The line $l_1$ has equation $x - 1 = 2 - y = -z$. The line $l_2$ passes through the point \(A(

Question

The line $l_1$ has equation $x - 1 = 2 - y = -z$. The line $l_2$ passes through the point $A(-2, 1, 7)$ and is parallel to the vector $-3\mathbf{i} + \mathbf{j} + 5\mathbf{k}$. Write down vector equations for $l_1$ and $l_2$.[2] Show that the lines $l_1$ and $l_2$ intersect and find the coordinates of $E$, the point of intersection of $l_1$ and $l_2$.[3] The acute angle between the lines $l_1$ and $l_2$ is denoted by $\theta$. By finding $\cos \theta$, show that $$\sin \theta = \frac{2\sqrt{2}}{\sqrt{35}}.$$ Hence find the shortest distance from $A$ to the line $l_1$, leaving your answer in exact form.[5]

Accepted Answer

(i) $l_1:\ \mathbf r=\begin{pmatrix}1\2\0\end{pmatrix}+\lambda\begin{pmatrix}1\-1\-1\end{pmatrix}$, $l_2:\ \mathbf r=\begin{pmatrix}-2\1\7\end{pmatrix}+\mu\begin{pmatrix}-3\1\5\end{pmatrix}$ (ii) $E=(4,-1,-3)$ (iii) $4\sqrt2$ units

Vectors - Angles Between Two Lines

Vectors - Angles Between Two Lines

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