Vectors: Question 4

Worked solution

Part (a): Showing $l_1$ and $l_2$ are not parallel

The direction vectors are $\mathbf{d_1} = \begin{pmatrix}2\\-1\\1\end{pmatrix}$ and $\mathbf{d_2} = \begin{pmatrix}1\\1\\-2\end{pmatrix}$.

If the lines were parallel, $\mathbf{d_1} = k\mathbf{d_2}$ for some constant $k$. From the first component, $2 = k(1) \implies k=2$. Checking against the second component: $k(1) = 2$, but the second component of $\mathbf{d_1}$ is $-1 \ne 2$.

Check independently using a different pair: from the first and third components, $k=2$ (first) but $k(-2) = 1 \implies k = -\dfrac12$ (third), these disagree too. Since no single constant $k$ works consistently, $\mathbf{d_1}$ and $\mathbf{d_2}$ are not scalar multiples of each other, so $l_1$ and $l_2$ are not parallel.

Part (b): Showing $l_1$ and $l_2$ intersect, and finding the point

Writing out each line in components: $l_1: \; x=1+2s,\;\; y=2-s,\;\; z=-3+s \qquad\qquad l_2: \; x=1+t,\;\; y=-1+t,\;\; z=2-2t$

Setting the $x$- and $y$-components equal to each other: $1+2s = 1+t \implies t = 2s \qquad (1)$ $2-s = -1+t \implies t = 3-s \qquad (2)$

Substituting (1) into (2): $2s = 3-s \implies 3s = 3 \implies s = 1, \quad t = 2$

Check the third (z) component is also consistent: $z_{l_1} = -3+s = -3+1 = -2, \qquad z_{l_2} = 2-2t = 2-2(2) = -2$

Both give $z=-2$, so all three components agree at $s=1$, $t=2$. The lines genuinely intersect (this is not guaranteed in 3D, so checking the third equation is essential).

Point of intersection: substituting $s=1$ into $l_1$: $\mathbf{r} = \begin{pmatrix}1+2(1)\\2-1\\-3+1\end{pmatrix} = \begin{pmatrix}3\\1\\-2\end{pmatrix}$

Check using $l_2$ with $t=2$: $\begin{pmatrix}1+2\\-1+2\\2-2(2)\end{pmatrix} = \begin{pmatrix}3\\1\\-2\end{pmatrix}$, the same point, confirming the intersection.

Part (c): Acute angle between $l_1$ and $l_2$

Scalar product of the direction vectors: $\mathbf{d_1}\cdot\mathbf{d_2} = (2)(1) + (-1)(1) + (1)(-2) = 2 - 1 - 2 = -1$

Magnitudes: $|\mathbf{d_1}| = \sqrt{2^2+(-1)^2+1^2} = \sqrt{6}, \qquad |\mathbf{d_2}| = \sqrt{1^2+1^2+(-2)^2} = \sqrt{6}$

Since $\mathbf{d_1}\cdot\mathbf{d_2}$ is negative, the angle between the vectors is obtuse; the angle between the lines is defined to be the acute angle, so we use the modulus of the scalar product:

$\cos\theta = \frac{|\mathbf{d_1}\cdot\mathbf{d_2}|}{|\mathbf{d_1}||\mathbf{d_2}|} = \frac{1}{\sqrt6\times\sqrt6} = \frac{1}{6}$

$\theta = \cos^{-1}\left(\frac16\right) \approx 80.4^\circ$

Check by recomputing the scalar product in a different order: $(-1)(1) + (1)(-2) + (2)(1) = -1-2+2 = -1$ (matches. Also $|\mathbf{d_1}|^2|\mathbf{d_2}|^2 = 6\times6=36$, so $|\mathbf{d_1}||\mathbf{d_2}|=6$, giving $\cos\theta = \tfrac16 \approx 0.1667$ and $\theta \approx 80.4^\circ$ again) the same result. (Note $\cos^{-1}(-\tfrac16)\approx99.6^\circ$ is the obtuse angle between the direction vectors as written; the acute angle between the lines is its supplement, $80.4^\circ$.)

Final answers

• (a) $\mathbf{d_1}$ and $\mathbf{d_2}$ are not scalar multiples, so $l_1 \nparallel l_2$.
• (b) $l_1$ and $l_2$ intersect at $\begin{pmatrix}3\\1\\-2\end{pmatrix}$
• (c) Acute angle between $l_1$ and $l_2 \approx 80.4^\circ$