2023TrialShort answer

Q13 (d) (ii)

2 marks

Question

Another line, $\ell_2$ , is defined by the vector equation

\vec r=\begin{pmatrix}0\\1\\2\end{pmatrix}+\lambda\begin{pmatrix}a\\1\\-1\end{pmatrix},\qquad \lambda,a\in\mathbb{R}.

Find the possible values of $a$ when the angle between $\ell_1$ and $\ell_2$ is $\frac{\pi}{4}$ .

Try the question before revealing the answer.

Solution

\ell_2:\quad \vec r=\begin{pmatrix}0\\1\\2\end{pmatrix}+\lambda\begin{pmatrix}a\\1\\-1\end{pmatrix},\quad \lambda,a\in\mathbb R

\begin{pmatrix}2\\1\\-1\end{pmatrix}\cdot\begin{pmatrix}a\\1\\-1\end{pmatrix}=\left|\begin{pmatrix}2\\1\\-1\end{pmatrix}\right|\left|\begin{pmatrix}a\\1\\-1\end{pmatrix}\right|\cos\theta

Case I, $\theta=\frac\pi4$ :

\begin{aligned} 2a+1+1&=\sqrt6\sqrt{a^2+2}\cdot\frac1{\sqrt2}\\ 2a+2&=\sqrt{3(a^2+2)}\\ (2a+2)^2&=3(a^2+2)\\ a^2+8a-2&=0\\ \therefore a&=-4+3\sqrt2\qquad(\text{reject negative}) \end{aligned}

Case II, $\theta=\frac{3\pi}{4}$ :

\begin{aligned} 2a+1+1&=\sqrt6\sqrt{a^2+2}\cdot\left(-\frac1{\sqrt2}\right)\\ 2a+2&=-\sqrt{3(a^2+2)}\\ (2a+2)^2&=3(a^2+2)\\ a^2+8a-2&=0\\ \therefore a&=-4-3\sqrt2\qquad(\text{reject positive}) \end{aligned}

\therefore\quad a=-4\pm3\sqrt2

Supplied marking guidance · human reviewed

Compare this with your working, not only your final sentence.

Syllabus alignment

Vectors

Vectors and vector equations of linesUse coordinates and vector equations for curves, lines, line segments and spheres in two and three dimensions.
Further operations with three-dimensional vectorsUse vector magnitude, unit vectors and scalar products to solve problems and prove geometric results.

Spot a problem? Send feedback