Vectors and Geometry

Vectors, EXT2 EQ-Bank 21

Let \(\mathbf{u}\) and \(\mathbf{v}\) be vectors in the plane, where \(\mathbf{v}\,\neq\, \mathbf{0}\).

For every real number \(t\), let \(P(t)=\abs{\mathbf{u} -t \mathbf{v}}^2\).

Show that \(P(t)=\abs{\mathbf{u}}^2-2 t( \mathbf{u} \cdot \mathbf{v} )+t^2 \abs{\mathbf{v}}^2\). (1 mark)

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Show that \(P(t)\) has a minimum value at \(t=\dfrac{ \mathbf{u} \cdot \mathbf{v} }{\abs{\mathbf{v}}^2}\). (2 marks)

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Hence prove the Cauchy-Schwarz inequality, \(\abs{\mathbf{u}\cdot\mathbf{v}} \leq \abs{\mathbf{u}}\abs{\mathbf{v}}\). (3 marks)

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

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

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

Show Worked Solution

a.	\(P(t)\)	\(=\abs{\mathbf{u} -t \mathbf{v}}^2\)
		\(=( \mathbf{u} -t \mathbf{v} ) \cdot( \mathbf{u} -t\mathbf{v} )\)
		\(= \mathbf{u} \cdot \mathbf{u} + \mathbf{u} \cdot(-t \mathbf{v} )+(-t \mathbf{v}) \cdot \mathbf{u} +t^2(\mathbf{v} \cdot \mathbf{v} )\)
		\(=\abs{\mathbf{u}}^2-2 t( \mathbf{u} \cdot \mathbf{v} )+t^2\abs{\mathbf{v}}^2\)

b. \(\text{Find}\ t\ \text{when}\ \ \dfrac{d P}{d t}=0:\)

\(2 t\abs{\mathbf{v}}^2-2 \mathbf{u} \cdot \mathbf{v}\)	\(=0\)
\(2 t\abs{\mathbf{v}}^2\)	\(=2 \mathbf{u} \cdot \mathbf{v}\)
\(t\)	\(=\dfrac{ \mathbf{u} \cdot \mathbf{v} }{\abs{\mathbf{v}}^2}\)

\(\dfrac{d^2 P}{d t^2}=2\abs{\mathbf{v}}^2>0\)

\(\therefore\ \text{Min SP at}\ \ t=\dfrac{ \mathbf{u} \cdot \mathbf{v} }{\abs{\mathbf{v}}^2}\)

c. \(P(t)=\abs{\mathbf{u} -t \mathbf{v}}^2\ \ \Rightarrow\ \ P(t) \geq 0\)

\(P\left(\dfrac{ \mathbf{u} \cdot \mathbf{v} }{\abs{\mathbf{v}}^2}\right)\)	\(=\left(\dfrac{ \mathbf{u} \cdot \mathbf{v} }{\abs{\mathbf{v}}^2}\right)^2\abs{\mathbf{v}}^2-2\left(\dfrac{ \mathbf{u} \cdot \mathbf{v} }{\abs{\mathbf{v}}^2}\right) \mathbf{u} \cdot \mathbf{v} +\abs{\mathbf{u}}^2\)
	\(=-\dfrac{( \mathbf{u} \cdot \mathbf{v} )^2}{\abs{\mathbf{v}}^2}+\abs{\mathbf{u}}^2\)

\(0\)	\( \leq -\dfrac{( \mathbf{u} \cdot \mathbf{v} )^2}{\abs{\mathbf{v}}^2}+\abs{\mathbf{u}}^2\)
\((\mathbf{u} \cdot \mathbf{v} )^2\)	\(\leq \abs{\mathbf{u}}^2\abs{\mathbf{v}}^2\)
\(\abs{ \mathbf{u} \cdot \mathbf{v} }\)	\(\leq\abs{\mathbf{u}}\abs{\mathbf{v}}\ \ \text{(i.e. the Cauchy-Schwarz inequality)}\)

Vectors, EXT2 EQ-Bank 18

Consider the triangle with vertices \(A(4,5,1), B(8,1,2)\) and the origin \(O(0,0,0)\). The triangle has three medians.

The median through \(B\) has vector equation

\(\lambda \, \overrightarrow{O M}+(1-\lambda) \overrightarrow{O B}=\left(\begin{array}{l}8 \\ 1 \\ 2\end{array}\right)+\lambda\left(\begin{array}{c}-6 \\ 1.5 \\ -1.5\end{array}\right)\) (Do NOT prove this.)

for a parameter \(\lambda \in[0,1]\) and where \(M\) is the midpoint of \(O A\).

Write down the median through \(A\) as a vector equation. (2 marks)

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The three medians of any triangle meet at a point known as the centroid.
Find the value of \(\lambda\) corresponding to the centroid. (2 marks)

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Into what ratio does the centroid divide the median \(B M\)? (1 mark)

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a. \(\left(\begin{array}{c}4 \\5 \\1\end{array}\right) + \mu \left(\begin{array}{c}0 \\-4.5 \\0\end{array}\right)\ \ \text{for}\ \ \mu \in[0,1]\)

b. \(\lambda=\dfrac{2}{3}.\)

c. \(2:1.\)

Show Worked Solution

a. \(\text{Let}\ N = \text{midpoint of}\ OB:\)

\(N \equiv \left(\dfrac{8+0}{2}, \dfrac{1+0}{2}, \dfrac{2+0}{2}\right) \equiv (4,0.5,1).\)

\(\text{Direction vector of the median through} \ A \ \text {is}\left(\begin{array}{c}4-4 \\0.5-5 \\1-1\end{array}\right) = \left(\begin{array}{c}0 \\-4.5 \\0\end{array}\right).\)

\(\text{Equation of median from}\ A:\)

\(\left(\begin{array}{c}4 \\5 \\1\end{array}\right) + \mu \left(\begin{array}{c}0 \\-4.5 \\0\end{array}\right)\ \ \text{for}\ \ \mu \in[0,1]\)

b. \(\text{Equation of median from}\ B:\)

\(\left(\begin{array}{c}8 \\1 \\2\end{array}\right) + \lambda \left(\begin{array}{c}-6 \\1.6 \\-1.5\end{array}\right)\ \ \text{for}\ \ \lambda \in[0,1]\)

\(\text{Medians intersect at centroid}\)

\(x\text{-coordinate of median through}\ B = 8-6\lambda\)

\(x\text{-coordinate of median through}\ A = 4\)

\(\text{Equating}\ x\text{-coordinates:}\)

\(8-6 \lambda=4\ \ \Rightarrow\ \ \lambda=\dfrac{2}{3}\)

\(\therefore\ \text{Point of intersection occurs at } \lambda=\dfrac{2}{3}.\)

c. \(\text{The centroid divides the median in the ratio of} \ 2:1.\)

Vectors, EXT2 V1 2025 HSC 15a

The adjacent sides of a parallelogram are represented by the vectors \(\underset{\sim}{a}=4 \underset{\sim}{i}+3 \underset{\sim}{j}-\underset{\sim}{k}\) and \(\underset{\sim}{b}=2 \underset{\sim}{i}-\underset{\sim}{j}+3 \underset{\sim}{k}\).

Show that the area of the parallelogram is \(6 \sqrt{10}\) square units. (4 marks)

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\(A=2 \times \left[\dfrac{1}{2}\abs{a}\abs{b} \sin \theta \right]\)

\(\underset{\sim}{a}=4\underset{\sim}{i}+3\underset{\sim}{j}-\underset{\sim}{k} \ \Rightarrow \ \abs{\underset{\sim}{a}}=\sqrt{16+9+1}=\sqrt{26}\)

\(\underset{\sim}{b}=2\underset{\sim}{i}-\underset{\sim}{j}+3 \underset{\sim}{k} \ \Rightarrow \ \abs{\underset{\sim}{b}}=\sqrt{4+1+9}=\sqrt{14}\)

\(\underset{\sim}{a} \cdot \underset{\sim}{b}=\left(\begin{array}{c}4 \\ 3 \\ -1\end{array}\right)\left(\begin{array}{c}2 \\ -1 \\ 3\end{array}\right)=8-3-3=2\)

\(\cos \theta=\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\abs{\underset{\sim}{a}}\abs{\underset{\sim}{b}}}=\dfrac{2}{\sqrt{26} \sqrt{14}}=\dfrac{1}{\sqrt{91}}\)

\(\sin \theta=\dfrac{\sqrt{90}}{\sqrt{91}}\)

\(\therefore A=2 \times \dfrac{1}{2} \times \sqrt{26} \times \sqrt{14} \times \dfrac{\sqrt{90}}{\sqrt{91}}=6 \sqrt{10} \ \text{units}^2\)

Show Worked Solution

\(A=2 \times \left[\dfrac{1}{2}\abs{a}\abs{b} \sin \theta \right]\)

\(\underset{\sim}{a}=4\underset{\sim}{i}+3\underset{\sim}{j}-\underset{\sim}{k} \ \Rightarrow \ \abs{\underset{\sim}{a}}=\sqrt{16+9+1}=\sqrt{26}\)

\(\underset{\sim}{b}=2\underset{\sim}{i}-\underset{\sim}{j}+3 \underset{\sim}{k} \ \Rightarrow \ \abs{\underset{\sim}{b}}=\sqrt{4+1+9}=\sqrt{14}\)

\(\underset{\sim}{a} \cdot \underset{\sim}{b}=\left(\begin{array}{c}4 \\ 3 \\ -1\end{array}\right)\left(\begin{array}{c}2 \\ -1 \\ 3\end{array}\right)=8-3-3=2\)

\(\cos \theta=\dfrac{\underset{\sim}{a} \cdot \underset{\sim}{b}}{\abs{\underset{\sim}{a}}\abs{\underset{\sim}{b}}}=\dfrac{2}{\sqrt{26} \sqrt{14}}=\dfrac{1}{\sqrt{91}}\)

\(\sin \theta=\dfrac{\sqrt{90}}{\sqrt{91}}\)

\(\therefore A=2 \times \dfrac{1}{2} \times \sqrt{26} \times \sqrt{14} \times \dfrac{\sqrt{90}}{\sqrt{91}}=6 \sqrt{10} \ \text{units}^2\)

Vectors, EXT2 V1 2024 HSC 16a

Consider the function \(y=\cos (k x)\), where \(k>0\). The value of \(k\) has been chosen so that a circle can be drawn, centred at the origin, which has exactly two points of intersection with the graph of the function and so that the circle is never above the graph of the function. The point \(P(a, b)\) is the point of intersection in the first quadrant, so \(a>0\) and \(b>0\), as shown in the diagram.

The vector joining the origin to the point \(P(a, b)\) is perpendicular to the tangent to the graph of the function at that point. (Do NOT prove this.)

Show that \(k>1\). (4 marks)

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\(y=\cos (k x) \ \Rightarrow \ \dfrac{dy}{dx}=-k \, \sin (k x)\)

\(P(a, b)=P(a, \cos (k a))\)

\(\text{Let}\ \ \underset{\sim}{p}=\overrightarrow{OP}.\)

\(\text{At}\ \ x=a, \ m_{\text{tang}} \perp \underset{\sim}{p} :\)

	\(m_{\text{tang}}\)	\(=-k \, \sin (ka)\)
	\(m_{\overrightarrow{OP}}\)	\(=\dfrac{\cos(ka)}{a}\)

\(-k \, \sin(ka) \times \dfrac{\cos(ka)}{a}=-1\)

	\(k\)	\(=\dfrac{a}{\sin(ka) \cos(ka)}\)
		\(=\dfrac{2a}{\sin(2ka)}\)

\(\text{For} \ \ \theta>0, \ \sin \theta<\theta\)

\(\Rightarrow k=\dfrac{2a}{\sin (2ka)}>\dfrac{2a}{2ka}>1\).

Show Worked Solution

\(y=\cos (k x) \ \Rightarrow \ \dfrac{dy}{dx}=-k \, \sin (k x)\)

\(P(a, b)=P(a, \cos (k a))\)

\(\text{Let}\ \ \underset{\sim}{p}=\overrightarrow{OP}.\)

\(\text{At}\ \ x=a, \ m_{\text{tang}} \perp \underset{\sim}{p} :\)

	\(m_{\text{tang}}\)	\(=-k \, \sin (ka)\)
	\(m_{\overrightarrow{OP}}\)	\(=\dfrac{\cos(ka)}{a}\)

♦♦♦ Mean mark 8%.

\(-k \, \sin(ka) \times \dfrac{\cos(ka)}{a}=-1\)

	\(k\)	\(=\dfrac{a}{\sin(ka) \cos(ka)}\)
		\(=\dfrac{2a}{\sin(2ka)}\)

\(\text{For} \ \ \theta>0, \ \sin \theta<\theta\)

\(\Rightarrow k=\dfrac{2a}{\sin (2ka)}>\dfrac{2a}{2ka}>1\).

Vectors, EXT2 V1 2024 HSC 14e

The diagram shows triangle \(O Q A\).

The point \(P\) lies on \(O A\) so that \(O P: O A=3: 5\).

The point \(B\) lies on \(O Q\) so that \(O B: O Q=1: 3\).

The point \(R\) is the intersection of \(A B\) and \(P Q\).

The point \(T\) is chosen on \(A Q\) so that \(O, R\) and \(T\) are collinear.

Let \(\underset{\sim}{a}=\overrightarrow{O A}, \ \underset{\sim}{b}=\overrightarrow{O B}\) and \(\overrightarrow{P R}=k \overrightarrow{P Q}\) where \(k\) is a real number.

Show that \(\overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b}\). (2 marks)

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Writing \(\overrightarrow{A R}=h \overrightarrow{A B}\), where \(h\) is a real number, it can be shown that \(\overrightarrow{O R}=(1-h) \underset{\sim}{a}+h \underset{\sim}{b}\). (Do NOT prove this.)

Show that \(k=\dfrac{1}{6}\). (2 marks)

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Find \(\overrightarrow{O T}\) in terms of \(\underset{\sim}{a}\) and \(\underset{\sim}{b}\). (2 marks)

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i. \(\underset{\sim}{a}=\overrightarrow{O A}, \ \underset{\sim}{b}=\overrightarrow{O B}, \ \overrightarrow{P R}=k \overrightarrow{P Q}\)

\(\overrightarrow{O P}: \overrightarrow{O A}=3: 5 \ \Rightarrow \ \overrightarrow{O P} = \dfrac{3}{5} \times \overrightarrow{O A} \)

\(\overrightarrow{O B}: \overrightarrow{O Q}=1: 3\ \Rightarrow \ \overrightarrow{O Q} = 3 \times \overrightarrow{OB} \)

\(\text {Show } \overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b}\)

	\(\overrightarrow{O R}\)	\(=\overrightarrow{O P}+k \overrightarrow{P Q}\)
		\(=\overrightarrow{O P}+k(\overrightarrow{O Q}-\overrightarrow{O P})\)
		\(=(1-k) \overrightarrow{O P}+k \overrightarrow{O Q}\)
		\(=\dfrac{3}{5}(1-k)\underset{\sim}{a}+3 k \underset{\sim}{b}\)

ii. \(\overrightarrow{A R}=h \overrightarrow{A B}, \quad h \in \mathbb{R}\)

\(\overrightarrow{OR}=(1-h) \underset{\sim}{a}+h \underset{\sim}{b} \ \ \text{(given)}\)

\(\overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b} \ \ \text{(part(i))}\)

\(\text{Vector }\underset{\sim}{a} \neq \lambda \underset{\sim}{b}\ (\lambda \in \mathbb{R})\ \Rightarrow \ \text{linearly independent and are basis vectors for } \overrightarrow{O R}\).

\(\text {Equating coefficients:}\)

\(\dfrac{3}{5}(1-k)=1-h\ \ldots\ (1)\)

\(h=3 k\ \ldots\ (2)\)

\(\text{Substituting (2) into (1)}\)

	\(\dfrac{3}{5}(1-k)\)	\(=1-3 k\)
	\(3-3 k\)	\(=5-15 k\)
	\(k\)	\(=\dfrac{1}{6}\)

iii. \(\overrightarrow{O T}=\dfrac{3}{4}(\underset{\sim}{a}+\underset{\sim}{b})\)

Show Worked Solution

i. \(\underset{\sim}{a}=\overrightarrow{O A}, \ \underset{\sim}{b}=\overrightarrow{O B}, \ \overrightarrow{P R}=k \overrightarrow{P Q}\)

\(\overrightarrow{O P}: \overrightarrow{O A}=3: 5 \ \Rightarrow \ \overrightarrow{O P} = \dfrac{3}{5} \times \overrightarrow{O A} \)

\(\overrightarrow{O B}: \overrightarrow{O Q}=1: 3\ \Rightarrow \ \overrightarrow{O Q} = 3 \times \overrightarrow{OB} \)

\(\text {Show } \overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b}\)

	\(\overrightarrow{O R}\)	\(=\overrightarrow{O P}+k \overrightarrow{P Q}\)
		\(=\overrightarrow{O P}+k(\overrightarrow{O Q}-\overrightarrow{O P})\)
		\(=(1-k) \overrightarrow{O P}+k \overrightarrow{O Q}\)
		\(=\dfrac{3}{5}(1-k)\underset{\sim}{a}+3 k \underset{\sim}{b}\)

ii. \(\overrightarrow{A R}=h \overrightarrow{A B}, \quad h \in \mathbb{R}\)

\(\overrightarrow{OR}=(1-h) \underset{\sim}{a}+h \underset{\sim}{b} \ \ \text{(given)}\)

\(\overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b} \ \ \text{(part(i))}\)

\(\text{Vector }\underset{\sim}{a} \neq \lambda \underset{\sim}{b}\ (\lambda \in \mathbb{R})\ \Rightarrow \ \text{linearly independent and are basis vectors for } \overrightarrow{O R}\).

\(\text {Equating coefficients:}\)

\(\dfrac{3}{5}(1-k)=1-h\ \ldots\ (1)\)

\(h=3 k\ \ldots\ (2)\)

\(\text{Substituting (2) into (1)}\)

	\(\dfrac{3}{5}(1-k)\)	\(=1-3 k\)
	\(3-3 k\)	\(=5-15 k\)
	\(k\)	\(=\dfrac{1}{6}\)

iii. \(\overrightarrow{O T}=\lambda \overrightarrow{O R}\)

\(\text {Using parts (i) and (ii):}\)

\(\overrightarrow{O R}=\dfrac{3}{5}\left(1-\dfrac{1}{6}\right)\underset{\sim}{a}+3\left(\dfrac{1}{6}\right) \underset{\sim}{b}=\dfrac{1}{2}\left(\underset{\sim}{a}+\underset{\sim}{b}\right)\)

\(\overrightarrow{O T}=\dfrac{\lambda}{2}(\underset{\sim}{a}+\underset{\sim}{b})\)

♦ Mean mark (iii) 45%.

\(\text {Find } \lambda:\)

	\(\overrightarrow{O T}\)	\(=\overrightarrow{O A}+\mu \overrightarrow{A Q}\)
	\(\dfrac{\lambda}{2}(\underset{\sim}{a}+\underset{\sim}{b})\)	\(=\underset{\sim}{a}+\mu(3 \underset{\sim}{b}-\underset{\sim}{a}) \quad \Big(\text{noting}\ \overrightarrow{A Q}=\overrightarrow{O Q}-\overrightarrow{O A}=3 \underset{\sim}{b}-\underset{\sim}{a}\Big)\)
		\(=\underset{\sim}{a}(1-\mu)+3 \mu \underset{\sim}{b}\)

\(\text {Equating coefficients:}\)

\(\dfrac{\lambda}{2}=3 \mu \ \Rightarrow \ \mu=\dfrac{\lambda}{6}\)

	\(1-\dfrac{\lambda}{6}\)	\(=\dfrac{\lambda}{2}\)
	\(6-\lambda\)	\(=3 \lambda\)
	\(\lambda\)	\(=\dfrac{3}{2}\)

\(\therefore \overrightarrow{O T}=\dfrac{3}{4}(\underset{\sim}{a}+\underset{\sim}{b})\)

Vectors, EXT2 V1 2023 HSC 10 MC

Consider any three-dimensional vectors \(\underset{\sim}{a}=\overrightarrow{O A}, \underset{\sim}{b}=\overrightarrow{O B}\) and \(\underset{\sim}{c}=\overrightarrow{O C}\) that satisfy these three conditions

\(\underset{\sim}{a} \cdot \underset{\sim}{b}=1\)

\(\underset{\sim}{b} \cdot \underset{\sim}{c}=2\)

\(\underset{\sim}{c} \cdot \underset{\sim}{a}=3\).

Which of the following statements about the vectors is true?

Two of \(\underset{\sim}{a}, \underset{\sim}{b}\) and \(\underset{\sim}{c}\) could be unit vectors.
The points \(A, B\) and \(C\) could lie on a sphere centred at \(O\).
For any three-dimensional vector \(\underset{\sim}{a}\), vectors \(\underset{\sim}{b}\) and \(\underset{\sim}{c}\) can be found so that \(\underset{\sim}{a}, \underset{\sim}{b}\) and \(\underset{\sim}{c}\) satisfy these three conditions.
\(\forall \ \underset{\sim}{a}, \underset{\sim}{b}\) and \(\underset{\sim}{c}\) satisfying the conditions, \(\exists \ r, s\) and \(t\) such that \(r, s\) and \(t\) are positive real numbers and \(r\underset{\sim}{a}+s \underset{\sim}{b}+t \underset{\sim}{c}=\underset{\sim}{0}\).

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\(B\)

Show Worked Solution

\(\text{By elimination}\)

\(\text{Option}\ A:\)

\(\text{If}\ \underset{\sim}{a}\ \text{and}\ \underset{\sim}{b}\ \text{are the two unit vectors,}\ \ \underset{\sim}{a} \cdot \underset{\sim}{b} = \abs{\underset{\sim}{a}} \abs{\underset{\sim}{b}} \cos\,\theta = \cos\,\theta\)

\(-1 \leq \cos\,\theta \leq1\ \ \Rightarrow \ \ -1 \leq \underset{\sim}{a} \cdot \underset{\sim}{b} \leq1 \)

\(\text{Given}\ \ \underset{\sim}{a} \cdot \underset{\sim}{b}=1\ \Rightarrow\ \underset{\sim}{a} = \underset{\sim}{b}\ \Rightarrow\ \underset{\sim}{b} \cdot \underset{\sim}{c} = \underset{\sim}{c} \cdot \underset{\sim}{a}\)

\(\text{Contradicts}\ \ \underset{\sim}{b} \cdot \underset{\sim}{c} = 2\ \ \text{and}\ \ \underset{\sim}{c} \cdot \underset{\sim}{a} = 3\)

\(\text{Similar reasoning rules out any pair satisfying all conditions (eliminate}\ A). \)

\(\text{Option}\ C:\ \text{If}\ \ \underset{\sim}{a} = \underset{\sim}{0}, \ \underset{\sim}{a} \cdot \underset{\sim}{b} = 0 \neq 1\ \text{(eliminate}\ C). \)

\(\text{Option}\ D:\)

\(\text{Consider the vectors below that satisfy the conditions,}\)

\[\underset{\sim}{a}=\left(\begin{array}{c} 1 \\ 1 \\ 1 \end{array}\right),\ \ \underset{\sim}{b}=\left(\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\right), \ \ \underset{\sim}{c}=\left(\begin{array}{c} 2 \\ 0 \\ 1 \end{array}\right) \]

\(\text{However,}\ r\underset{\sim}{a}+s \underset{\sim}{b}+t \underset{\sim}{c}=\underset{\sim}{0}\ \text{requires}\ \ r=s=t=0\ \ \text{which are not}\)

\(\text{positive constants (eliminate}\ D).\)

\(\Rightarrow B\)

♦♦♦ Mean mark 15%.

Vectors, EXT2 V1 2023 HSC 15b

On the triangular pyramid \(A B C D, L\) is the midpoint of \(A B, M\) is the midpoint of \(A C, N\) is the midpoint of \(A D, P\) is the midpoint of \(C D, Q\) is the midpoint of \(B D\) and \(R\) is the midpoint of \(B C\).

Let \(\underset{\sim}{b}=\overrightarrow{A B}, \underset{\sim}{c}=\overrightarrow{A C}\) and \(\underset{\sim}{d}=\overrightarrow{A D}\).

Show that \(\overrightarrow{L P}=\dfrac{1}{2}(-\underset{\sim}{b}+\underset{\sim}{c}+\underset{\sim}{d})\). (1 mark)

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It can be shown that

\(\overrightarrow{M Q}=\dfrac{1}{2}(\underset{\sim}{b}-\underset{\sim}{c}+\underset{\sim}{d})\) and

\(\overrightarrow{N R}=\dfrac{1}{2}(\underset{\sim}{b}+\underset{\sim}{c}-\underset{\sim}{d})\). (Do NOT prove these.)

Prove that

\( \Big{|}\overrightarrow{A B}\Big{|}^2+\Big{|}\overrightarrow{A C}\Big{|}^2+\Big{|}\overrightarrow{A D}\Big{|}^2+\Big{|}\overrightarrow{B C}\Big{|}^2+\Big{|}\overrightarrow{B D}\Big{|}^2+\Big{|}\overrightarrow{C D}\Big{|}^2 \)

\(=4\left(\Big{|}\overrightarrow{L P}\Big{|}^2+\Big{|}\overrightarrow{M Q}\Big{|}^2+\Big{|}\overrightarrow{N R}\Big{|}^2\right)\) (3 marks)

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

Show Worked Solution

i.	\(\overrightarrow{LP}\)	\(=\overrightarrow{LA}+\overrightarrow{AC}+\overrightarrow{CP} \)
		\(=\dfrac{1}{2} \overrightarrow{BA}+\overrightarrow{AC}+\dfrac{1}{2}\overrightarrow{CD} \)
		\(=-\dfrac{1}{2} \underset{\sim}b +\underset{\sim}c + \dfrac{1}{2}(\underset{\sim}d-\underset{\sim}c) \)
		\(=-\dfrac{1}{2} \underset{\sim}b +\underset{\sim}c + \dfrac{1}{2}\underset{\sim}d-\dfrac{1}{2}\underset{\sim}c\)
		\(=\dfrac{1}{2} (-\underset{\sim}b+\underset{\sim}c+\underset{\sim}d) \)

ii. \(\overrightarrow{M Q}=\dfrac{1}{2}(\underset{\sim}{b}-\underset{\sim}{c}+\underset{\sim}{d})\)

\(\overrightarrow{N R}=\dfrac{1}{2}(\underset{\sim}{b}+\underset{\sim}{c}-\underset{\sim}{d})\).

\(\text{RHS}\)	\(=4\left(\Big{\|}\overrightarrow{L P}\Big{\|}^2+\Big{\|}\overrightarrow{M Q}\Big{\|}^2+\Big{\|}\overrightarrow{N R}\Big{\|}^2\right)\)
	\(=4\left(\overrightarrow{L P}\cdot \overrightarrow{L P} +\overrightarrow{MQ}\cdot \overrightarrow{MQ} +\overrightarrow{NR}\cdot \overrightarrow{NR}\right)\)
	\(=4\Bigg{(}\dfrac{1}{4}(-\underset{\sim}b+\underset{\sim}c+\underset{\sim}d) \cdot(-\underset{\sim}b+\underset{\sim}c+\underset{\sim}d) + \)
	\( \dfrac{1}{4}(\underset{\sim}{b}-\underset{\sim}{c}+\underset{\sim}{d}) \cdot(\underset{\sim}{b}-\underset{\sim}{c}+\underset{\sim}{d}) + \)
	\( \dfrac{1}{4}(\underset{\sim}{b}+\underset{\sim}{c}-\underset{\sim}{d}) \cdot(\underset{\sim}{b}+\underset{\sim}{c}-\underset{\sim}{d}) \Bigg{)}\)
	\(=(\underset{\sim}c+(\underset{\sim}d-\underset{\sim}b)) \cdot(\underset{\sim}c+(\underset{\sim}d-\underset{\sim}b))+(\underset{\sim}d+(\underset{\sim}b-\underset{\sim}c))\cdot (\underset{\sim}d+(\underset{\sim}b-\underset{\sim}c)) + \)
	\( (\underset{\sim}b+(\underset{\sim}c-\underset{\sim}d))\cdot (\underset{\sim}b+(\underset{\sim}c-\underset{\sim}d)) \)
	\(=\|\underset{\sim}c\|^2+2\underset{\sim}c(\underset{\sim}d-\underset{\sim}b) + \|\underset{\sim}d-\underset{\sim}b\|^2 + \)
	\(\|\underset{\sim}d\|^2+2\underset{\sim}d(\underset{\sim}b-\underset{\sim}c) + \|\underset{\sim}b-\underset{\sim}c\|^2 + \)
	\( \|\underset{\sim}b\|^2+2\underset{\sim}b(\underset{\sim}c-\underset{\sim}d) + \|\underset{\sim}c-\underset{\sim}d\|^2 \)
	\(=\|\underset{\sim}b\|^2+\|\underset{\sim}c\|^2+\|\underset{\sim}d\|^2+\|\underset{\sim}b-\underset{\sim}c\|^2+\|\underset{\sim}d-\underset{\sim}b\|^2+\|\underset{\sim}c-\underset{\sim}d\|^2 + \)
	\( 2(\underset{\sim}c \cdot\underset{\sim}d-\underset{\sim}c \cdot\underset{\sim}b+\underset{\sim}d \cdot\underset{\sim}b-\underset{\sim}d \cdot\underset{\sim}c+\underset{\sim}b \cdot\underset{\sim}c-\underset{\sim}b \cdot\underset{\sim}d) \)
	\(=\Big{\|}\overrightarrow{A B}\Big{\|}^2+\Big{\|}\overrightarrow{A C}\Big{\|}^2+\Big{\|}\overrightarrow{A D}\Big{\|}^2+\Big{\|}\overrightarrow{B C}\Big{\|}^2+\Big{\|}\overrightarrow{B D}\Big{\|}^2+\Big{\|}\overrightarrow{C D}\Big{\|}^2 + 0\)
	\(=\ \text{LHS} \)

♦♦ Mean mark (ii) 34%.

Vectors, EXT2 V1 2023 HSC 11d

The quadrilaterals \(A B C D\) and \(A B E F\) are parallelograms.

By considering \(\overrightarrow{A B}\), show that \(C D F E\) is also a parallelogram. (2 marks)

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

Show Worked Solution

\(\text{Parrallelogram}\ \Rightarrow\ \text{Show opposite sides are equal} \)

\( \Big{|}\overrightarrow{A B} \Big{|} = \Big{|}\overrightarrow{CD} \Big{|} \ \ (ABCD\ \text{is a parallelogram}) \)

\( \Big{|}\overrightarrow{A B} \Big{|} = \Big{|}\overrightarrow{EF} \Big{|} \ \ (ABEF\ \text{is a parallelogram}) \)

\( \Rightarrow \Big{|}\overrightarrow{CD} \Big{|} = \Big{|}\overrightarrow{EF} \Big{|} \)

\( \Big{|}\overrightarrow{CE} \Big{|} = \Big{|}\overrightarrow{BE} \Big{|}-\Big{|}\overrightarrow{BC} \Big{|} \)

\(\text{Similarly,} \)

\( \Big{|}\overrightarrow{DF} \Big{|} = \Big{|}\overrightarrow{AF} \Big{|}-\Big{|}\overrightarrow{AD} \Big{|} \)

\( \text{Since}\ \ \Big{|}\overrightarrow{BE} \Big{|} = \Big{|}\overrightarrow{AF} \Big{|}\ \ \text{and}\ \ \Big{|}\overrightarrow{BC} \Big{|} = \Big{|}\overrightarrow{AD} \Big{|}\)

\( (ABCD\ \text{and}\ ABEF\ \text{are parallelograms}) \)

\( \Rightarrow\ \Big{|}\overrightarrow{DF} \Big{|} = \Big{|}\overrightarrow{CE} \Big{|} \)

\(\therefore CDFE\ \text{is a parallelogram} \)

Vectors, EXT2 V1 EQ-Bank 31

Point `O` is the circumcentre of triangle `ABC` which is the centre of the circle that passes through each of its vertices.

Point `P` is the centroid of triangle `ABC` where the bisectors of each angle within the triangle intersect.

Point `Q` is such that `vec(OQ)=3vec(OP)`.

Prove that `vec(CQ) ⊥ vec(AB)` (5 marks)

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`text{See Worked Solution}`

Show Worked Solution

`vec(OP)=(vec(OA)+vec(OB)+vec(OC))/3`

`vec(OQ)`	`=3\ vec(OP)`
	`=3((vec(OA)+vec(OB)+vec(OC))/3)`
	`=vec(OA)+vec(OB)+vec(OC)`
	`=((x_1+x_2+x_3),(y_1+y_2+y_3))`

`vec(CQ)`	`=vec(OQ)-vec(OC)`
	`=((x_1+x_2+x_3),(y_1+y_2+y_3))-((x_3),(y_3))`
	`=((x_1+x_2),(y_1+y_2))`

`vec(AB)=vec(OB)-vec(OA)=((x_2),(y_2))-((x_1),(y_1))=((x_2-x_1),(y_2-y_1))`

`text{Test}\ \ vec(CQ)*vec(AB)=0:`

`text{LHS}`	`=((x_1+x_2),(y_1+y_2))((x_2-x_1),(y_2-y_1))`
	`=(((x_1+x_2)(x_2-x_1)),((y_1+y_2)(y_2-y_1)))`
	`=(x_2)^2-(x_1)^2+(y_2)^2-(y_1)^2`
	`=(x_2)^2+(y_2)^2-((x_1)^2+(y_1)^2)`
	`=abs(vec(OB))^2-abs(vec(OA))^2`
	`=OB^2-OA^2\ \ \ text{(radii)}`
	`=0`

`:.vec(CQ) ⊥ vec(AB)\ \ text{… as required}`

Vectors, EXT2 V1 EQ-Bank 11

Use vector methods to find the coordinates of the point that divides the interval joining the points `A(7,-3,0)` and `B(2,2,-10)` in the ratio `2:3`. (3 marks)

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`(5,-1,-4)`

Show Worked Solution

`text{Method 1}`

`text{Let}\ \ underset~c =\ text{position vector of the dividing point}`

`underset~c`	`=((7),(-3),(0)) + 2/5((2-7),(2-(-3)),(-10-0))`
	`=((7),(-3),(0)) + ((-2),(2),(-4))`
	`=((5),(-1),(-4))`

`text{Method 2}`

`underset~c`	`=((2),(2),(-10)) + 3/5((7-2),(-3-2),(0-(-10)))`
	`=((2),(2),(-10)) + ((3),(-3),(6))`
	`=((5),(-1),(-4))`

Vectors, EXT2 V1 EQ-Bank 22

Given `underset~a=(3,-2,1)` and `underset~b=(4,3,-4)`, verify numerically that

`underset~a*underset~b=abs(underset~a)abs(underset~b)cos theta=x_1x_2+y_1y_2+z_1z_2`

where `underset~a=(x_1,y_1,z_1)` and `underset~b=(x_2,y_2,z_2)` (4 marks)

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`text{See Worked Solution}`

Show Worked Solution

`text{Scalar Product result 1:}`

`underset~a*underset~b`	`=x_1x_2+y_1y_2+z_1z_2`
	`=3xx4+(-2)xx3+1xx(-4)`
	`=2`

`text{Scalar Product result 2:}`

`underset~a*underset~b=abs(underset~a)abs(underset~b)cos theta`

`text{Let}\ \ underset~c=underset~b-underset~a`

`underset~c=((4),(3),(-4))-((3),(-2),(1))=((1),(5),(-5))`

`abs(underset~c)=sqrt(1^2+5^2+(-5)^2)=sqrt(51)`

`abs(underset~a)=sqrt(3^2+(-2)^2+1^2)=sqrt(14)`

`abs(underset~b)=sqrt(4^2+3^2+(-4)^2)=sqrt(41)`

`text{Using cosine rule:}\ \ c^2=a^2+b^2-2ab\ cosC`

`=>\ \ ab\ cosC=(a^2+b^2-c^2)/2`

`underset~a*underset~b`	`=abs(underset~a)abs(underset~b)cos theta`
	`=(abs(underset~a)^2+abs(underset~b)^2-abs(underset~c)^2)/2`
	`=(14+41-51)/2`
	`=2`

`:.underset~a*underset~b=abs(underset~a)abs(underset~b)cos theta=x_1x_2+y_1y_2+z_1z_2`

Vectors, EXT2 V1 EQ-Bank 23

In a rectangular prism, `M` is the midpoint of `AD`.

Use vector methods to find

`angle HBD` to 1 decimal place. (2 marks)

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`angle HBM` to 1 decimal place. (2 marks)

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a. `40.5°`

b. `51.5°`

Show Worked Solution

a. `text{Consider the position vectors of points using}\ B\ text{as origin:}`

`vec(BH)=((0),(9),(6)), \ abs(vec(BH))=sqrt(9^2+6^2)=sqrt117`

`vec(BD)=((4),(9),(0)), \ abs(vec(BD))=sqrt(4^2+9^2)=sqrt97`

`cos angleHBD`	`=(vec(BH)vec(BD))/(abs(vec(BH))abs(vec(BD)))`
	`=(0xx4+9xx9+6xx0)/(sqrt117 xx sqrt97)`
`angleHBD`	`=cos^(-1)(81/(sqrt117 xx sqrt97))`
	`=40.5°`

b. `vec(BM)=((4),(9/2),(0)), \ abs(vec(BM))=sqrt(4^2+(9/2)^2)=sqrt(145/4)`

`cos angleHBM`	`=(vec(BH) * vec(BM))/(abs(vec(BH))*abs(vec(BM)))`
	`=(0xx4+9xx9/2+6xx0)/(sqrt117 xx sqrt(145/4))`
`angleHBM`	`=cos^(-1)(40.5/(sqrt117 xx sqrt(145/4)))`
	`=51.5°\ \text{(1 d.p.)}`

Vectors, EXT2 V1 EQ-Bank 15

A parallelogram is formed by joining the points `P(-2,1,4), Q(1,4,5), R(0,2,3)` and `S(a,b,c)`.

Use vector methods to find `a,b` and `c`. (2 marks)

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`a=-3, b=-1, c=2`

Show Worked Solution

`text{Opposite sides of a parallelogram are equal and parallel.}`

`=> vec(PQ)=vec(SR)`

`vec(PQ)=((1+2),(4-1),(5-4))=((3),(3),(1))`

`vec(SR)=((-a),(2-b),(3-c))`

`text{Equating coordinates:}`

`-a=3\ \ =>\ \ a=-3`

`2-b=3\ \ =>\ \ b=-1`

`3-c=1\ \ =>\ \ c=2`

Vectors, EXT2 V1 EQ-Bank 24

Classify the triangle formed by joining the points `A(3,1,0), B(-2,4,3)` and `C(3,3,-2)`. (4 marks)

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`text{ΔABC is a right-angled (scalene) triangle with a right-angle at B.}`

`text{(See Worked Solutions)}`

Show Worked Solution

`text{Calculating the side lengths:}`

`abs(AB)=sqrt((3+2)^2+(1-4)^2+(0-3)^2)=sqrt43`

`abs(BC)=sqrt((-2-3)^2+(4-3)^2+(3+2)^2)=sqrt51`

`abs(AC)=sqrt((3-3)^2+(1-3)^2+(0+2)^2)=sqrt8`

`=>\ text{Triangle is scalene.}`

`text{From above,}`

`abs(BC)^2=abs(AB)^2+abs(AC)^2\ \ (angleBAC=90°)`

`text{Consider scalar product of direction vectors:}`

`vec(AB)=((3),(1),(0))+lambda((-2-3),(4-1),(3-0))=((3),(1),(0))+lambda((-5),(3),(3))`

`vec(AC)=((3),(1),(0))+mu((3-3),(3-1),(-2-0))=((3),(1),(0))+mu((0),(2),(-2))`

`vec(AB)*vec(AC)=((-5),(3),(3))((0),(2),(-2))=0+6-6=0`

`:.vec(AB) ⊥ vec(AC)`

`:.\ text{ΔABC is a right-angled (scalene) triangle with a right-angle at A.}`

Vectors, EXT2 V1 EQ-Bank 14

In triangle `ABC`, `M` is the midpoint of `AC` and `N` is the midpoint of `AB`.

Use vector methods to prove that

`MN=1/2CB` (2 marks)

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`MN` is parallel to `CB` (1 mark)

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a. `text(See Worked Solution)`

b. `text(See Worked Solution)`

Show Worked Solution

`vec(AC)+vec(CB)+vec(BA)=0\ \ \ text{(resultant vector ends at starting point)}`

`vec(CB)=-(vec(AC)+vec(BA))`

`text(S)text(ince)\ M\ text (and)\ N\ text(are midpoints:)`

`vec(MN)`	`=-1/2vec(AC)-1/2vec(BA)`
	`=-1/2(vec(AC)+vec(AB))`
	`=1/2vec(CB)\ \ text(… as required)`

b. `text{If}\ \ vec(u) = kvec(v)\ \ (k\ text{scalar})\ \ =>\ \ vec(u)\ text{||}\ vec(v)`

`vec(MN)=1/2vec(CB)\ \ text{(see (i))`

`:.MN\ text{||}\ RCB`

Vectors, EXT2 V1 EQ-Bank 13

Use vector methods to find the coordinates of the point that divides the interval joining the points `A(-1,3,2)` and `B(7,-1,-6)` in the ratio `1:3`. (3 marks)

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`(1,2,0)`

Show Worked Solution

`text{Method 1}`

`text{Let}\ \ underset~c =\ text{position vector of the dividing point}`

`underset~c`	`=((-1),(3),(2)) + 1/4((7-(-1)),(-1-3),(-6-2))`
	`=((-1),(3),(2)) + ((2),(-1),(-2))`
	`=((1),(2),(0))`

`text{Method 2}`

`underset~c`	`=((7),(-1),(-6)) + 3/4((-1-7),(3-(-1)),(2-(-6)))`
	`=((7),(-1),(-6)) + ((-6),(3),(6))`
	`=((1),(2),(0))`

Vectors, EXT2 V1 EQ-Bank 12

Use two vector methods to locate the midpoint of the interval joining the points `A(3,-2,1)` and `B(5,4,-3)`. (3 marks)

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`(4,-1,-1)`

Show Worked Solution

`text{Method 1}`

`text{Let}\ \ underset~c =\ text{position vector of the midpoint}`

`underset~c`	`=((3),(-2),(1)) + 1/2((5-3),(4-(-2)),(-3-1))`
	`=((3),(-2),(1)) + ((1),(3),(-2))`
	`=((4),(1),(-1))`

`text{Method 2}`

`underset~c`	`=((5),(4),(-3)) + 1/2((3-5),(-2-4),(1-(-3)))`
	`=((5),(4),(-3)) + ((-1),(-3),(2))`
	`=((4),(1),(-1))`

Vectors, EXT2 V1 2022 HSC 9 MC

Let \(A\) and \(B\) be two distinct points in three-dimensional space. Let \(M\) be the midpoint of \(A B\).

Let \(S_1\) be the set of all points \(P\) such that \(\overrightarrow{AP} \cdot \overrightarrow{BP}=0\).

Let \(S_2\) be the set of all points \(N\) such that \(\Big|\overrightarrow{AN}\Big|=\Big| \overrightarrow{MN} \Big| \).

The intersection of \(S_1\) and \(S_2\) is the circle \(S\).

What is the radius of the circle \(S\) ?

\(\dfrac{\Big| \overrightarrow{AB} \Big|}{2} \)
\(\dfrac{\Big| \overrightarrow{AB} \Big|}{4} \)
\(\dfrac{\sqrt3 \Big| \overrightarrow{AB} \Big|}{2} \)
\(\dfrac{\sqrt3 \Big| \overrightarrow{AB} \Big|}{4} \)

Show Answers Only

\(D\)

Show Worked Solution

`text{Diagram below is a 2-D sliced image of the 3-D geometry:}`

\(\overrightarrow{AP} \cdot \overrightarrow{BP}=0\).

\(r=\dfrac{\Big| \overrightarrow{AB} \Big|}{2} \)

\(S_2\ \text{includes}\ N\ \text{where}\ \Big| \overrightarrow{AN} \Big|=\Big| \overrightarrow{MN} \Big| = \dfrac{r}{2} \)

\(\text{Let}\ r_s=\ \text{radius of}\ S\)

\(\text{Point}\ P\ \text{is intersection of}\ S_1\ \text{and}\ S_2 \)

\(\text{By Pythagoras (see diagram):}\)

\(r_s^2\)	\(=r^2-(\dfrac{r}{2})^2 \)
	\(=\dfrac{3r^2}{4}\)
\(r_s\)	\(=\dfrac{\sqrt3}{2} \times r\)
	\(=\dfrac{\sqrt3}{2} \cdot \dfrac{\Big\| \overrightarrow{AB} \Big\|}{2} \)
	\(=\dfrac{\sqrt3 \Big\| \overrightarrow{AB} \Big\|}{4} \)

\(=>D\)

♦♦♦ Mean mark 20%.

Vectors, EXT2 V1 2022 HSC 14a

The two non-parallel vectors `\vec{u}` and `\vec{v}` satisfy `lambda vec(u)+mu vec(v)= vec(0)` for some real numbers `\lambda` and `\mu`.
Show that `\lambda=\mu=0`. (2 marks)

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The two non-parallel vectors `\vec{u}` and `\vec{v}` satisfy `\lambda_1 \vec{u}+\mu_1 \vec{v}=\lambda_2 \vec{u}+\mu_2 \vec{v}` for some real numbers `\lambda_1, \lambda_2, \mu_1` and `\mu_2`.
Using part (i), or otherwise, show that `\lambda_1=\lambda_2` and `\mu_1=\mu_2`. (1 mark)

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The diagram below shows the tetrahedron with vertices `A, B, C` and `S`.

The point `K` is defined by `vec(SK)=(1)/(4) vec(SB)+(1)/(3) vec(SC)`, as shown in the diagram.

The point `L` is the point of intersection of the straight lines `S K` and `B C`.

Using part (ii), or otherwise, determine the position of `L` by showing that `vec(BL)=(4)/(7) vec(BC)`. (2 marks)

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The point `P` is defined by `vec(AP)=-6 vec(AB)-8 vec(AC)`.
Does `P` lie on the line `A L`? Justify your answer. (2 marks)

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`text{Proof (See Worked Solution)}`
`text{Proof (See Worked Solution)}`
`text{Proof (See Worked Solution)}`
`P\ text{lies on}\ vec(AL)\ \ text{(See Worked Solution)}`

Show Worked Solution

i.	`lambda vecu+mu vecv`	`=0`
	`lambda vecu`	`=-mu vecv`

`lambda=0\ \ text{or}\ \ vecu=-(mu/lambda)vecv=k vecv\ \ (kinRR)`

`text{S}text{ince}\ \ vecu and vecv\ \ text{are not parallel}`

`=> lambda=mu=0`

ii. `\lambda_1 \vec{u}+\mu_1 \vec{v}=\lambda_2 \vec{u}+\mu_2 \vec{v}`

`\lambda_1 \vec{u}-\lambda_2 \vec{u}+\mu_1 \vec{v}-\mu_2 \vec{v}`	`=vec0`
`(\lambda_1-\lambda_2)\vec{u}+(\mu_1-\mu_2)\vec{v}`	`=vec0`

`text{Using part (i):}`

`(\lambda_1-\lambda_2)=0 and (\mu_1-\mu_2)=0`

`:.\lambda_1=\lambda_2 and \mu_1=\mu_2\ …\ text{as required}`

Mean mark (ii) 56%.

iii.	`vec(BL)`	`=lambda vec(BC)`
		`=lambda(vec(BS)+vec(SC))\ \ \ …\ (1)`

`vec(BL)`	`=vec(BS)+mu vec(SK)`
	`=-vec(SB)+mu(1/4vec(SB)+1/3vec(SC))`
	`=-vec(SB)+mu/4vec(SB)+mu/3vec(SC)`
	`=mu/3vec(SC)+(mu/4-1)vec(SB)\ \ \ …\ (2)`

♦♦♦ Mean mark (iii) 25%.

`text{Using}\ \ (1) = (2):`

`lambda vec(BS)+ lambda vec(SC)=mu/3vec(SC)+(mu/4-1)vec(SB)`

`mu/3=lambda, \ \ 1-mu/4=lambda`

`mu/3`	`=1-mu/4`
`(4mu+3mu)/12`	`=1`
`(7mu)/12`	`=1`
`mu`	`=12/7`

`lambda=(12/7)/3=4/7`

`:.vec(BL)=(4)/(7) vec(BC)\ \ text{… as required}`

iv. `vec(AP)=-6 vec(AB)-8 vec(AC)`

`text{If}\ P\ text{lies on}\ AL, ∃k\ text{such that}\ \ vec(AP)=k vec(AL)`

♦♦♦ Mean mark (iv) 23%.

`-6 vec(AB)-8 vec(AC)`	`=k vec(AL)`
`-6 vec(AB)-8 (vec(AB)+vec(BC))`	`=k(vec(AB)+vec(BL))`
`-14 vec(AB)-8vec(BC)`	`=k(vec(AB)+4/7vec(BC))\ \ text{(see part (iii))}`
`-14 vec(AB)-8vec(BC)`	`=kvec(AB)+(4k)/7vec(BC)`

`k=-14, \ \ (4k)/7=-8\ \ =>\ \ k=-14`

`:.\ P\ text{lies on}\ \ AL.`

Vectors, EXT2 V1 2021 HSC 16a

The point `P(x, y, z)` lies on the sphere of radius 1 centred at the origin `O`.
Using the position vector of `P, overset->{OP} = x underset~i + y underset~j + z underset~k` , and the triangle inequality, or otherwise, show that `| x | + | y | + |z | ≥ 1`. (2 marks)

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Given the vectors `underset~a = ((a_1),(a_2),(a_3))` and `underset~b = ((b_1),(b_2),(b_3))` , show that
`|a_1 b_1 + a_2 b_2 + a_3 b_3 | ≤ sqrt{a_1^2 + a_2^2 + a_3^2 }\ sqrt{b_1^2 + b_2^2 + b_3^2}`. (3 marks)

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As in part (i), the point `P (x, y, z)` lies on the sphere of radius 1 centred at the origin `O`.
Using part (ii), or otherwise, show that `| x | + | y | + | z | ≤ sqrt3`. (2 marks)

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`text{See Worked Solution}`
`text{See Worked Solution}`
`text{See Worked Solution}`

Show Worked Solution

i. `text{Triangle inequality:} \ |x| + |y| ≥ |x + y|`

♦ Mean mark (i) 50%.

`\|x\| + \|y\| + \|z\|`	` = \|x_underset~i\| + \|y_underset~j\| + \|z_underset~k\|`
	`≥ \|x_underset~i + y_underset~j\| + \|z_underset~k\|`
	`≥ \|x_underset~i + y_underset~j + z_underset~k\|`
	`≥ 1\ \ (\|overset->{OP}\| = \| x_underset~i + y_underset~j + z_underset~k \| = 1)`

ii. `text{Using the dot product:}`

♦ Mean mark (ii) 42%.

`underset~a * underset~b = a_1 b_1 + a_2 b_2 + a_3 b_3`

`underset~a * underset~b = |underset~a| |underset~b| cos theta`

`a_1 b_1 + a_2 b_2 + a_3 b_3`	`= sqrt{a_1^2 + a_2^2 + a_3^2} * sqrt{b_1^2 + b_2^2 + b_3^2} * cos theta`
`\|a_1 b_1 + a_2 b_2 + a_3 b_3\|`	`= sqrt{a_1^2 + a_2^2 + a_3^2} * sqrt{b_1^2 + b_2^2 + b_3^2} * \|cos theta\|`

`text{S} text{ince} \ -1 ≤ cos theta ≤ 1 \ => \ |cos theta| ≤ 1`

`:. \ | a_1 b_1 + a_2 b_2 + a_3 b_3 | ≤ sqrt{a_1^2 + a_2^3 + a_3^2} * sqrt{b_1^2 + b_2^2 + b_3^2}`

iii. `text{Using part (ii) with vectors:}`

♦♦♦ Mean mark (iii) 14%.

`underset~a = ((1),(1),(1)) \ , \ underset~b = (( | x| ),( |y| ),( |z| ))`

`\|\ \|x\| + \|y\| + \|z\|\ \|`	`≤ sqrt{1^2 + 1^2 + 1^2} * sqrt{ x^2 + y^2 + z^2}`
`\|x\| + \|y\| + \|z\|`	`≤ sqrt3`

Vectors, EXT2 V1 2021 HSC 12e

The diagram shows the pyramid `ABCDS` where `ABCD` is a square. The diagonals of the square bisect each other at `H`.

Show that `overset->{HA} + overset->{HB} + overset->{HC} + overset->{HD} = underset~0` (1 mark)

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Let `G` be the point such that `overset->{GA} + overset->{GB} + overset->{GC} + overset->{GD} + overset->{GS} = underset~0`.

Using part (i), or otherwise, show that `4 overset->{GH} + overset->{GS} = underset~0`. (2 marks)

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Find the value of `λ` such that `overset->{HG} = λ overset->{HS}` (1 mark)

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`text{See Worked Solution}`
`text{See Worked Solution}`
`1/5`

Show Worked Solution

i. `text{S} text{ince diagonal} \ overset->{AC} \ text{is bisected by H:}`

`overset->{HA} =- overset->{HC}`

`text{Similarly for diagonal} \ overset->{BD}`

`overset->{HB} = – overset->{HD}`

`:. \ overset->{HA} + overset->{HB} + overset->{HC} + overset->{HD}`	`= – overset->{HC} – overset->{HD} + overset->{HC} + overset->{HD}`
	`= underset~0`

ii. `overset->{GA} = overset->{GH} + overset->{HA} \ , \ overset->{GB} = overset->{GH} + overset->{HB}`

`overset->{GC} = overset->{GH} + overset->{HC} \ , \ overset->{GD} = overset->{GH} + overset->{HD}`

`overset->{GA} + overset->{GB} + overset->{GC} + overset->{GD} + overset->{GS}`

`= overset->{GH} + overset->{HA} + overset->{GH} + overset->{HB} + overset->{GH} + overset->{HC} + overset->{GH} + overset->{HD} + overset->{GS}`

`= 4 overset->{GH} + (overset->{HA} + overset->{HB} + overset->{HC} + overset->{HD} + overset->{GS})`

`= 4 overset->{GH} + overset->{GS}`

`overset->{GA} + overset->{GB} + overset->{GC} + overset->{GD} + overset->{GS} = underset~0\ \ \ text{(given)}`

`:. 4 overset->{GH} + overset->{GS} = underset~0`

iii. `overset->{HS} = overset->{HG} + overset->{GS}`

`overset->{GS} = overset->{HS} + overset->{GH}`

♦ Mean mark 48%.

`text{Using part (ii):}`

`4 overset->{GH} + overset->{HS} + overset->{GH}`	`= underset~0`
`5 overset->{GH}`	`= overset->{SH}`
`overset->{GH}`	`= 1/5 overset->{SH}`
`overset->{HG}`	`= 1/5 overset->{HS}`

`:. \ λ = 1/5`

Vectors, EXT2 V1 2021 HSC 1 MC

Four cubes are placed in a line as shown on the diagram.

Which of the following vectors is equal to `overset->{AB} + overset->{CQ}`

`overset->{AQ}`
`overset->{CP}`
`overset->{PB}`
`overset->{RA}`

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`B`

Show Worked Solution

`overset->{AB} \ + \ overset->{CQ}`	`= overset->{CD} + overset->{DP}`
	`= overset->{CP}`

`=>\ B`

Vectors, EXT2 V1 2020 HSC 15b

The point `C` divides the interval `AB` so that `frac{CB}{AC} = frac{m}{n}`. The position vectors of `A` and `B` are `underset~a` and `underset~b` respectively, as shown in the diagram.

Show that `overset->(AC) = frac{n}{m + n} (underset~b - underset~a)`. (2 marks)

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Prove that `overset->(OC) = frac{m}{m + n} underset~a + frac{n}{m + n} underset~b`. (1 mark)

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Let `OPQR` be a parallelogram with `overset->(OP) = underset~p` and `overset->(OR) = underset~r`. The point `S` is the midpoint of `QR` and `T` is the intersection of `PR` and `OS`, as shown in the diagram.

Show that `overset->(OT) = frac{2}{3} underset~r + frac{1}{3} underset~p`. (3 marks)

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Using parts (ii) and (iii), or otherwise, prove that `T` is the point that divides the interval `PR` in the ratio 2 :1. (1 mark)

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`text{See Worked Solutions}`
`text{See Worked Solutions}`
`text{See Worked Solutions}`
`text{See Worked Solutions}`

Show Worked Solution

`frac{overset->(AC)}{overset->(AB)}`	`= frac{n}{m + n}`
`overset->(AC)`	`= frac{n}{m + n} * overset->(AB)`
	`= frac{n}{m + n} (underset~b – underset~a)`

ii.	`overset->(OC)`	`= overset->(OA) + overset->(AC)`
		`= underset~a + frac{n}{m + n} (underset~b – underset~a)`
		`= underset~a – frac{n}{m + n} underset~a + frac{n}{m + n} underset~b`
		`= (1 – frac{n}{m + n}) underset~a + frac{n}{m + n} underset~b`
		`= (frac{m + n – n}{m + n}) underset~a + frac{n}{m + n} underset~b`
		`= frac{m}{m + n} underset~a + frac{n}{m + n} underset~b`

iii. `text{Show} \ \ overset->(OT) = frac{2}{3} underset~r + frac{1}{3} underset~p`

♦ Mean mark part (iii) 50%.

`text{Consider} \ \ Delta PTO \ \ text{and} \ \ Delta RTS:`

`angle PTO = angle RTS \ (text{vertically opposite})`

`angle OPT = angle SRT \ (text{vertically opposite})`

`therefore \ Delta PTO \ text{|||} \ Delta RTS\ \ text{(equiangular)}`

`OT : TS = OP : SR = 2 : 1`

`(text{corresponding sides in the same ratio})`

`frac{overset->(OT)}{overset->(OS)}`	`= frac{2}{3}`
`overset->(OT)`	`= frac{2}{3} overset->(OS)`
	`= frac{2}{3} ( underset~r + frac{1}{2} underset~p)`
	`= frac{2}{3} underset~r + frac{1}{3} underset~p`

Mean mark part (iv) 51%.

iv. `text{Let} \ \ overset->(OT) \ text{divide} \ PR\ text{so that}\ \ frac{TR}{PT} = frac{m}{n}`

`text{Using part (ii):}`

`overset->(OT)`	`= frac{m}{m + n} underset~p + frac{n}{m + n} c`
`overset->(OT)`	`= frac{1}{3} underset~p + frac{2}{3} underset~r \ \ \ (text{part (iii)})`
`frac{m}{m + n}`	`= frac{1}{3} , frac{n}{m + n} = frac{2}{3}`

`=> \ m = 1 \ , \ n = 2`

`therefore \ T \ text{divides} \ PR \ text{in ratio 2 : 1}.`

Vectors, EXT2 V1 2019 SPEC2 4

The base of a pyramid is the parallelogram `ABCD` with vertices at points `A(2,−1,3), B(4,−2,1), C(a,b,c)` and `D(4,3,−1)`. The apex (top) of the pyramid is located at `P(4,−4,9)`.

Find the values of `a, b` and `c`. (2 marks)

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Find the cosine of the angle between the vectors `overset(->)(AB)` and `overset(->)(AD)`. (2 marks)

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Find the area of the base of the pyramid. (2 marks)

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a. `a = 6, b = 2, c = −3`

b. `4/9`

c. `2sqrt65\ text(u²)`

Show Worked Solution

a.	`overset(->)(AB)`	`= (4 – 2)underset~i + (−2 + 1)underset~j + (1 – 3)underset~k`
		`= 2underset~i – underset~j – 2underset~k`

`text(S)text(ince)\ ABCD\ text(is a parallelogram)\ => \ overset(->)(AB)= overset(->)(DC)`

`overset(->)(DC) = (a – 4)underset~i + (b – 3)underset~j + (c + 1)underset~k`

`a – 4 = 2 \ => \ a = 6`

`b – 3 = −1 \ => \ b = 2`

`c + 1 = −2 \ => \ c = −3`

b. `overset(->)(AB) = 2underset~i – underset~j – 2underset~k`

`overset(->)(AD) = 2underset~i + 4underset~j – 4underset~k`

`cos angleBAD`	`= (overset(->)(AB) · overset(->)(AD))/(\|overset(->)(AB)\| · \|overset(->)(AD)\|)`
	`= (4 – 4 + 8)/(sqrt(4 + 1 + 4) · sqrt(4 + 16 + 16))`
	`= 4/9`

c.	`1/2 xx text(Area)_(ABCD)`	`= 1/2 ab sin c`
	`text(Area)_(ABCD)`	`= \|overset(->)(AB)\| · \|overset(->)(AD)\| *sin(cos^(−1)\ 4/9)`

`:. text(Area)_(ABCD)= 3 · 6 · sqrt65/9= 2sqrt65\ text(u)^2`

Vectors, EXT2 V1 2019 NHT 5

A triangle has vertices `A(sqrt3 + 1, –2, 4), \ B(1, –2, 3)` and `C(2, –2, sqrt3 + 3)`.

Find angle `ABC` (3 marks)

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Find the area of the triangle. (2 marks)

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a. `∠ABC = (pi)/(6)`

b. `1 \ text(u)^2`

Show Worked Solution

a. `overset(->)(BA) = ((sqrt3 + 1), (-2), (4)) – ((1), (-2), (3)) = ((sqrt3), (0), (1))`

`overset(->)(BC) = ((2), (-2), (sqrt3 + 3)) – ((1), (-2), (3)) = ((1), (0), (sqrt3))`

`cos ∠ABC= (overset(->)(BA) · overset(->)(BC))/ (|overset(->)(BA)| |overset(->)(BC)|) = (2 sqrt3)/(sqrt4 sqrt4)= (sqrt3)/(2)`

`:. \ ∠ABC = (pi)/(6)`

b.	`text(Area)`	`= (1)/(2) · \|overset(->)(BA)\| \|overset(->)(BC)\| \ sin ∠ABC`
		`= (1)/(2) xx 2 xx 2 xx sin \ (pi)/(6)`
		`= 1 \ text(u)^2`

Vectors, EXT2 V1 EQ-Bank 25

Let `OABCD` be a right square pyramid where `underset ~a = vec(OA),\ underset ~b = vec(OB),\ underset ~c = vec(OC)` and `underset ~d = vec(OD)`.

Show that `underset~a + underset~c = underset~b + underset~d`. (3 marks)

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`text(See Worked Solution)`

Show Worked Solution

`text(Let)\ \ A=(p,–p,–k),`

`underset ~a`	`= overset(->)(OA) = punderset~i – punderset~j – qunderset~k`
`underset ~b`	`= overset(->)(OB) = punderset~i + punderset~j – qunderset~k`
`underset ~c`	`= overset(->)(OC) = −punderset~i + punderset~j – qunderset~k`
`underset ~d`	`= overset(->)(OA) = −punderset~i – punderset~j – qunderset~k`

`underset~a + underset~c`	`= −2qunderset~k`
`underset ~b + underset ~d`	`= −2qunderset~k`

`:. underset~a + underset~c = underset~b + underset~d`

Vectors, EXT2 V1 EQ-Bank 19

A cube with side length 3 units is pictured below.

Calculate the magnitude of vector `vec(AG)`. (1 mark)

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Find the acute angle between the diagonals `vec(AG)` and `vec(BH)`. (3 marks)

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a. `3 sqrt 3\ text(units)`

b. `70^@32′`

Show Worked Solution

a. `A(3, 0 , 0), \ \ G(0, 3, 3)`

	`vec(AG)`	`= ((0), (3), (3))-((3), (0), (0)) = ((text{−3}), (3), (3))`
	`\|\ vec(AG)\ \|`	`= sqrt (9 + 9 + 9)= 3 sqrt 3\ text(units)`

b.	`H (3, 3, 3)`
	`vec(BH) = ((3), (3), (3))`

`vec(AG) ⋅ vec(BH)`	`= \|\ vec(AG)\ \| ⋅ \|\ vec(BH)\ \|\ cos theta`
`((text{−3}), (3), (3)) ⋅ ((3), (3), (3))`	`= sqrt (9 + 9 + 9) ⋅ sqrt (9 + 9 + 9) cos theta`
`-9 + 9 + 9`	`= 27 cos theta`
`cos theta`	`= 1/3`
`theta`	`= 70.52…= 70^@32^{′}`

Vectors, EXT2 V1 EQ-Bank 28

`ABCDEFGH` are the vertices of a rectangular prism.

Show that the internal diagonals of the prism, `AG` and `DF`, intersect. (2 marks)

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Calculate the acute angle, `theta`, between the diagonals, to the nearest minute. (2 marks)

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

b. `83^@37^{′}`

Show Worked Solution

a.	`A(2, text{−2}, 0),`	`G(text{−2}, 2, 2)`
	`D(2, 2, 0),`	`F (text{−2}, text{−2}, 2)`

`text(Midpoint)\ AG = ((1/2 (2-2)),(1/2 (text{−2} + 2)),(1/2 (0 + 2))) = ((0), (0), (1))`

`text(Midpoint)\ DF = ((1/2 (2-2)),(1/2 (2-2)),(1/2 (0 + 2))) = ((0), (0), (1))`

`text(S) text(ince midpoints are the same), AG and DF\ text(intersect.)`

b.	`vec(AG) = ((text{−2}), (2), (2))-((2), (text{−2}), (0)) = ((text{−4}), (4), (2))`
	`vec(DF) = ((text{−2}), (text{−2}), (2))-((2), (2), (0)) = ((text{−4}), (text{−4}), (2))`

`vec (AG) ⋅ vec (DF) = |\ vec (AG)\ | ⋅ |\ vec(DF)\ |\ cos theta`

`((text{−4}), (4), (2)) ⋅ ((text{−4}), (text{−4}), (2)) = sqrt 36 sqrt 36 cos theta`

`16-16 + 4`	`= 36 cos theta`
`cos theta`	`= 1/9`
`theta`	`= 83.62…= 83^@37^{′}`

Vectors, EXT2 V1 EQ-Bank 26

Point `B` sits on the arc of a semi-circle with diameter `AC`.

Using vectors, show `angle ABC` is a right angle. (2 marks)

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`text(See Worked Solution)`

Show Worked Solution

`text(Let)\ \ vec (OA) = underset~a, \ text(and)\ \ vec(OC)=underset~c`

`text(Prove) \ overset(->)(AB) ⊥ overset(->)(BC)`

`|underset~a| = |underset~b| = |underset~c|\ \ \ text{(radii)}`

`underset~c = – underset~a `

`overset(->)(AB) ⋅ overset(->)(BC)`	`= (underset~b-underset~a) (underset~c-underset~b)`
	`= underset~b · underset~c-\|underset~b\|^2-underset~a · underset~c + underset~a · underset~b`
	`= underset~b · underset~c-\|underset~b\|^2 + underset~c · underset~c-underset~c · underset~b`
	`= \|underset~c\|^2-\|underset~b\|^2`
	`= 0`

`:. \ ∠ABC \ text(is a right angle.)`

Vectors, EXT2 V1 EQ-Bank 27

`OABD` is a trapezium in which `overset(->)(OA) = underset~a` and `overset(->)(OB) = underset~b`.

`OC` is parallel to `AB` and `DC : CB = 1:2`

Using vectors, express `overset(->)(DA)` in terms of `underset~a` and `underset~b`. (3 marks)

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`text(See Worked Solution)`

Show Worked Solution

`overset(->)(DA) = overset(->)(DB) + overset(->)(BA)`

`overset(->)(BA) = underset~a-underset~b`

`text(S) text(ince) \ OC || AB \ \ text(and) \ \ OA || CB`

`=> OABC \ text(is a parallelogram)`

`overset(->)(OA)`	`= overset(->)(CB) = underset~a`
`overset(->)(DC)`	`= (1)/(2) \ underset~a `
`overset(->)(DB)`	`= overset(->)(DC) + overset(->)(CB)`
	`= (3)/(2) \ underset~a`

`:. \ overset(->)(DA)`	`= (3)/(2) \ underset~a + (underset~a-underset~b)`
	`= (5)/(2) \ underset~a-underset~b`

Vectors, EXT2 V1 2015 VCE 1

Consider the rhombus `OABC` shown below, where `vec (OA) = a underset ~i` and `vec (OC) = underset ~i + underset ~j + underset ~k`, and `a` is a positive real constant.

Find `a.` (1 mark)

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Show that the diagonals of the rhombus `OABC` are perpendicular. (2 marks)

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a. `sqrt 3`

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

Show Worked Solution

a. `|\ vec(OA)\ | = |\ vec(OC)\ |,`

`:. a`	`= sqrt (1^2 + 1^2 + 1^2)`
	`= sqrt 3`

b. `overset(->)(CB) = overset(->)(OA), \ \ overset(->)(AB) = overset(->)(OC)`

MARKER’S COMMENT: Vector notation was poor in many answers.

`overset(->)(OB) = overset(->)(OC) + overset(->)(CB)`

`= underset~i + underset~j + underset~k + sqrt3 underset~i`

`= (sqrt3 + 1)underset~i + underset~j + underset~k`

`overset(->)(AC) = overset(->)(AO) + overset(->)(OC)`

`= −sqrt3underset~i + underset~i + underset~j + underset~k`

`= (1 – sqrt3)underset~i + underset~j + underset~k`

`overset(->)(AC) · overset(->)(OB)`	`= (1 + sqrt3)(1 – sqrt3) + 1 + 1`
	`= 1 – 3 + 1 + 1`
	`= 0`

`:. overset(->)(AC) ⊥ overset(->)(OB)`

`\|x\| + \|y\| + \|z\|`	` = \|x_underset~i\| + \|y_underset~j\| + \|z_underset~k\|`
	`≥ \|x_underset~i + y_underset~j\| + \|z_underset~k\|`
	`≥ \|x_underset~i + y_underset~j + z_underset~k\|`
	`≥ 1\ \ (\|overset->{OP}\| = \| x_underset~i + y_underset~j + z_underset~k \| = 1)`