Vectors: TJC 2024 Promo Q11

(a)

$$\overrightarrow{AB} =\begin{pmatrix}0\\k\\0\end{pmatrix}-\begin{pmatrix}-2\\0\\-1\end{pmatrix} =\begin{pmatrix}2\\k\\1\end{pmatrix}$$ Since $l_1$ makes an angle of $\dfrac\pi3$ with $l_2$, we have $\overrightarrow{AB}$ making an acute angle of $\dfrac\pi3$ with line parallel to $\begin{pmatrix}3\\0\\4\end{pmatrix}$. $$\begin{aligned} \therefore\ \cos\frac\pi3 &=\frac{\begin{pmatrix}2\\k\\1\end{pmatrix}\cdot\begin{pmatrix}3\\0\\4\end{pmatrix}} {\sqrt{2^2+k^2+1^2}\sqrt{3^2+4^2}}\\ \frac12&=\frac{10}{\sqrt{5+k^2}(5)}\\ \sqrt{5+k^2}&=4\\ 5+k^2&=16\\ k^2&=11\\ k&=\sqrt{11}\text{ or }-\sqrt{11}\ (\text{rejected since }k\text{ is +ve}) \end{aligned}$$

Q11(b)

Method 1:

Since $N$ is a point on $l_1$, we have $$\overrightarrow{ON}=\begin{pmatrix}-2\\0\\-1\end{pmatrix} +\lambda\begin{pmatrix}3\\0\\4\end{pmatrix} =\begin{pmatrix}-2+3\lambda\\0\\-1+4\lambda\end{pmatrix}.$$ $$\therefore\ \overrightarrow{BN} =\begin{pmatrix}-2+3\lambda\\0\\-1+4\lambda\end{pmatrix} -\begin{pmatrix}0\\\sqrt{11}\\0\end{pmatrix} =\begin{pmatrix}-2+3\lambda\\-\sqrt{11}\\-1+4\lambda\end{pmatrix}.$$ Since $BN$ is perpendicular to $l_1$, we have $\overrightarrow{BN}\cdot\begin{pmatrix}3\\0\\4\end{pmatrix}=0$ $$\begin{aligned} &\Longrightarrow \begin{pmatrix}-2+3\lambda\\-\sqrt{11}\\-1+4\lambda\end{pmatrix} \cdot\begin{pmatrix}3\\0\\4\end{pmatrix}=0\\ &\Longrightarrow-6+9\lambda-4+16\lambda=0\\ &\Longrightarrow\lambda=\frac25 \end{aligned}$$

$$\Longrightarrow\ \overrightarrow{ON} =\begin{pmatrix}-2+3(\frac25)\\0\\-1+4(\frac25)\end{pmatrix} =\begin{pmatrix}-\frac45\\0\\\frac35\end{pmatrix}$$

11(c)

Method 1: $$\overrightarrow{BN}=\begin{pmatrix}-\frac45\\-\sqrt{11}\\\frac35\end{pmatrix} \Longrightarrow |\overrightarrow{BN}|=\sqrt{\left(-\frac45\right)^2+(-\sqrt{11})^2+\left(\frac35\right)^2} =\sqrt{12}$$ $$\overrightarrow{AN} =\begin{pmatrix}-\frac45\\0\\\frac35\end{pmatrix} -\begin{pmatrix}-2\\0\\-1\end{pmatrix} =\begin{pmatrix}\frac65\\0\\\frac85\end{pmatrix} \Longrightarrow |\overrightarrow{AN}|=\sqrt{\left(\frac65\right)^2+\left(\frac85\right)^2}=2$$ $$\text{Area of triangle }ABN =\frac12|\overrightarrow{AN}||\overrightarrow{BN}| =\frac12(2)(\sqrt{12})=2\sqrt3$$

Method 2: $$\text{Area of triangle }ABN =\frac12|\overrightarrow{AN}\times\overrightarrow{BN}| =\frac12\left| \begin{pmatrix}\frac65\\0\\\frac85\end{pmatrix} \times \begin{pmatrix}-\frac45\\-\sqrt{11}\\\frac35\end{pmatrix} \right|$$ $$=\frac12\left| \begin{pmatrix}\frac{8\sqrt{11}}5\\2\\-\frac{6\sqrt{11}}5\end{pmatrix} \right| =\frac12\sqrt{\left(\frac{8\sqrt{11}}5\right)^2+2^2+ \left(-\frac{6\sqrt{11}}5\right)^2}=2\sqrt3$$