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Vectors: NYJC 2024 Promo Q5

Nanyang Junior College2024 PromoPromo Q5●●○ Standard8 marks

What this question tests

Vectors, lines and planes.

Question

The position vectors of points AA, BB and CC, relative to the origin OO are a\mathbf{a}, b\mathbf{b} and c\mathbf{c} respectively, where a=3i+2j−k,b=4i−3j+2k,c=3i−j−k.\mathbf{a} = 3\mathbf{i} + 2\mathbf{j} - \mathbf{k}, \qquad \mathbf{b} = 4\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}, \qquad \mathbf{c} = 3\mathbf{i} - \mathbf{j} - \mathbf{k}.

  1. Find a×b\mathbf{a} \times \mathbf{b} and deduce the exact area of triangle OABOAB. [3]

  2. Hence, find the volume of the tetrahedron OABCOABC. [3]

    [The volume of a tetrahedron is 13×base area×perpendicular height.]\left[\text{The volume of a tetrahedron is } \dfrac{1}{3} \times \text{base area} \times \text{perpendicular height}.\right]

  3. Find the perpendicular distance from BB to line OAOA. [2]

Show full worked solution▾

(a)

a×b=(32−1)×(4−32)=(1−10−17)\begin{aligned} \mathbf a\times\mathbf b&=\begin{pmatrix}3\\2\\-1\end{pmatrix}\times \begin{pmatrix}4\\-3\\2\end{pmatrix}=\begin{pmatrix}1\\-10\\-17\end{pmatrix} \end{aligned} Area of △AOB\triangle AOB =12∣a×b∣=12∣(1−10−17)∣=121+100+289=12390 units2\begin{aligned} &=\frac12\lvert\mathbf a\times\mathbf b\rvert\\ &=\frac12\left\lvert\begin{pmatrix}1\\-10\\-17\end{pmatrix}\right\rvert\\ &=\frac12\sqrt{1+100+289}\\ &=\frac12\sqrt{390}\text{ units}^2 \end{aligned}

(b)

Method 1:

Perpendicular ht of tetrahedron = 11+100+289OC→⋅(1−10−17)=1390(3−1−1)⋅(1−10−17)=30390\begin{aligned} \frac{1}{\sqrt{1+100+289}}\overrightarrow{OC}\mathbin{\boldsymbol\cdot} \begin{pmatrix}1\\-10\\-17\end{pmatrix} &=\frac{1}{\sqrt{390}}\begin{pmatrix}3\\-1\\-1\end{pmatrix}\mathbin{\boldsymbol\cdot} \begin{pmatrix}1\\-10\\-17\end{pmatrix}\\ &=\frac{30}{\sqrt{390}} \end{aligned} Vol of tetrahedron OABC=13(3902)(30390)=5 units3\text{Vol of tetrahedron }OABC=\frac13\left(\frac{\sqrt{390}}2\right) \left(\frac{30}{\sqrt{390}}\right)=5\text{ units}^3

Method 2

Let XX be the foot of perpendicular from CC to plane OABOAB. OX→=(3−1−1)+μ(1−10−17), for some μ∈R\overrightarrow{OX}=\begin{pmatrix}3\\-1\\-1\end{pmatrix} +\mu\begin{pmatrix}1\\-10\\-17\end{pmatrix},\text{ for some }\mu\in\mathbb R Equation of plane OABOAB: r⋅(1−10−17)=0\mathbf r\mathbin{\boldsymbol\cdot}\begin{pmatrix}1\\-10\\-17\end{pmatrix}=0 Since the point XX also lies on the plane OABOAB, ((3−1−1)+μ(1−10−17))⋅(1−10−17)=03+μ+10+100μ+17+289μ=0390μ=−30μ=−113\begin{aligned} \left(\begin{pmatrix}3\\-1\\-1\end{pmatrix}+\mu\begin{pmatrix}1\\-10\\-17\end{pmatrix}\right) \mathbin{\boldsymbol\cdot}\begin{pmatrix}1\\-10\\-17\end{pmatrix}&=0\\ 3+\mu+10+100\mu+17+289\mu&=0\\ 390\mu&=-30\\ \mu&=-\frac1{13} \end{aligned} CX→=OX→−OC→=(3−1−1)−113(1−10−17)−(3−1−1)=−113(1−10−17)\begin{aligned} \overrightarrow{CX}&=\overrightarrow{OX}-\overrightarrow{OC}\\ &=\begin{pmatrix}3\\-1\\-1\end{pmatrix}-\frac1{13}\begin{pmatrix}1\\-10\\-17\end{pmatrix} -\begin{pmatrix}3\\-1\\-1\end{pmatrix}\\ &=-\frac1{13}\begin{pmatrix}1\\-10\\-17\end{pmatrix} \end{aligned} perpendicular height, CX=1131+(−10)2+(−17)2=39013=30390\begin{aligned} \text{perpendicular height, }CX&=\frac1{13}\sqrt{1+(-10)^2+(-17)^2}\\ &=\frac{\sqrt{390}}{13}\\&=\frac{30}{\sqrt{390}} \end{aligned}

(c)

Method 1

Perpendicular distance from BB to line OAOA =∣b×a^∣=3909+4+1=1957or 5.28\begin{aligned} &=\lvert\mathbf b\times\hat{\mathbf a}\rvert\\ &=\frac{\sqrt{390}}{\sqrt{9+4+1}}\\ &=\sqrt{\frac{195}{7}}\quad\text{or }5.28 \end{aligned}

Method 2

Area of OAB=12390=12OA(h)\text{Area of }OAB=\frac12\sqrt{390}=\frac12OA(h) Thus, perpendicular distance from BB to line OAOA h=39014=1957.h=\frac{\sqrt{390}}{\sqrt{14}}=\sqrt{\frac{195}{7}}.

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