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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)

Show Answers Only

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 \(\underset{\sim}{r}\):}\)

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

 
\(\Rightarrow D\)