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Vectors, SPEC2 2025 VCAA 5

Consider three planes defined by the equations  \(\Pi_1: \ 2 x+9 z=8, \Pi_2: \ 3 x+6 y+5 z=7\)  and  \(\Pi_3: \ x+9 y-3 z=7\).

  1. Find the point of intersection of the three planes.   (1 mark)

    1. Find a vector that gives the direction of the line of intersection of the planes \(\Pi_2\) and \(\Pi_3\).   (2 marks)

    2. Find a set of parametric equations that give the coordinates of the points that lie on this line of intersection.   (1 mark)

  3. Find the shortest distance from the point \((1,1,2)\) to the plane \(\Pi_3\).   (2 marks)

  4. Consider a family of planes, \(\Psi\), with equation  \(6 x+27 z=m\), where \(m \in N\).
    1. Show that the plane \(\Pi_1\) is parallel to each member of \(\Psi\).   (1 mark)

    2. Find all values of \(m\) for which the shortest distance between plane \(\Pi_1\) and the plane of the form  \(6 x+27 z=m\)  is  \(\dfrac{23}{3 \sqrt{85}}\).   (3 marks)

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a.    \(\text{Intersection at}\ (-5,2,2)\)
 

b.i.  \(\underset{\sim}{d}=\left(\begin{array}{c}-63 \\ 14 \\ 21\end{array}\right)\)
 

b.ii. \(x=-5-63 \lambda\)

\(y=2 + 14 \lambda\)

\(z=2+21 \lambda\)
 

c.    \(\dfrac{3}{\sqrt{91}}\)
 

d.i.  \(\Psi: \ 6 x+27 z=m\)

\(\text{Normal to} \ \Psi: \ 6 \underset{\sim}{i}+27 \underset{\sim}{k}\)

\(\text{Normal to} \ \Pi_2: \ 2 \underset{\sim}{i}+9 \underset{\sim}{k}\)

\(\displaystyle \binom{6}{27}=3\binom{2}{9} \ \Rightarrow \ \text{Planes are parallel}\)

d.ii.  \(m=1,47\)

Show Worked Solution

a.    \(\Pi_1: \ 2 x+9 z=8\)

\(\Pi_2: \ 3 x+6 y+5 z=7\)

\(\Pi_3: \ x+9 y-3 z=7\)

\(\text {Solve simultaneously (by CAS):}\)

\(\text{Intersection at}\ (-5,2,2)\)
 

b.i.  \(\underset{\sim}{d} \ \text{is} \perp \text{to normals of} \ \Pi_2 \ \text{and}\  \Pi_3.\)

\(\underset{\sim}{d}=n_2 \times n_3=\left|\begin{array}{ccc}i & j & k \\ 3 & 6 & 5 \\ 1 & 9 & -3\end{array}\right|=\left(\begin{array}{c}-63 \\ 14 \\ 21\end{array}\right)\)
 

b.ii. \((-5,2,2)\ \text{lies on both planes (from part a.)}\)

\(x=-5-63 \lambda\)

\(y=2 + 14 \lambda\)

\(z=2+21 \lambda\)

♦ Mean mark (b.ii) 51%.

c.    \(\text{Find shortest distance from (1, 1, 2) to \(\Pi_3\).}\)

\(\Pi_3: \ x+9 y-3 z=7 \ \ \Rightarrow \ \ (7,0,0)\ \text{lies on plane.}\)

\(\text {Spanning vector to} \ (1,1,2) \ \text{is} \ (-6 \underset{\sim}{i}+\underset{\sim}{j}+2 \underset{\sim}{k})\)

\(\text{Distance}=\dfrac{1}{\sqrt{91}} \times \abs{\left(\begin{array}{c}-6 \\ 1 \\ 2\end{array}\right)\left(\begin{array}{c}1 \\ 9 \\ -3\end{array}\right)}=\dfrac{3}{\sqrt{91}}\)
 

d.i.  \(\Psi: \ 6 x+27 z=m\)

\(\text{Normal to} \ \Psi: \ 6 \underset{\sim}{i}+27 \underset{\sim}{k}\)

\(\text{Normal to} \ \Pi_2: \ 2 \underset{\sim}{i}+9 \underset{\sim}{k}\)

\(\displaystyle \binom{6}{27}=3\binom{2}{9} \ \Rightarrow \ \text{Planes are parallel}\)
  

d.ii. \(\text{Point on}\ \Pi_1: (4,0,0)\)

\(\text{Point on} \ \Psi:\left(\dfrac{m}{6}, 0,0\right)\)

\(\text{Spanning vector}=\left(4-\dfrac{m}{6}\right) \underset{\sim}{i}\)

\(\text{Distance}=\abs{\left(4-\dfrac{m}{6}\right) \underset{\sim}{i} \cdot \dfrac{(2 \underset{\sim}{i}+9\underset{\sim}{k})}{\sqrt{85}}}=\dfrac{23}{3 \sqrt{85}}\)

\(\text{Solve}\ \ \abs{8-\dfrac{m}{3}}=\dfrac{23}{3} \ \ \text{for} \ m:\)

\(m=1,47\)

♦ Mean mark (d.ii) 45%.

Vectors, SPEC2 2025 VCAA 19 MC

The plane with equation  \(x+y+z=a\), where \(a \in R\), intersects the coordinate axes at three points that form the vertices of a triangle.

The area of this triangle is given by

  1. \(\dfrac{a^2 \sqrt{3}}{4}\)
  2. \(\dfrac{a^2 \sqrt{3}}{2}\)
  3. \(\dfrac{a^2}{4}\)
  4. \(a^2 \sqrt{3}\)
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\(B\)

Show Worked Solution

\(\text{Plane:}\ \  x+y+z=a\)

\(\text{Axis intercepts:}\ \ (a, 0,0),(0, a, 0),(0,0, a)\)
 

\(\text{Distance between \((a, 0,0)\) and \((0, a, 0)\) :}\)

\(d=\sqrt{(a-0)^2+(0-a)^2+(0-0)^2}=\sqrt{2} a\)
 

\(\text{Similarly for the other two sides}\)

\(\Rightarrow \ \text{Equilateral triangle with side length} \ \sqrt{2} a\)

\(A=\dfrac{1}{2} a b \sin C=\dfrac{1}{2}(\sqrt{2} a)^2 \times \sin 60^{\circ}=\dfrac{1}{2} \times 2 a^2 \times \dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3} a^2}{2}\)

\(\Rightarrow B\)

Vectors, SPEC2 2025 VCAA 15 MC

Consider the two planes described by the equations  \(2 x+2 y+z=2\)  and  \(a x+4 z=1\), where \(a\) is a positive constant.

The angle between the two planes is  \(\cos ^{-1}\left(\dfrac{2}{3}\right)\).

The value of \(a\) satisfies the equation

  1. \(a+2=\sqrt{a^2+16}\)
  2. \(\dfrac{2 a+4}{3 \sqrt{a^2+16}}=\dfrac{3}{2}\)
  3. \(2 a+4=3 \sqrt{a^2+16}\)
  4. \(\dfrac{2 a+4}{\sqrt{a^2+16}}=\dfrac{2}{3}\)
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\(A\)

Show Worked Solution

\(\text{Plane 1:}\ \ 2 x+2 y+z=2\)

\(\text{Plane 2:}\ \ a x+4 z=1\)

\(\text{Angle between planes = angle between normals}\)

♦ Mean mark 52%.

\(\displaystyle {\underset{\sim}{n}}_1=\left(\begin{array}{l}2 \\ 2 \\ 1\end{array}\right), \quad {\underset{\sim}{n}}_2=\left(\begin{array}{l}a \\ 0 \\ 4\end{array}\right)\)

\({\underset{\sim}{n}}_1 \cdot {\underset{\sim}{n}}_2\) \(=\abs{{\underset{\sim}{n}}_1}\abs{{\underset{\sim}{n}}_2}\cos(\theta)\)
\(2 a+4\) \(=\sqrt{4+4+1} \times \sqrt{a^2+16} \times \dfrac{2}{3}\)
\(2 a+4\) \(=3 \sqrt{a^2+16} \times \dfrac{2}{3}\)
\(a+2\) \(=\sqrt{a^2+16}\)

 

\(\Rightarrow A\)

Vectors, SPEC2 2024 VCAA 5

Consider the points  \(A(1,-2,3)\) and \(B(2,-5,-1)\).

  1. Find a vector equation, in terms of the components \(\underset{\sim}{i}, \underset{\sim}{j}\) and \(\underset{\sim}{k}\), for the line passing through these points.   (1 mark)

  2. Consider the different line  \(L_1: r _1(t)=2 \underset{\sim}{ i }+\underset{\sim}{ j }-3 \underset{\sim}{k}+t(-\underset{\sim}{ i }+2\underset{\sim}{ j }+\underset{\sim}{ k }), t \in R\).
  3. Find the shortest distance from \(L_1\) to point \(A\).
  4. Give your answer in the form  \(\dfrac{a \sqrt{b}}{c}\)  where  \(a, b\) and \(c\) are positive integers.   (3 marks)

  5. Let \(C\) be the point \((0,2,-5)\).
  6. Find the Cartesian equation of the plane that contains the points \(A, B\) and \(C\).   (3 marks)

  7. Another plane has the Cartesian equation  \(2 x-3 y+4 z=12\).
  8. This plane intersects the coordinate axes at three points, which form the vertices of a triangle.
    1. Find the coordinates of these three points.   (1 mark)

    2. Find the area of the triangle.
    3. Give your answer in the form  \(m \sqrt{n}\)  where \(m\) and \(n\) are integers.   (2 marks)

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a.    \(\underset{\sim}{r}(t)=\underset{\sim}{i}-2 \underset{\sim}{j}+3 \underset{\sim}{k}+t(\underset{\sim}{i}-3\underset{\sim}{j}-4 \underset{\sim}{k}), \quad(t \in R)\)

\(\text{or}\)

\(\underset{\sim}{r}(t)=2\underset{\sim}{i}-5 \underset{\sim}{j}-\underset{\sim}{k}+t(\underset{\sim}{i}-3\underset{\sim}{j}-4 \underset{\sim}{k}), \quad (t \in R)\)

b.  \(\text{Shortest distance}=\dfrac{5 \sqrt{66}}{6}\)

c.   \(40x+12 y+z=19\)

d.i.  \(X(6,0,0), Y(0,-4,0), Z(0,0,3)\)

d.ii.  \(\text{Area}=3 \sqrt{29}\)

Show Worked Solution

a.    \(\overrightarrow{AB}=\left(\begin{array}{c}2 \\ -5 \\ -1\end{array}\right)-\left(\begin{array}{c}1 \\ -2 \\ 3\end{array}\right)=\left(\begin{array}{c}1 \\ -3 \\ -4\end{array}\right)\)

\(\underset{\sim}{r}(t)=\underset{\sim}{i}-2 \underset{\sim}{j}+3 \underset{\sim}{k}+t(\underset{\sim}{i}-3\underset{\sim}{j}-4 \underset{\sim}{k}), \quad(t \in R)\)

\(\text{or}\)

\(\underset{\sim}{r}(t)=2\underset{\sim}{i}-5 \underset{\sim}{j}-\underset{\sim}{k}+t(\underset{\sim}{i}-3\underset{\sim}{j}-4 \underset{\sim}{k}), \quad (t \in R)\)
 

b.    \({\underset{\sim}{r}}_\text{A}=\underset{\sim}{i}-2\underset{\sim}{j}+3 \underset{\sim}{k}\)

\({\underset{\sim}{r}}_1=(2-t) \underset{\sim}{i}+(1+2 t) \underset{\sim}{j}+(t-3) \underset{\sim}{k}\)

\({\underset{\sim}{r}}_1-{\underset{\sim}{r}}_\text{A}=(1-t)\underset{\sim}{i}+(2 t+3) \underset{\sim}{j}+(t-6) \underset{\sim}{k}\)

\(\abs{{\underset{\sim}{r}}_1-{\underset{\sim}{r}}_\text{A}}=\sqrt{(1-t)^2+(2 t+3)^2+(t-6)^2}\)

\(\text{Find \(t\) when} \ \ \dfrac{d}{dt}\left(\abs{{\underset{\sim}{r}}_1-{\underset{\sim}{r}}_\text{A}}\right)=0:\)

  \(t=\dfrac{1}{6}\)

\(\therefore\ \text{Shortest distance}=\dfrac{5 \sqrt{66}}{6}\)
 

c.    \(C(0,2,-5)\)

\(\text{Plane equation:} \ \dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\)

\(\text{Solve:} \ \ \dfrac{1}{a}-\dfrac{2}{b}+\dfrac{3}{c}=1, \quad \dfrac{2}{a}-\dfrac{5}{b}-\dfrac{1}{c}=1, \quad \dfrac{0}{a}+\dfrac{2}{b}-\dfrac{5}{c}=1\)

\(a=\dfrac{19}{40}, \quad b=\dfrac{19}{12}, \quad c=19\)

\(\text{Equation of plane:}\)

\(\dfrac{40x}{19}+\dfrac{12y}{19}+\dfrac{z}{19}=1 \ \Rightarrow \ 40x+12 y+z=19\)
 

d.i.   \(2 x-3 y+4 z=12\)

\(2x=12\ \) \(\ \Rightarrow \ x=6\ \) \(\Rightarrow\  X(6,0,0)\)
\(-3 y=12\ \) \(\ \Rightarrow \ y=-4\ \) \(\Rightarrow\  Y(0,-4,0)\)
\(4 z=12\ \) \(\ \Rightarrow \ z=3\ \) \(\Rightarrow\  Z(0,0,3)\)
 

d.ii  \(\overrightarrow{XY}=\left(\begin{array}{c}0 \\ -4 \\ 0\end{array}\right)-\left(\begin{array}{l}6 \\ 0 \\ 0\end{array}\right)=\left(\begin{array}{c}-6 \\ -4 \\ 0\end{array}\right)\)

\(\overrightarrow{X Z}=\left(\begin{array}{l}0 \\ 0 \\ 3\end{array}\right)-\left(\begin{array}{l}6 \\ 0 \\ 0\end{array}\right)=\left(\begin{array}{c}-6 \\ 0 \\ 3\end{array}\right)\)
 
\(\text{Area}=\dfrac{1}{2}\abs{\overrightarrow{XY} \times \overrightarrow{XZ}}=\dfrac{1}{2}\abs{-12\underset{\sim}{i}+18 \underset{\sim}{j}-24 \underset{\sim}{k}}=\sqrt{261}=3 \sqrt{29}\)

Vectors, SPEC2 2024 VCAA 18 MC

The point of intersection of the line  \(\underset{\sim}{ r }=\underset{\sim}{ i }+\underset{\sim}{ j }-2 \underset{\sim}{ k }+t(-2 \underset{\sim}{ i }+\underset{\sim}{ j }+3 \underset{\sim}{ k })\), where  \(t \in R\)  and the plane  \(3 x-2 y+4 z=5\)  is

  1. \((-5,-1,2)\)
  2. \((-1,2,1)\)
  3. \((3,4,1)\)
  4. \((-5,4,7)\)
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\(D\)

Show Worked Solution

\(\text{Find \(t\) such that \(\underset{\sim}{r}\) is a point on the plane.}\)

\(\text{Plane equation} \ \ 3x-2 y+4z=5\)

\(\text{Substitute} \ \ x=1-2t, \ y=1+t, \ z=-2+3t\)

\(\text{Solve for} \ t:\)

\(3(1-2 t)-2(1+t)+4(-2+3 t)=5\)

\(t=3\)

\(\text{Substitute  \(t=3\)  into \(\underset{\sim}{r}\):}\)

\(\underset{\sim}{r}(3)\) \(=(1-6,1+3,-2+9)\)
  \(=(-5, 4, 7)\)

 
\(\Rightarrow D\)