Equations of Lines — Question 4

Question 4

For a real parameter $k$, consider $$\begin{aligned} L_1&:\ \vec r=\left\langle 1,2,0\right\rangle+s\left\langle 2,-1,3\right\rangle,\\ L_2(k)&:\ \vec r=\left\langle 4,-2,7\right\rangle+t\left\langle 1,k,-1\right\rangle. \end{aligned}$$

Tasks

1. Find the value of $k$ for which the lines intersect.

2. Find the intersection point.

3. For that value of $k$, find the acute angle between the lines.

Question 4 – Solution

Strategy The $x$- and $z$-coordinate equations determine $s$ and $t$ without $k$; the $y$-coordinate then determines the required parameter.

Intersection Equating $x$- and $z$-coordinates gives $$1+2s=4+t,\qquad 3s=7-t.$$ The unique solution is $s=2$, $t=1$. The $y$-coordinates then require $$2-s=-2+kt\quad\Longrightarrow\quad 0=-2+k,$$ so $\boxed{k=2}$. Both lines meet at $\boxed{Q=(5,0,6)}$.

Angle The directions are $\vec d_1=\left\langle 2,-1,3\right\rangle$ and $\vec d_2=\left\langle 1,2,-1\right\rangle$. Thus $$\cos\theta=\frac{|\vec d_1\cdot\vec d_2|}{\|\vec d_1\|\|\vec d_2\|} =\frac{3}{\sqrt{14}\sqrt 6}=\frac{3}{2\sqrt{21}}.$$ Hence $\boxed{\theta=\cos^{-1}\!\left(\frac{3}{2\sqrt{21}}\right)\approx 70.9^\circ}$.

Verification Substitution of $s=2$, $t=1$, and $k=2$ produces $(5,0,6)$ in both equations.