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Vectors, SPEC1 2024 VCAA 4

Consider the vectors  \(\underset{\sim}{ a }=3 \underset{\sim}{ j }+3 \underset{\sim}{ k }, \ \underset{\sim}{ b }=2 \underset{\sim}{ i }-\underset{\sim}{ j }-2 \underset{\sim}{ k }\)  and  \(\underset{\sim}{ c }=n \underset{\sim}{ i }+2 \underset{\sim}{ j }+\underset{\sim}{ k }\),  where  \(n \in Z\).

  1. Find the angle between \(\underset{\sim}{ a }\) and \(\underset{\sim}{ b }\).   (2 marks)

  2. Find all possible values of \(n\) such that the dot product of \(\underset{\sim}{ a }\) and \(\underset{\sim}{ c }\) is equal to the magnitude of the cross product of \(\underset{\sim}{ a }\) and \(\underset{\sim}{c}\).   (2 marks)

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a.      \(\theta=\dfrac{3 \pi}{4}\left(\text{or} \ 135^{\circ}\right)\)

b.    \(n= \pm 2\)

Show Worked Solution

a.  \(\underset{\sim}{a}=\left(\begin{array}{l}0 \\ 3 \\ 3\end{array}\right) \Rightarrow \abs{\underset{\sim}{a}}=\sqrt{18}=3 \sqrt{2}\)

     \(\underset{\sim}{b}=\left(\begin{array}{c}2 \\ -1 \\ -2\end{array}\right) \Rightarrow\abs{\underset{\sim}{b}}=\sqrt{9}=3\)

     \(\underset{\sim}{c}=\left(\begin{array}{l}n \\ 2 \\ 1\end{array}\right) \Rightarrow \abs{\underset{\sim}{c}}=\sqrt{n^2+5}\)

     \(\cos \theta=\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\abs{\underset{\sim}{a}} \cdot \abs{\underset{\sim}{b}}}=\dfrac{-3-6}{3 \sqrt{2} \times 3}=-\dfrac{1}{\sqrt{2}}\)

    \(\therefore \theta=\cos ^{-1}\left(-\dfrac{1}{\sqrt{2}}\right)=\dfrac{3 \pi}{4}\left(\text{or} \ 135^{\circ}\right)\)
 

b.  \(\underset{\sim}{a} \times \underset{\sim}{c}=\left|\begin{array}{ccc}\underset{\sim}{i} & \underset{\sim}{j} & \underset{\sim}{k} \\ 0 & 3 & 3 \\ n & 2 & 1\end{array}\right|=-3\underset{\sim}{i}+3 n\underset{\sim}{j}-3 n \underset{\sim}{k}\)

     \(\abs{\underset{\sim}{a} \times \underset{\sim}{c}}=\sqrt{9+9n^2+9n^2}=\sqrt{18 n^2+9}\)

     \(\underset{\sim}{a} \cdot \underset{\sim}{c}=0 \times n + 3 \times 2 + 3 \times 1=9\)

     \(\text{Find \(n\) given:}\)

  \(\sqrt{18 n^2+9}\) \(=9\)
  \(18 n^2+9\) \(=81\)
  \(n^2\) \(=4\)
  \(n\) \(= \pm 2\)

Vectors, SPEC2 2023 VCAA 17 MC

Consider the vectors  \(\underset{\sim}{\text{a}}=\alpha \underset{\sim}{\text{i}}+\underset{\sim}{\text{j}}-\underset{\sim}{\text{k}}, \ \underset{\sim}{\text{b}}=3 \underset{\sim}{\text{i}}+\beta \underset{\sim}{\text{j}}+4 \underset{\sim}{\text{k}}\)  and  \(\underset{\sim}{\text{c}}=2 \underset{\sim}{\text{i}}-7 \underset{\sim}{\text{j}}+\gamma \underset{\sim}{\text{k}}\), where \(\alpha, \beta, \gamma \in R\). If  \(\underset{\sim}{\text{a}} \times \underset{\sim}{\text{b}}=\underset{\sim}{\text{c}}\), then

  1. \(\alpha=-2, \ \beta=-1, \ \gamma=-5\)
  2. \(\alpha=-1, \ \beta=2, \ \gamma=-1\)
  3. \(\alpha=1, \ \beta=-2, \ \gamma=-5\)
  4. \(\alpha=-2, \ \beta=-1, \ \gamma=-1\)
  5. \(\alpha=1, \ \beta=-2, \ \gamma=5\)
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\(C\)

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\begin {aligned}\underset{\sim}{\text{a}} \times \underset{\sim}{\text{b}}=\left|\begin{array}{ccc}
\underset{\sim}{\text{i}} & \underset{\sim}{\text{j}} & \underset{\sim}{\text{k}} \\ \alpha & 1 & -1 \\ 3 & \beta & 4\end{array}\right|\ & =(4+\beta)\underset{\sim}{\text{i}}-(4\alpha+3)\underset{\sim}{\text{j}}+(\alpha \beta-3) \underset{\sim}{\text{k}} \end{aligned}

\(\text{Equating coefficients given}\ \ \underset{\sim}{\text{a}} \times \underset{\sim}{\text{b}}=\underset{\sim}{\text{c}}:\)

\(4+\beta =2\ \ \Rightarrow \ \beta=-2 \)

\(4\alpha +3 = 7 \ \ \Rightarrow \ \alpha = 1 \)

\(\alpha \beta-3=\gamma \ \ \Rightarrow \ \gamma=-5 \)

\(\Rightarrow C\)

Vectors, SPEC1 2023 VCAA 9

A plane contains the points \( A(1,3,-2), B(-1,-2,4)\) and \( C(a,-1,5)\), where \(a\) is a real constant. The plane has a \(y\)-axis intercept of 2 at the point \(D\).

  1. Write down the coordinates of point \(D\).   (1 mark)
  1. Show that \(\overrightarrow{A B}\) and \(\overrightarrow{A D}\) are \(-2 \underset{\sim}{\text{i}}-5 \underset{\sim}{\text{j}}+6 \underset{\sim}{\text{k}}\)  and  \(-\underset{\sim}{\text{i}}-\underset{\sim}{\text{j}}+2 \underset{\sim}{\text{k}}\), respectively.   (1 mark)
  1. Hence find the equation of the plane in Cartesian form.  (2 marks)
  1.  Find \(a\).   (1 mark)
  1.  \(\overline{A B}\) and \(\overline{A D}\) are adjacent sides of a parallelogram. Find the area of this parallelogram.   (1 mark)
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a.    \(D(0,2,0)\)

b.    \(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\left(\begin{array}{c}
-1 \\
-2 \\
4
\end{array}\right)-\left(\begin{array}{c}
1 \\
3 \\
-2
\end{array}\right)=\left(\begin{array}{c}
-2 \\
-5 \\
6
\end{array}\right)=-2 \underset{\sim}{\text{i}}-5\underset{\sim}{\text{j}}+6 \underset{\sim}{\text{k}}\)

\(\overrightarrow{A D}=\overrightarrow{O D}-\overrightarrow{O A}=\left(\begin{array}{c}0 \\ 2 \\ 0\end{array}\right)-\left(\begin{array}{c}1 \\ 3 \\ -2\end{array}\right)=\left(\begin{array}{c}-1 \\ -1 \\ 2\end{array}\right)=-\underset{\sim}{\text{i}}-\underset{\sim}{\text{j}}+2\underset{\sim}{\text{k}}\)

c.    \(4 x+2 y+3 z =4\)

d.    \(a=-\dfrac{9}{4}\)

e.    \(A=\sqrt{29}\)

Show Worked Solution

a.    \(D(0,2,0)\)

 
b.    \(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\left(\begin{array}{c}
-1 \\
-2 \\
4
\end{array}\right)-\left(\begin{array}{c}
1 \\
3 \\
-2
\end{array}\right)=\left(\begin{array}{c}
-2 \\
-5 \\
6
\end{array}\right)=-2 \underset{\sim}{\text{i}}-5\underset{\sim}{\text{j}}+6 \underset{\sim}{\text{k}}\)

\(\overrightarrow{A D}=\overrightarrow{O D}-\overrightarrow{O A}=\left(\begin{array}{c}0 \\ 2 \\ 0\end{array}\right)-\left(\begin{array}{c}1 \\ 3 \\ -2\end{array}\right)=\left(\begin{array}{c}-1 \\ -1 \\ 2\end{array}\right)=-\underset{\sim}{\text{i}}-\underset{\sim}{\text{j}}+2\underset{\sim}{\text{k}}\)

 
c.    \(\overrightarrow{A B} \times \overrightarrow{A D}=\left|\begin{array}{ccc}
\underset{\sim}{\text{i}} & \underset{\sim}{\text{j}} & \underset{\sim}{\text{k}} \\ -2 & -5 & 6 \\ -1 & -1 & 2\end{array}\right|\)

\(\ \quad \quad \quad \quad \   \begin{aligned} & =(-10+6)\underset{\sim}{\text{i}}-(-4+6)\underset{\sim}{\text{j}}+(2-5) \underset{\sim}{\text{k}} \\ & =-4 \underset{\sim}{i}-2 \underset{\sim}{j}-3 \underset{\sim}{k}\end{aligned}\)

\(\text{Plane:}\ \ -4 x-2 y-3 z=k\)

\(\text{Substitute}\  D(0,2,0)\  \text{into equation}\ \ \Rightarrow \ \text{k}=-4\)

\begin{aligned}
\therefore-4 x-2 y-3 z & =-4 \\
4 x+2 y+3 z & =4
\end{aligned}

 
d.    \(\text{Find}\ a\ \Rightarrow \ \text{substitute}\ (a, 1,-5)\ \text{into plane equation:}\)

\begin{aligned}
-4 & =-4a+2-15 \\
4 a & =-9 \\
a & =-\dfrac{9}{4}
\end{aligned}

e.     \(\text{Area}\) \(=|\overrightarrow{A B} \times \overrightarrow{A D}|\)
    \(=|-4\underset{\sim}{i}-2\underset{\sim}{j}-3\underset{\sim}{k}|\)
    \(=\sqrt{16+4+9} \)
    \(=\sqrt{29}\)