{\sim}{k}+\lambda\left(\underset{\sim}{i}-\underset{\sim}{j}+4\underset{\sim}{k}\right)\)

\({\underset{\sim}{r}}_2(\mu)=\underset{\sim}{i}+0 \underset{\sim}{j}+s \underset{\sim}{k}+\mu(\underset{\sim}{i}+\underset{\sim}{j}-\underset{\sim}{k})\)

♦ Mean mark 49%.

\(\text{Since intersection occurs at}\ (4,3, t):\)

\(4=2+\lambda \ \Rightarrow \ \lambda=2\)

\(3=r-\lambda \ \Rightarrow \ 3=r-2\ \Rightarrow\ r=5\)

\(t=-3+4 \lambda=-3+4(2)=5\)

\(4=1+\mu \ \Rightarrow \ \mu=3\)

\(t=s-\mu \ \Rightarrow \ 5=s-3\ \Rightarrow\ s=8\)

\(\Rightarrow D\)

Vectors, EXT2 V1 2025 SPEC1 2

Consider the following two lines, \(L_1\) and \(L_2\).

\(L_1\) passes through the point \(A_1(2,3,1)\) and has direction  \(\underset{\sim}{u}=\underset{\sim}{i}+2 \underset{\sim}{j}-\underset{\sim}{k}\).

\(L_2\) passes through the point \(A_2(1,3,2)\) and has direction  \(\underset{\sim}{v}=-\underset{\sim}{i}-\underset{\sim}{j}+\underset{\sim}{k}\).

Find the coordinates of the point of intersection of the two lines.   (3 marks)

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\(\text{Intersection at}\ (3,5,0).\)

Show Worked Solution

\(\underset{\sim}{u}=\underset{\sim}{i}+2 \underset{\sim}{j}-\underset{\sim}{k} \ \ \text {passes through} \ A_1(2,3,1)\)

\(L_1:(2+t) \underset{\sim}{i}+(3+2 t) \underset{\sim}{j}+(1-t) \underset{\sim}{k}\)

\(\underset{\sim}{v}=\underset{\sim}{-i}-\underset{\sim}{j}+\underset{\sim}{k} \ \ \text {passes through}\  A_2(1,3,2)\)

\(L_2:(1-s) \underset{\sim}{i}+(3-s)\underset{\sim}{j}+(2+s) \underset{\sim}{k}\)

\(\text{Equating components for intersection:}\)

\(2+t=1-s \ \ \Rightarrow \ \ s+t=-1\ \ldots\ (1)\)

\(3+2 t=3-s \ \ \Rightarrow \ \ s+2 t=0\ \ldots\ (2)\)

\(\text{Subtract: (2) – (1)}\)

\(t=1 \ \Rightarrow \ s=-2\)

\(\text{Point of intersection:}\)

\((2+1,3+2,1-1) \equiv(3,5,0)\)

Vectors, EXT2 V1 2025 HSC 16c

Consider the point \(B\) with three-dimensional position vector \(\underset{\sim}{b}\) and the line  \(\ell: \underset{\sim}{a}+\lambda \underset{\sim}{d}\), where \(\underset{\sim}{a}\) and \(\underset{\sim}{d}\) are three-dimensional vectors, \(\abs{\underset{\sim}{d}}=1\) and \(\lambda\) is a parameter.

Let \(f(\lambda)\) be the distance between a point on the line \(\ell\) and the point \(B\).

  1. Find \(\lambda_0\), the value of \(\lambda\) that minimises \(f\), in terms of \(\underset{\sim}{a}, \underset{\sim}{b}\) and \(\underset{\sim}{d}\).   (2 marks)

  2. Let \(P\) be the point with position vector  \(\underset{\sim}{a}+\lambda_0 \underset{\sim}{d}\).
  3. Show that \(PB\) is perpendicular to the direction of the line \(\ell\).   (1 mark)

  4. Hence, or otherwise, find the shortest distance between the line \(\ell\) and the sphere of radius 1 unit, centred at the origin \(O\), in terms of \(\underset{\sim}{d}\) and \(\underset{\sim}{a}\).
  5. You may assume that if \(B\) is the point on the sphere closest to \(\ell\), then \(O B P\) is a straight line.   (3 marks)

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i.    \(\lambda_0=\underset{\sim}{d}(\underset{\sim}{b}-\underset{\sim}{a})\)

ii.   \(\text{See Worked Solutions.}\)

iii.  \(d_{\min }= \begin{cases}\sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}-1, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}>1 \\ 0, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2} \leqslant 1 \ \ \text{(i.e. it touches sphere) }\end{cases}\)

Show Worked Solution

i.    \(\ell=\underset{\sim}{a}+\lambda \underset{\sim}{d}, \quad\abs{\underset{\sim}{d}}=1\)

\(\text{Vector from point \(B\) to a point on \(\ell\)}:\ \underset{\sim}{a}+\lambda \underset{\sim}{d}-\underset{\sim}{b}\)

\(f(\lambda)=\text{distance between \(\ell\) and \(B\)}\)

\(f(\lambda)=\abs{\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d}}\)

\(\text{At} \ \ \lambda_0, f(\lambda) \ \ \text{is a min}\ \Rightarrow \ f(\lambda)^2 \ \ \text {is also a min}\)

♦♦ Mean mark (i) 33%.
\(f(\lambda)^2\) \(=\abs{\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d}}^2\)
  \(=(\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d})(\underset{\sim}{a}-\underset{\sim}{b}+\lambda \underset{\sim}{d})\)
  \(=(\underset{\sim}{a}-\underset{\sim}{b})\cdot (\underset{\sim}{a}-\underset{\sim}{b})+2\lambda (\underset{\sim}{a}-\underset{\sim}{b}) \cdot \underset{\sim}{d}+\lambda^2 \underset{\sim}{d} \cdot  \underset{\sim}{d}\)
  \(=\lambda^2|\underset{\sim}{d}|^2+2 \underset{\sim}{d} \cdot (\underset{\sim}{a}-\underset{\sim}{b}) \lambda +\abs{\underset{\sim}{a}-\underset{\sim}{b}}^2\)
  \(=\lambda^2+2 \underset{\sim}{d} \cdot (\underset{\sim}{a}-\underset{\sim}{b}) \lambda+\abs{\underset{\sim}{a}-\underset{\sim}{b}}^2\)

 

\(f(\lambda)^2 \ \ \text{is a concave up quadratic.}\)

\(f(\lambda)_{\text {min}}^2 \ \ \text{occurs at the vertex.}\)

\(\lambda_0=-\dfrac{b}{2 a}=-\dfrac{2 \underset{\sim}{d} \cdot (\underset{\sim}{a}-\underset{\sim}{b})}{2}=\underset{\sim}{d} \cdot (\underset{\sim}{b}-\underset{\sim}{a})\)
 

ii.    \(P \ \text{has position vector} \ \ \underset{\sim}{a}+\lambda_0 \underset{\sim}{d}\)

\(\text{Show} \ \ \overrightarrow{PB} \perp \ell:\)

♦♦♦ Mean mark (ii) 22%.

\(\overrightarrow{PB}=\underset{\sim}{b}-\underset{\sim}{p}=\underset{\sim}{b}-\underset{\sim}{a}-\lambda_0 \underset{\sim}{d}\)

\(\overrightarrow{P B} \cdot \underset{\sim}{d}\) \(=\left(\underset{\sim}{b}-\underset{\sim}{a}-\lambda_0 \underset{\sim}{d}\right) \cdot \underset{\sim}{d}\)
  \(=(\underset{\sim}{b}-\underset{\sim}{a}) \cdot \underset{\sim}{d}-\lambda_0 \underset{\sim}{d} \cdot \underset{\sim}{d}\)
  \(=\lambda_0-\lambda_0\abs{\underset{\sim}{d}}^2\)
  \(=0\)

 

\(\therefore \overrightarrow{PB}\ \text{is perpendicular to the direction of the line}\ \ell. \)
 

iii.   \(\text{Shortest distance between} \ \ell \ \text{and sphere (radius\(=1\))}\)

\(=\ \text{(shortest distance \(\ell\) to \(O\))}-1\)

♦♦♦ Mean mark (iii) 4%.

\(f\left(\lambda_0\right)=\text{shortest distance \(\ell\) to point \(B\)}\)

\(\text{Set} \ \ \underset{\sim}{b}=0 \ \Rightarrow \ f\left(\lambda_0\right)=\text{shortest distance \(\ell\) to \(0\)}\)

\(\Rightarrow \lambda_0=\underset{\sim}{d} \cdot (\underset{\sim}{b}-\underset{\sim}{a})=-\underset{\sim}{d} \cdot \underset{\sim}{a}\)

\(f\left(\lambda_0\right)\) \(=\abs{\underset{\sim}{a}-\underset{\sim}{b}-(\underset{\sim}{d} \cdot \underset{\sim}{a})\cdot \underset{\sim}{d}}=\abs{\underset{\sim}{a}-(\underset{\sim}{d} \cdot \underset{\sim}{a})\cdot \underset{\sim}{d}}\)
\(f\left(\lambda_0\right)^2\) \(=\abs{\underset{\sim}{a}}^2-2( \underset{\sim}{a}\cdot \underset{\sim}{d})^2+(\underset{\sim}{d} \cdot \underset{\sim}{a})^2\abs{\underset{\sim}{d}}^2\)
  \(=\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2\)
\(f\left(\lambda_0\right)\) \(=\sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}\)

 

\(\text {Shortest distance of \(\ell\) to sphere \(\left(d_{\min }\right)\):}\)

\(d_{\min }= \begin{cases}\sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}-1, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2}>1 \\ 0, & \sqrt{\abs{\underset{\sim}{a}}^2-(\underset{\sim}{a} \cdot \underset{\sim}{d})^2} \leqslant 1 \ \ \text{(i.e. it touches sphere) }\end{cases}\)

Vectors, EXT2 V1 2025 HSC 10 MC

Which of the following gives the same curve as  \(\left(\begin{array}{c}\cos (t) \\ -t \\ \sin (t)\end{array}\right)\) for  \(t \in \mathbb{R}\) ?

  1. \(\left(\begin{array}{c}\cos (2 t) \\ 2 t \\ \sin (2 t)\end{array}\right)\)
  2. \(\left(\begin{array}{c}\cos \left(t^2+\dfrac{\pi}{2}\right) \\ t^2+\dfrac{\pi}{2} \\ \sin \left(t^2+\dfrac{\pi}{2}\right)\end{array}\right)\)
  3. \(\left(\begin{array}{c}\cos \left(t^2\right) \\ -t^2 \\ \sin \left(t^2\right)\end{array}\right)\)
  4. \(\left(\begin{array}{c}\cos \left(2 t+\dfrac{\pi}{2}\right) \\ 2 t+\dfrac{\pi}{2} \\ -\sin \left(2 t+\dfrac{\pi}{2}\right)\end{array}\right)\)
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\(D\)

Show Worked Solution

\(\text{Find which option is a re-parametrisation of the given curve.}\)

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

\(\text{Let}\ \ u=- \left(2t + \dfrac{\pi}{2}\right)\)

\(\text{Since}\ t \in \mathbb{R}\ \ \Rightarrow \ \ u \in \mathbb{R}\)

\(\cos \left(2 t+\dfrac{\pi}{2}\right) = \cos\left(- \left(2 t+\dfrac{\pi}{2}\right) \right) = \cos\,u\)

\(2 t+\dfrac{\pi}{2} = -\left( -\left(2 t+\dfrac{\pi}{2} \right) \right) = -u\)

\(-\sin \left(2 t+\dfrac{\pi}{2}\right) = \sin\left(- \left(2 t+\dfrac{\pi}{2}\right) \right) = \sin\,u\)

\(\Rightarrow D\)

♦ Mean mark 52%.

Vectors, EXT2 V1 2025 HSC 1 MC

Points \(A\) and \(B\) are \((-3,1)\) and \((1,4)\) respectively.

Which of the following is a vector equation of the line \(A B\) with parameter \(\lambda\) ?

  1. \(\displaystyle \binom{x}{y}=\binom{1}{4}+\lambda\binom{3}{4}\)
  2. \(\displaystyle\binom{x}{y}=\binom{3}{4}+\lambda\binom{1}{4}\)
  3. \(\displaystyle\binom{x}{y}=\binom{4}{3}+\lambda\binom{-3}{1}\)
  4. \(\displaystyle\binom{x}{y}=\binom{-3}{1}+\lambda\binom{4}{3}\)
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\(D\)

Show Worked Solution

\(\overrightarrow{AB}=\displaystyle \binom{1}{4}-\binom{-3}{1}=\binom{4}{3}\)

\(\text{Line} \ \ AB:\)

\(\displaystyle \binom{x}{y}=\binom{-3}{1}+\lambda\binom{4}{3}\)

\(\Rightarrow D\)

Vectors, EXT2 V1 2024 HSC 15a

Consider the three vectors  \(\underset{\sim}{a}=\overrightarrow{O A}, \underset{\sim}{b}=\overrightarrow{O B}\)  and  \(\underset{\sim}{c}=\overrightarrow{O C}\), where \(O\) is the origin and the points \(A, B\) and \(C\) are all different from each other and the origin.

The point \(M\) is the point such that  \(\dfrac{1}{2}(\underset{\sim}{a}+\underset{\sim}{b})=\overrightarrow{O M}\).

  1. Show that \(M\) lies on the line passing through \(A\) and \(B\).   (1 mark)

  2. The point \(G\) is the point such that  \(\dfrac{1}{3}(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c})=\overrightarrow{O G}\).
  3. Show that \(G\) lies on the line passing through \(M\) and \(C\), and lies between \(M\) and \(C\).   (2 marks)

  4. The complex numbers \(x, w\) and \(z\) are all different and all have modulus 1.
  5. Using part (ii), or otherwise, show that  \(\dfrac{1}{3}(x+w+z)\) is never a cube root of \(x w z\).   (2 marks)

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i.    \(\text{Equation of line through \(A\) and \(B\)}\)

\(\Rightarrow \ell_1=\overrightarrow{O A}+\lambda \overrightarrow{A B}\)

  \(\overrightarrow{O M}\) \(=\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}\)
    \(=\underset{\sim}{a}-\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}\)
    \(=\overrightarrow{O A}+\dfrac{1}{2}(\underset{\sim}{b}-\underset{\sim}{a})\)
    \(=\overrightarrow{O A}+\dfrac{1}{2} \overrightarrow{A B}\)

 
\(\therefore \overrightarrow{OM} \ \text{lies on} \ \ell_1\).
 

ii.    \(\text{Equation of line through \(M\) and \(C\)}\)

\(\Rightarrow \ell_2=\overrightarrow{OC}+\lambda \overrightarrow{CM}\)

  \(\overrightarrow{O G}\) \(=\dfrac{1}{3}(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c})\)
    \(=\underset{\sim}{c}-\dfrac{2}{3} \underset{\sim}{c}+\dfrac{1}{3} \underset{\sim}{a}+\dfrac{1}{3} \underset{\sim}{b}\)
    \(=\overrightarrow{OC}+\dfrac{2}{3}\left(\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}-\underset{\sim}{c}\right)\)
    \(=\overrightarrow{OC}+\dfrac{2}{3} \overrightarrow{CM}\)

 
\(\therefore \overrightarrow{O G} \ \ \text{lies on} \ \ \ell_2\)

\(\ \ \overrightarrow{O G} \neq \overrightarrow{O C}  \ \ \text{and} \ \ \overrightarrow{O G} \neq \overrightarrow{O M}\)

\(\therefore G \ \ \text{lies between} \ \ C \ \text{and} \ M\).
 

iii.  \(\text{Place}\ x, w,\ \text{and}\ z\ \text{on unit circle.}\)
 

\(\abs{w}=\abs{x}=\abs{z}=1\)

\(\text{Using part (ii):}\)

\(G \equiv \dfrac{1}{3}(x+w+z)\)

\(G \ \text{lies on} \ CM \Rightarrow G \ \text{is inside the unit circle.}\)

\(\Rightarrow\left|\dfrac{1}{3}(x+w+z)\right|<1\)

\(\text{Since}\ \ \abs{xwz}=\abs{x}\abs{w}\abs{z}=1\)

\(\Rightarrow \ \text{All cube roots have modulus = 1.}\)

\(\therefore \dfrac{1}{3}(x+w+z) \ \ \text{cannot be a cube root of  \(xwz\).}\)

Show Worked Solution

i.    \(\text{Equation of line through \(A\) and \(B\)}\)

\(\Rightarrow \ell_1=\overrightarrow{O A}+\lambda \overrightarrow{A B}\)

  \(\overrightarrow{O M}\) \(=\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}\)
    \(=\underset{\sim}{a}-\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}\)
    \(=\overrightarrow{O A}+\dfrac{1}{2}(\underset{\sim}{b}-\underset{\sim}{a})\)
    \(=\overrightarrow{O A}+\dfrac{1}{2} \overrightarrow{A B}\)

 
\(\therefore \overrightarrow{OM} \ \text{lies on} \ \ell_1\).
 

ii.    \(\text{Equation of line through \(M\) and \(C\)}\)

\(\Rightarrow \ell_2=\overrightarrow{OC}+\lambda \overrightarrow{CM}\)

  \(\overrightarrow{O G}\) \(=\dfrac{1}{3}(\underset{\sim}{a}+\underset{\sim}{b}+\underset{\sim}{c})\)
    \(=\underset{\sim}{c}-\dfrac{2}{3} \underset{\sim}{c}+\dfrac{1}{3} \underset{\sim}{a}+\dfrac{1}{3} \underset{\sim}{b}\)
    \(=\overrightarrow{OC}+\dfrac{2}{3}\left(\dfrac{1}{2} \underset{\sim}{a}+\dfrac{1}{2} \underset{\sim}{b}-\underset{\sim}{c}\right)\)
    \(=\overrightarrow{OC}+\dfrac{2}{3} \overrightarrow{CM}\)

 
\(\therefore \overrightarrow{O G} \ \ \text{lies on} \ \ \ell_2\)

\(\ \ \overrightarrow{O G} \neq \overrightarrow{O C}  \ \ \text{and} \ \ \overrightarrow{O G} \neq \overrightarrow{O M}\)

\(\therefore G \ \ \text{lies between} \ \ C \ \text{and} \ M\).

♦ Mean mark (ii) 43%.

iii.  \(\text{Place}\ x, w,\ \text{and}\ z\ \text{on unit circle.}\)
 

♦♦♦ Mean mark (iii) 10%.

\(\abs{w}=\abs{x}=\abs{z}=1\)

\(\text{Using part (ii):}\)

\(G \equiv \dfrac{1}{3}(x+w+z)\)

\(G \ \text{lies on} \ CM \Rightarrow G \ \text{is inside the unit circle.}\)

\(\Rightarrow\left|\dfrac{1}{3}(x+w+z)\right|<1\)

\(\text{Since}\ \ \abs{xwz}=\abs{x}\abs{w}\abs{z}=1\)

\(\Rightarrow \ \text{All cube roots have modulus = 1.}\)

\(\therefore \dfrac{1}{3}(x+w+z) \ \ \text{cannot be a cube root of  \(xwz\).}\)

Vectors, EXT2 V1 2024 HSC 13a

The point \(A\) has position vector  \(8 \underset{\sim}{i}-6 \underset{\sim}{j}+5 \underset{\sim}{k}\). The line \(\ell\) has vector equation

\(x \underset{\sim}{i}+y \underset{\sim}{j}+z \underset{\sim}{k}=t(\underset{\sim}{i}+\underset{\sim}{j}+2 \underset{\sim}{k})\).

The point \(B\) lies on \(\ell\) and has position vector  \(p \underset{\sim}{i}+p \underset{\sim}{j}+2 p \underset{\sim}{k}\).

  1. Show that  \(\abs{A B}^2=6 p^2-24 p+125\).   (1 mark)

  2. Hence, or otherwise, determine the shortest distance between the point \(A\) and the line \(\ell\).  (2 marks)

Show Answers Only

i.     \(A\left(\begin{array}{c}8 \\ -6 \\ 5\end{array}\right), \quad \ell: \left(\begin{array}{l}x \\ y \\ z\end{array}\right)+t\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right), \quad B\left(\begin{array}{c}p \\ p \\ 2 p\end{array}\right)\)
 

\(\overrightarrow{A B}=\left(\begin{array}{l}p \\ p \\ 2 p\end{array}\right)-\left(\begin{array}{c}8 \\ -6 \\ 5\end{array}\right)=\left(\begin{array}{c}p-8 \\ p+6 \\ 2 p-5\end{array}\right)\)
 

  \(\abs{AB}^2\) \(=(p-8)^2+(p+6)^2+(2 p-5)^2\)
    \(=p^2-16 p+64+p^2+12 p+36+4 p^2-20 p+25\)
    \(=6 p^2-24 p+125\)

 

ii.    \(\abs{AB}_{\text {min}}=\sqrt{101} \text { units}\)

Show Worked Solution

i.     \(A\left(\begin{array}{c}8 \\ -6 \\ 5\end{array}\right), \quad \ell: \left(\begin{array}{l}x \\ y \\ z\end{array}\right)+t\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right), \quad B\left(\begin{array}{c}p \\ p \\ 2 p\end{array}\right)\)
 

\(\overrightarrow{A B}=\left(\begin{array}{l}p \\ p \\ 2 p\end{array}\right)-\left(\begin{array}{c}8 \\ -6 \\ 5\end{array}\right)=\left(\begin{array}{c}p-8 \\ p+6 \\ 2 p-5\end{array}\right)\)
 

  \(\abs{AB}^2\) \(=(p-8)^2+(p+6)^2+(2 p-5)^2\)
    \(=p^2-16 p+64+p^2+12 p+36+4 p^2-20 p+25\)
    \(=6 p^2-24 p+125\)

  

ii.    \(\text{Find shortest distance between \(A\) and \(\ell\).}\)

\(\Rightarrow \text { Find \(p\) when \(\abs{A B}\) is a minimum:}\)

\(\text{Minimum occurs when}\ \ p=\dfrac{-b}{2 a}=\dfrac{24}{2 \times 6}=2\)

\(\therefore \abs{AB}_{\text {min}}=\sqrt{6(2)^2-24(2)+125}=\sqrt{101} \text { units}\)

♦ Mean mark (ii) 45%.

Vectors, EXT2 V1 2024 HSC 12e

The line \(\ell\) passes through the points \(A(3,5,-4)\) and \(B(7,0,2)\).

  1. Find a vector equation of the line \(\ell\).   (1 mark)

  2. Determine, giving reasons, whether the point \(C(10,5,-2)\) lies on the line \(\ell\).   (2 marks)

Show Answers Only

i.    \(\text{ Equation of line:} \ \left(\begin{array}{c}3 \\ 5 \\ -4\end{array}\right)+\lambda\left(\begin{array}{c}4 \\ -5 \\ 6\end{array}\right),\ (\lambda \in \mathbb{R} )\)

ii.    \(\text{If \(C\) lies on the line}, \ \exists \lambda \in \mathbb{R}\  \ \text{such that:}\)

\(\left(\begin{array}{c}10 \\ 5 \\ -2\end{array}\right)=\left(\begin{array}{c}3 \\ -5 \\ 6\end{array}\right)+\lambda\left(\begin{array}{c}4 \\ -5 \\ 6\end{array}\right)\)

 
\(\text{Find \(\lambda\) for \(x\)-component: }\)

\(10=3+4 \lambda \ \Rightarrow \ \lambda=\dfrac{7}{4}\)

\(\text{Check \(\lambda=\dfrac{7}{4}\) for \(y\)-component:}\)

\(5=-5+\dfrac{7}{4}(-5) \ \ \Rightarrow\ \ \text{not correct}\)

\(\therefore C\  \text{does not lie on the line.}\)

Show Worked Solution

i.     \(A(3,5,-4), \quad B(7,0,2)\)

\(\overrightarrow{A B}=\left(\begin{array}{l}7 \\ 0 \\ 2\end{array}\right)-\left(\begin{array}{c}3 \\ 5 \\ -4\end{array}\right)=\left(\begin{array}{c}4 \\ -5 \\ 6\end{array}\right)\)

\(\therefore \text{ Equation of line:} \ \left(\begin{array}{c}3 \\ 5 \\ -4\end{array}\right)+\lambda\left(\begin{array}{c}4 \\ -5 \\ 6\end{array}\right),\ (\lambda \in \mathbb{R} )\)
 

ii.    \(\text{If \(C\) lies on the line}, \ \exists \lambda \in \mathbb{R}\  \ \text{such that:}\)

\(\left(\begin{array}{c}10 \\ 5 \\ -2\end{array}\right)=\left(\begin{array}{c}3 \\ -5 \\ 6\end{array}\right)+\lambda\left(\begin{array}{c}4 \\ -5 \\ 6\end{array}\right)\)

 
\(\text{Find \(\lambda\) for \(x\)-component: }\)

\(10=3+4 \lambda \ \Rightarrow \ \lambda=\dfrac{7}{4}\)

\(\text{Check \(\lambda=\dfrac{7}{4}\) for \(y\)-component:}\)

\(5=-5+\dfrac{7}{4}(-5) \ \ \Rightarrow\ \ \text{not correct}\)

\(\therefore C\  \text{does not lie on the line.}\)

Vectors, EXT2 V1 2023 HSC 5 MC

Which of the following is a true statement about the lines  \(\ell_1={\displaystyle\left(\begin{array}{cc}-1 \\ 2 \\ 5\end{array}\right)+\lambda\left(\begin{array}{c}-1 \\ 3 \\ 1\end{array}\right)}\)  and  \(\ell_2=\left(\begin{array}{c}3 \\ -10 \\ 1\end{array}\right)+\mu\left(\begin{array}{c}1 \\ -3 \\ -1\end{array}\right) ?\)

  1. \(\ell_1\) and \(\ell_2\) are the same line.
  2. \(\ell_1\) and \(\ell_2\) are not parallel and they intersect.
  3. \(\ell_1\) and \(\ell_2\) are parallel and they do not intersect.
  4. \(\ell_1\) and \(\ell_2\) are not parallel and they do not intersect.
Show Answers Only

\(A\)

Show Worked Solution

\(\text{Since}\ \ \left(\begin{array}{c}-1 \\ 3 \\ 1\end{array}\right) = -1 \left(\begin{array}{c}1 \\ -3 \\ -1\end{array}\right), \ \ell_1\ \text{is parallel to}\ \ell_2 \)

\(\text{Test if point}\ (3,-10,1)\ \text{lies on}\ \ell_1: \)

\(\text{i.e.}\ \ \exists \lambda\ \ \text{such that} \)

♦ Mean mark 49%.

\( \left(\begin{array}{cc}3 \\ -10 \\ 1\end{array}\right) = \left(\begin{array}{cc}-1 \\ 2 \\ 5\end{array}\right) + \lambda \left(\begin{array}{cc}-1 \\ 3 \\ 1\end{array}\right)\)

\(\lambda = -4\ \ \text{satisfies equation} \)

\(\therefore\ \ell_1\ \text{and}\ \ell_2\ \text{are the same line.}\)

\(\Rightarrow A\)

Vectors, EXT2 V1 2023 HSC 15c

A curve \( \mathcal{C}\) spirals 3 times around the sphere centred at the origin and with radius 3, as shown.

A particle is initially at the point \((0,0,-3)\) and moves along the curve \(\mathcal{C}\) on the surface of the sphere, ending at the point \((0,0,3)\).
 

By using the diagram below, which shows the graphs of the functions  \(f(x)=\cos (\pi x)\)  and  \(g(x)=\sqrt{9-x^2}\), and considering the graph  \(y=f(x)g(x)\), give a possible set of parametric equations that describe the curve \( \mathcal{C}\).  (3 marks)
 

Show Answers Only

\(x= \cos{(\pi t)}\sqrt{9-t^2} \)

\(y= -\sin{(\pi t)}\sqrt{9-t^2} \)

\( z=t \)

Show Worked Solution

\(\text{Since the curve lies on a sphere with radius 3:}\)

\(x^2+y^2+z^2=3^3 \)

\(\text{Considering the graph}\ \ y=\cos (\pi t)\sqrt{9-t^2}\ \ \text{(as per hint)} \)

\(\Big(\cos (\pi t)\sqrt{9-t^2}\Big)^2+\Big(\sin (\pi t)\sqrt{9-t^2}\Big)^2+t^2=3^2 \ \ …\ (1) \)

\(\text{Since}\ z\ \text{increases and}\ x\ \text{and}\ y\ \text{change signs} \)

\( \Rightarrow z=t \)
 

\(\text{In order to satisfy the equation in (1): } \)

\( x,y\ \text{must be one of }\ \ \pm \cos{(\pi t)}\sqrt{9-t^2}\ \ \text{or}\ \ \pm \sin{(\pi t)}\sqrt{9-t^2} \)
 

\(\text{At}\ \ z=0,\ t=0, \ x=3\ \ \text{(from graph):} \)

\( \Rightarrow x= \cos{(\pi t)}\sqrt{9-t^2} \)
 

\(\text{At}\ \ z=0+\epsilon,\ t=0+\epsilon, \ y \lt 0\ \ \text{(from graph):} \)

\( \Rightarrow y= -\sin{(\pi t)}\sqrt{9-t^2} \)

♦♦♦ Mean mark 22%.

Vectors, EXT2 V1 2023 HSC 11c

Find a vector equation of the line through the points  \(A(-3,1,5)\)  and  \(B(0,2,3)\).  (2 marks)

Show Answers Only

\[\underset{\sim}{v}=\left(\begin{array}{c} -3 \\1 \\ 5 \end{array}\right) + \lambda \left(\begin{array}{c} 3 \\ 1 \\ -2 \end{array}\right),\ \ \ \text{for some}\ \ \lambda \in \mathbb{R} \]

Show Worked Solution

\[\overrightarrow{AB}=\left(\begin{array}{c} 0-(-3) \\2-1 \\ 3-5 \end{array}\right) = \left(\begin{array}{c} 3 \\ 1 \\ -2 \end{array}\right)\]

\[\therefore \underset{\sim}{v}=\left(\begin{array}{c} -3 \\1 \\ 5 \end{array}\right) + \lambda \left(\begin{array}{c} 3 \\ 1 \\ -2 \end{array}\right),\ \ \ \text{for some}\ \ \lambda \in \mathbb{R} \]

Vectors, EXT2 V1 2022 HSC 11e

Let `ℓ_(1)` be the line with equation `([x],[y])=([-1],[7])+lambda([3],[2]),lambda inRR`.

The line `ℓ_(2)` passes through the point  `A(-6,5)`  and is parallel to `ℓ_(1)`.

Find the equation of the line `ℓ_(2)` in the form  `y=mx+c`.  (2 marks)

Show Answers Only

`y=2/3x+9`

Show Worked Solution

`m_(ℓ_(1))=2/3`

`text{Equation of}\ ℓ_(2)\ text{has}\ m=2/3\ text{and passes through}\ (-6,5):`

`y-5` `=2/3(x+6)`  
`y` `=2/3x+9`  

Vectors, EXT2 V1 2021 HSC 12c

Two lines are given by  `text(r)_1 = ((-2),(1),(3)) + lambda((1),(0),(2))`  and  `text(r)_2 = ((4),(-2),(q)) + mu ((p),(3),(-1))` , where `p` and `q` are real numbers. These lines intersect and are perpendicular.

Find the values of `p` and `q`.  (3 marks)

Show Answers Only

`p = 2 \ , \ q = 20`

Show Worked Solution

`text{S} text{ince vectors are perpendicular}`

`((1),(0),(2)) ((p),(3),(-1))` `= 0`
`p – 2` `= 0`
`p` `= 2`
 

`text{S} text{ince lines intersect, equate}\ y text{-coordinates:} `

`1 + lambda 0` `= -2 + 3 mu`
`mu` `= 1`

 

`text{Find} \ lambda \ text{by equating}\ xtext{-coordinates:}`

`-2 + lambda` `= 4 + 1 xx 2`
`lambda` `= 8`

 
`text{Equating}\ ztext{-coordinates:}`

`3 + 8 xx 2` `= q – 1 xx 1`
`q` `= 20`

Vectors, EXT2 V1 2021 HSC 7 MC

Which diagram best shows the curve described by the position vector

`underset~r (t) = -5 text{cos}(t) underset~i + 5 text{sin}(t) underset~j + t underset~k`  for  `0 ≤ t ≤ 4 pi` ?

 

Show Answers Only

`D`

Show Worked Solution

`text{By elimination}`

`text{Check graph coordinates for specific values of}\ t:`

`text{When} \ \ t = 4 pi \ , \ underset~r = -5 underset~i + 0 underset~j + 4 pi underset~k`

`-> \ text{Eliminate A and B}`

`text{When} \ t = pi/4 \ , \ underset~r = (-5)/sqrt(2) underset~i + 5/sqrt(2) underset~j + pi/4 underset~k`

`-> \ text{Eliminate C}`
 

`=>\ D`

Vectors, EXT2 V1 2021 HSC 3 MC

Which of the following is a vector equation of the line joining the points  `A (4, 2, 5)`  and  `B (–2, 2, 1)`?

  1. `underset~r = ((4), (2), (5)) + λ ((1),(2),(3))`
  2. `underset~r = ((4), (2), (5)) + λ ((3),(0),(2))`
  3. `underset~r = ((1), (2), (3)) + λ ((4),(2),(5))`
  4. `underset~r = ((3), (0), (2)) + λ ((4),(2),(5))`
Show Answers Only

`B`

Show Worked Solution
`overset->{AB}` `= ((-2),(2),(1)) – ((4),(2),(5)) = ((-6),(0),(-4))`  
`underset~r` `= ((4),(2),(5)) + λ_1 ((-6),(0),(-4))`  
  `= ((4), (2), (5)) + λ_2 ((3),(0),(2))`  

 
`=>\ B`

Vectors, EXT2 V1 2020 HSC 13b

Consider the two lines in three dimensions given by
 

`underset~r = ((3),(-1),(7)) + λ_1 ((1),(2),(1))`  and  `underset~r = ((3),(-6),(2)) + λ_2 ((-2),(1),(3))`.
 

By equating components, find the point of intersection of the two lines.   (3 marks)

Show Answers Only

`((1),(-5),(5))`

Show Worked Solution

`underset~(r_1) = ((3),(-1),(7)) + λ_1 ((1),(2),(1)) = ((3 + λ_1),(-1 + 2λ_1),(7 + λ_1))`

`underset~(r_2) = ((3),(-6),(2)) + λ_2 ((-2),(1),(3)) = ((3 – 2λ_2),(-6 + λ_2),(2 + 3λ_2))`

 
`text{Intersection occurs when:}`

`3 + λ_1 ` `= 3 – 2λ_2 \ … \ (1)`
`-1 + 2λ_1` `= -6 + λ_2 \ … \ (2)`
`7 + λ_1` `= 2 + 3λ_2 \ … \ (3)`

 
`text{Subtract} \ (3) – (1):`

`4` `= -1 + 5 λ_2`
`λ_2` `=1`

 
`text{Substitute} \ \ λ_2 = 1\ \ text{into} \ (1):`

`3 + λ_1` `= 1`
`λ_1` `= -2`

 

`text{Test that}\  \ λ_1 = -2 \ , \  λ_2 = 1\ \ text{satisfies} \ (2):`

`-1 + 2 xx  – 2` `= -6 + 1`
`-5` `= -5`

 
`∃ \ λ_1,  λ_2, \ text{that satisfy all equations}`

`=> \ text{3 lines intersect at a point}`

`:.\ text{Point of intersection}`

`= ((3),(-6),(2)) + 1 ((-2),(1),(3)) = ((1),(-5),(5))`

Vectors, EXT2 V1 2020 HSC 3 MC

 What is the Cartesian equation of the line  `underset~r = ((1),(3)) + lambda ((-2),(4))`?

  1. `2y + x = 7`
  2. `y - 2x = -5`
  3. `y + 2x = 5`
  4. `2y - x = -1`
Show Answers Only

`C`

Show Worked Solution

`((x),(y)) = ((1),(3)) + λ ((-2),(4))`
 

`x = 1 – 2λ \ => \ λ = frac{1- x}{2}`

`y = 3 + 4λ \ => \ λ = frac{y- 3}{4}`
 

`frac{y – 3}{4}` `= frac{1 – x}{2}`
`y – 3` `= 2 – 2x`
`y + 2x` `= 5`

  
`=> \ C`

Vectors, EXT2 V1 EQ-Bank 16

Show that the points `A(2, 1, text{−1}), \ B(4, 2, text{−3})` and `C(text{−4}, text{−2}, 5)` are collinear.   (2 marks)

Show Answers Only

`text(Proof)\ \ text{(See Worked Solutions)}`

Show Worked Solution

`vec(AB) = ((4), (2), (text{−3})) – ((2), (1), (text{−1})) = ((2), (1), (text{−2}))`

`vec(AC) = ((text{−4}), (text{−2}), (5)) – ((2), (1), (text{−1})) = ((text{−6}), (text{−3}), (6)) = -3((2), (1), (text{−2}))`

 

`text(S) text(ince)\ vec(AB)\ text(||)\ vec(AC) and A\ text(lies on both lines,)`

`A, B, C\ \ text(are collinear).`

Vectors, EXT2 V1 EQ-Bank 15

  1. Find the equation of the vector line `underset~v` that passes through  `Atext{(5, 2, 3)}` and `B(7, 6, 1)`.   (1 mark)

  2. A sphere has centre `underset~c` at `text{(2, 3, 5)}` and a radius of `5sqrt2`  units.
    Find the points where the vector line `underset~v` meets the sphere.   (3 marks)

Show Answers Only

a.    `underset~v = ((5),(2),(3)) + lambda((1),(2),(−1))`

b.    `((2),(–4),(6)), \ ((7),(6),(1))`

Show Worked Solution

a.    `overset(->)(BA) = ((7-5),(6-2),(1-3)) = ((2),(4),(−2)) = 2((1),(2),(−1))`

`underset~v = ((5),(2),(3)) + lambda((1),(2),(−1))`
 

b.    `text(General point)\ underset~v:`

`x = 5 + lambda`

`y = 2 + 2lambda`

`z = 3-lambda`
 

`text(Equation of sphere,)\ underset~c = (2, 3, 5),\ text(radius)\ 5sqrt2:`

`(x-2)^2 + (y-3)^2 + (z -5)^2` `= (5sqrt2)^2`
`(lambda + 3)^2 + (2lambda-1)^2 + (−lambda-2)^2` `= 50`
`lambda^2 + 6lambda + 9 + 4lambda^2-4lambda + 1 + lambda^2 + 4lambda + 4` `= 50`
`6lambda^2 + 6lambda + 14` `= 50`
`6lambda^2 + 6lambda-36` `= 0`
`6(lambda + 3)(lambda-2)` `= 0`
`lambda` `= –3\ text(or)\ 2`

 
`text(When)\ \ lambda = –3,`

`text(Intersection) = ((5),(2),(3))-3((1),(2),(−1)) = ((2),(–4),(6))`

`text(When)\ \ lambda = 2,`

`text(Intersection) = ((5),(2),(3)) + 2((1),(2),(−1)) = ((7),(6),(1))`

Vectors, EXT2 V1 EQ-Bank 14

A sphere is represented by the equation

`x^2-4x + y^2 + 8y + z^2-3z + 2 = 0`

  1. Determine the centre  `underset~c`  and radius of the sphere.   (2 marks)

  2. Find the vector equation of the sphere.   (1 mark)

Show Answers Only

a.    `underset~c = ((2),(-4),({3}/{2})) \ , \ text(radius) = (9)/(2)`

b.    `| \ underset~r-((2),(-4),({3}/{2})) | = (9)/(2)`

Show Worked Solution

a.    `x^2-4x + y^2 + 8y + z^2-3z + 2 = 0`

`(x-2)^2 + (y+4)^2 + (z-{3}/{2})^2 + 2-(89)/(4) = 0`

`(x-2)^2 + (y+4)^2 + (z-{3}/{2})^2 = (81)/(4)`

`:. \ underset~c = ((2),(-4),({3}/{2})) \ , \ text(radius) = (9)/(2)`
  

b.    `text(Vector equation:)`

`| \ underset~r-((2),(-4),({3}/{2})) | = (9)/(2)`

Vectors, EXT2 V1 EQ-Bank 22

Consider the two vector line equations

`underset~(v_1) = ((1),(4),(−2)) + lambda_1((3),(0),(−1)), qquad underset~(v_2) = ((3),(2),(2)) + lambda_2((4),(2),(−6))`

  1. Show that `underset~(v_1)` and `underset~(v_2)` intersect and determine the point of intersection.   (2 marks)

  2. What is the acute angle between the vector lines, to the nearest minute.   (2 marks)

Show Answers Only

a.    `text(See Worked Solutions)`

b.    `40°29’\ \ (text(nearest minute))`

Show Worked Solution

a.    `text(Solve simultaneously:)`

`1 + 3lambda_1` `= 3 + 4lambda_2` `\ \ …\ (1)`
`4 + 0lambda_1` `= 2 + 2lambda_2` `\ \ …\ (2)`
`−2 – lambda_1` `= 2 – 6lambda_2` `\ \ …\ (3)`

 
`=> lambda_2 = 1\ \ \ text{(from (2))}`

`=>lambda_1 = 2\ \ \ text{(from (1) and (3))}`

`:.\ text(vector lines intersect)`
  

`text(P.O.I.) = ((1),(4),(−2)) + 2((3),(0), (−1)) = ((7),(4),(−4))`
 

b.    `underset~(v_1) = underset~(a_1) + lambda_1*underset~(b_1)`

`underset~(v_2) = underset~(a_2) + lambda_2*underset~(b_2)`

`costheta` `= (underset~(b_1) · underset~(b_2))/(|underset~b_1||underset~b_2|)`
  `= (12 + 0 + 6)/(sqrt10 sqrt56)`
  `= 0.7606…`

 

`theta` `= 40.479…`
  `= 40°29’\ \ (text(nearest minute))`

Vectors, EXT2 V1 EQ-Bank 19

  1. What vector line equation, `underset~r`, corresponds to the Cartesian equation
  2. `qquad (x + 2)/5 = (y-5)/4`   (1 mark)

  3. Express   `underset~v`  in Cartesian form where,
  4. `qquad underset~v = ((1),(−4)) + lambda((3),(1))`   (1 mark)

Show Answers Only

 a.    `underset~r = ((−2),(5)) + lambda((5),(4))`

 b.    `y = (x-13)/3`

Show Worked Solution

a.    `underset~r = ((−2),(5)) + lambda((5),(4))`

COMMENT: Ensure you know the format for conversion from Cartesian to vector form (part i)!

 

b.    `((x),(y)) = ((1),(−4)) + lambda((3),(1))`

`x = 1 + 3lambda \ \ => \ lambda = (x-1)/3`
`y = −4 + lambda\ \ => \ lambda = y + 4`

 
`y + 4 = (x-1)/3`

`:. y = (x-13)/3`

Vectors, EXT2 V1 EQ-Bank 18

Determine the equation of the line vector `underset~r`, given it passes through the point `(7, 1, 0)` and is parallel to the line joining `P(2, −1, 2)` and `Q(3, 4, 1)`.   (2 marks)

Show Answers Only

`underset~r = ((7),(1),(0)) + lambda((1),(5),(−1))`

Show Worked Solution

`overset(->)(PQ) = ((3),(4),(1)) – ((2),(−1),(2)) = ((1),(5),(−1))`
 

`:. underset~r = ((7),(1),(0)) + lambda((1),(5),(−1))`

Vectors, EXT2 V1 EQ-Bank 17

Find the value of `n` given

`underset~v = ((5),(2),(n)) + lambda_1((2),(1),(3))` 

is perpendicular to

`underset~u = ((2),(0),(1)) + lambda_2((2),(n),(1))`.   (2 marks)

Show Answers Only

`−7`

Show Worked Solution

`text(Given)\ \ underset~v ⊥ underset~u:`

`((2),(1),(3)) · ((2),(n),(1)) = 0`

`4 + n + 3` `= 0`
`n` `= −7`

Vectors, EXT2 V1 EQ-Bank 23

  1. Find values of `a, b, c` and `d` such that  `underset~v = ((a),(b)) + 2((c),(d))`  is a vector equation of a line that passes through `((3),(1))` and `((−3),(−3))`.   (2 marks)

  2. Determine whether  `underset~u = ((4),(6)) + lambda((−2),(3))`  is perpendicular to `underset~v`.   (1 mark)

  3. Express `underset~u` in Cartessian form.   (1 mark)

Show Answers Only

a.    `a = 3, b = 1, c = −3, d = −2,\ text{or}\ a = −3, b = −3, c = 3, d = 2`

b.    `text(See worked solutions)`

c.    `y = −3/2x + 12`

Show Worked Solution

a.    `text(Method 1)`

`overset(->)(OA) = underset~a = ((3),(1)),\ \ overset(->)(OB) = underset~b = ((−3),(−3))`

`overset(->)(AB)` `= overset(->)(OB)-overset(->)(OA)`
  `= ((−3),(−3))-((3),(1))`
  `= ((−6),(−4))`

 

`underset~v` `= underset~a + lambdaunderset~b`
  `= ((3),(1)) + lambda((−6),(−4))`
  `= ((3),(1)) + 2((−3),(−2))`

 
`:. a = 3, b = 1, c = −3, d = −2`
 

`text(Method 2)`

`overset(->)(BA)` `= overset(->)(OA)-overset(->)(OB)`
  `= ((3),(1))-((−3),(−3))`
  `= ((6),(4))`

 
`underset~v = ((−3),(−3)) + 2((3),(2))`

`:. a = −3, b = −3, c = 3, d = 2`
 

b.   `underset~u = ((4),(6)) + lambda((−2),(3))`

`underset~v = ((3),(1)) + 2((−3),(−2))`

`((−2),(3)) · ((−3),(−2)) = 6-6 = 0`

`:. underset~u ⊥ underset~v`
 

c.   `((x),(y))= ((4),(6)) + lambda((−2),(3))`

`x = 4-2lambda\ \ \ …\ (1)`

`y = 6 + 3lambda\ \ \ …\ (2)`

`text(Substitute)\ \ lambda = (4-x)/2\ \ text{from (1) into (2):}`

`y` `= 6 + 3((4-x)/2)`
`y` `= 6 + 6-(3x)/2`
`y` `= −3/2x + 12`

Vectors, EXT2 V1 EQ-Bank 13

Consider the vectors  `underset~u = a underset~i - b underset~j + c underset~k`  and  `underset~v = underset~i - 8underset~j + 4underset~k`.

Find all possible values of `a, b` and `c` if `underset~u` is parallel to `underset~v` and  has a magnitude of 3.   (3 marks)

Show Answers Only

`1/3 , 8/3 , 4/3`

`text(or)`

` -1/3 , – 8/3 , – 4/3`

Show Worked Solution
`|underset~v|` `= sqrt(1 + 64 + 16) = 9`
`underset~overset^v` `= underset~v /|underset~v| =  (1)/(9) underset~i – (8)/(9) underset~j + (4)/(9) underset~k \ \ text{(magnitude of 1)}`

 
`text(S) text(ince) \ underset~u  \ text(has a magnitude of 3:)`

`underset~u` `= ± 3 ((1)/(9) underset~i – (8)/(9) underset~j + (4)/(9) underset~k)`
  `= ± (1/3 underset~i – 8/3 underset~j + 4/3 underset~k)`

 

`:. \ a, b, c` `= 1/3 , – 8/3 , 4/3\ \ text{or}\ \ -1/3 , 8/3 , – 4/3`

Vectors, EXT1 V1 EQ-Bank 5 MC

Which pair of line segments intersect at exactly one point

A.    `{(underset ~u = ((3), (2)) + lambda ((text{−1}),(2)) text{,} quad qquad 0 <= lambda <= 1), (underset ~v = ((2), (1)) + lambda ((2), (text{−4})) text{,} quad qquad 0 <= lambda <= 1):}`
   
B.    `{(underset ~u = ((4), (1)) + lambda ((3), (text{−1})) text{,} quad qquad 0 <= lambda <= 1), (underset ~v = ((3), (2)) + lambda ((2), (2)) text{,} quad qquad 0 <= lambda <= 1):}`
   
C.    `{(underset ~u = ((4), (0)) + lambda ((text{−3}),(6)) text{,} quad qquad 0 <= lambda <= 1), (underset ~v = ((0), (1)) + lambda ((1), (text{−2})) text{,} quad qquad 0 <= lambda <= 1):}`
   
D.    `{(underset ~u = ((0), (2)) + lambda ((3), (text{−2})) text{,} quad qquad 0 <= lambda <= 1), (underset ~v = ((0), (1)) + lambda ((1), (1)) text{,} quad qquad 0 <= lambda <= 1):}`
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`D`

Show Worked Solution

`text(S) text(ince)\ ((2), (text{−4})) = -2((text{−1}), (2)) and ((text{−3}), (6)) = -3((text{−1}), (2))`

`=> A and C\ text(are parallel lines.)`
 

`text(Consider)\ D:`

`3 lambda_1` `= lambda_2` `\ text{… (1)}`
`2 – 2 lambda_1` `= 1 + lambda_2` `\ text{… (2)}`

 
`text(Substitute)\ text{(1) into (2)}`

`2 – 2 lambda_1` `= 1 + 3 lambda_1`
`lambda_1` `= 1/5`
`lambda_2` `= 3/5`

 
`text(Similarly,)\ lambda_1, lambda_2\ text(in)\ B\ \ text(can be calculated)`

`text(and found to be outside)\ \ 0 <= lambda <= 1.`

`=> D`

Vectors, EXT2 V1 EQ-Bank 24

  1. Determine the point of intersection of `underset ~a` and `underset ~b` given.

`qquad underset ~a = ((3), (5), (1)) + lambda ((1), (3), (text{−2})),`  and

`qquad underset ~b = ((text{−2}), (2), (text{−1})) + mu ((1), (text{−1}), (2))`   (2 marks)

  1. Determine if the point `(2, text{−2}, 5)` lies on `underset ~b`.   (1 mark)
Show Answers Only

a.    `((1), (text{−1}), (5))`

b.    `text(See Worked Solutions)`

Show Worked Solution

a.    `text(At point of intersection:)`

`3 + lambda` `= -2 + mu\ \ text{… (1)}`
`5 + 3 lambda` `= 2-mu\ \ text{… (2)}`
`1-2 lambda` `= -1 + 2 mu\ \ text{… (3)}`

 
`(1) + (2)`

`8 + 4 lambda` `= 0`
`lambda` `= -2,\ \ mu = 3`

 
`text{Intersection (using}\ lambda = –2 text{)}:`
 

`((x), (y), (z)) = ((3-2 xx 1), (5-2 xx 3), (1-2 xx text{−2})) = ((1), (text{−1}), (5))`

 

b.    `text(If)\ (2, text{−2}, text{−10})\ text(lies on)\ underset ~b, ∃ mu\ \ text(that satisfies:)`

`-2 + mu` `= 2\ \ text{… (1)}\ => \ mu = 4`
`2-mu` `= ­text{−2}\ \ text{… (2)}\ => \ mu = 4`
`-1 + 2 mu` `= 5\ \ text{… (3)}\ => \ mu = 3`

 
`=>\  text(No solution)`

`:. (2, text{−2}, 5)\ \ text(does not lie on)\ underset ~b.`

Vectors, EXT2 V1 EQ-Bank 11

  1. Find the equation of line vector  `underset ~r`, given it passes through  `(1, 3, –2)`  and  `(2, –1, 2)`.   (2 marks)

  2. Determine if  `underset ~r`  passes through  `(4, –9, 10)`.   (1 mark)

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a.    `underset ~r = ((1), (3), (-2)) + lambda ((1), (-4), (4)) or`

`underset ~r = ((2), (-1), (2)) + lambda ((-1), (4), (-4))`

b.    text(See Worked Solutions)`

Show Worked Solution

a.    `text(Method 1)`

`text(Let)\ \ A(1, 3, –2) and B(2, –1, 2)`

`vec (AB)` `= ((2), (-1), (2))-((1), (3), (-2)) = ((1), (-4), (4))`
`underset ~r` `= ((1), (3), (-2)) + lambda ((1), (-4), (4))`

 
`text (Method 2)`

`vec (BA)` `= ((1), (3), (-2))-((2), (-1), (2)) = ((-1), (4), (-4))`
`underset ~r` `= ((2), (-1), (2)) + lambda ((-1), (4), (-4))`

 

b.    `text(If)\ \ (4, –9, 10)\ \ text(lies on the vector line,)`

`∃ lambda\ \ text(that satisfies:)`

`1 + lambda` `= 4\ \ text{… (1)}`
`3-4 lambda` `= -9\ \ text{… (2)}`
`-2 + 4 lambda` `= 10\ \ text{… (3)}`

 
`lambda = 3\ \ text(satisfies all equations)`

`:. (4, –9, 10)\ \ text(lies on the line.)`

Vectors, EXT2 V1 EQ-Bank 30

Use the vector form of the linear equations

`3x-2y = 4`  and  `3y + 2x-6 = 0`

to show they are perpendicular.   (3 marks)

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`text(Proof)\ text{(See Worked Solutions)}`

Show Worked Solution
`3x-2y` `= 4`
`3x` `= 2y + 4`
`3/2 x` `=y+2`
`x/(2/3)` `= y + 2`

 
`underset ~(v_1) = ((0), (-2)) + lambda ((2/3), (1))`

`3y + 2x-6` `= 0`
`2x` `= -3y + 6`
`-2/3 x` `= y-2`
`x/(-3/2)` `= y-2`

 
`underset ~(v_2) = ((0), (2)) + lambda ((-3/2), (1))`

`((2/3), (1)) ((-3/2), (1)) = -1 + 1 = 0`

`:. underset ~(v_1) _|_ underset ~(v_2)`

Vectors, EXT2 V1 EQ-Bank 29

Find the value of `x` and `y`, given

`underset ~r = ((5), (-1), (2)) + lambda ((x), (y), (-3))`

and  `underset ~r`  is perpendicular to both `underset ~v` and `underset ~w`, where

`underset ~v = ((1), (2), (1)) + mu_1 ((3), (-3), (-1))`  and  `underset ~w = ((-3), (1), (1)) + mu_2 ((-4), (-1), (-2))`.   (2 marks)

Show Answers Only

`x = 1, y = 2`

Show Worked Solution

`text(S) text(ince)\ underset ~r\ text(is perpendicular to)\ underset ~v and underset ~w:`
 

`((x), (y), (-3))((3), (-3), (-1)) = 0`

`3x-3y + 3` `= 0`
`x-y` `= -1\ text{… (1)}`

 
`((x), (y), (-3))((-4), (-1), (-2)) = 0`

`-4x-y + 6` `= 0`
`4x + y` `= 6\ text{… (2)}`

 
`(1) + (2)`

`5x` `= 5`
`x` `= 1`
`y` `= 2`