2023TrialShort answer

Q13 (d) (iii)

2 marks

Question

$\ell_1$ and $\ell_2$ have a unique point of intersection when $a\ne2$ . Find the point of intersection in terms of $a$ .

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Solution

\begin{aligned} \left(\begin{matrix}2+2\lambda\\1+\lambda\\2+2\lambda\end{matrix}\right)\cdot \left(\begin{matrix}1\\2\\1\end{matrix}\right)&=0\\ 4+4\lambda+1+\lambda+2+2\lambda&=0\\ 9\lambda+7&=0\\ \lambda&=-\frac{7}{9} \end{aligned}

Equation of sphere is

\left|\vec r-\left(\begin{matrix}1\\0\\1\end{matrix}\right)\right|=|\overrightarrow{CP}|

\left|\vec r-\left(\begin{matrix}1\\0\\1\end{matrix}\right)\right|= \left|\left(\begin{matrix}4/9\\2/9\\-5/9\end{matrix}\right)\right|

\left|\vec r-\left(\begin{matrix}1\\0\\1\end{matrix}\right)\right|=\frac{\sqrt{5}}{3}

(x-1)^2+y^2+(z-1)^2=\frac{5}{9}

\begin{aligned} \begin{pmatrix}-1\\0\\3\end{pmatrix}+\lambda\begin{pmatrix}2\\1\\-1\end{pmatrix} &=\begin{pmatrix}0\\1\\2\end{pmatrix}+\mu\begin{pmatrix}a\\1\\-1\end{pmatrix}\\ -1+2\lambda&=a\mu \qquad (1)\\ \lambda&=1+\mu \qquad (2)\\ 3-\lambda&=2-\mu \qquad (3) \end{aligned}

Sub (2) into (1):

\begin{aligned} -1+2(1+\mu)&=a\mu\\ 1+2\mu&=a\mu\\ \therefore\quad \mu&=\frac{1}{a-2}\qquad(a\ne2) \end{aligned}

\begin{aligned} \therefore\ \text{The point of intersection} &=\left(\frac{a}{a-2},\ 1+\frac{1}{a-2},\ 2-\frac{1}{a-2}\right)\\ &=\left(\frac{a}{a-2},\frac{a-1}{a-2},\frac{2a-5}{a-2}\right) \end{aligned}

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