Vectors: Question 3

Worked solution

Step 1: Compare the direction vectors

$l_1$ has direction $\mathbf{d_1} = \begin{pmatrix}4\\-2\\6\end{pmatrix}$ and $l_2$ has direction $\mathbf{d_2} = \begin{pmatrix}-6\\3\\-9\end{pmatrix}$.

Testing whether $\mathbf{d_2} = k\mathbf{d_1}$ for some constant $k$, using the first component: $-6 = 4k \implies k = -\dfrac{3}{2}$.

Checking this value of $k$ against the other two components: $-\frac{3}{2}\times(-2) = 3 \checkmark \qquad -\frac{3}{2}\times 6 = -9 \checkmark$

All three components agree with $k = -\dfrac{3}{2}$, so $\mathbf{d_2}$ is a scalar multiple of $\mathbf{d_1}$: the lines are parallel (not skew, and not intersecting at a single point, since parallel, non-identical lines never meet).

Step 2: Recompute the scalar multiple independently as a check

Working from the second and third components instead: $\dfrac{3}{-2} = -\dfrac{3}{2}$ and $\dfrac{-9}{6} = -\dfrac{3}{2}$, the same constant $k=-\dfrac32$ found both ways. This confirms $\mathbf{d_1}$ and $\mathbf{d_2}$ are genuinely parallel.

Step 3: Determine if the lines are the same line or distinct

Since the lines are parallel, they are either identical or never meet. Take the point $\begin{pmatrix}2\\1\\-1\end{pmatrix}$ on $l_1$ (at $s=0$) and test whether it lies on $l_2$:

$\begin{pmatrix}2\\1\\-1\end{pmatrix} = \begin{pmatrix}1\\-3\\2\end{pmatrix} + t\begin{pmatrix}-6\\3\\-9\end{pmatrix}$

• $x$: $1 - 6t = 2 \implies t = -\dfrac{1}{6}$
• $y$: $-3 + 3t = 1 \implies t = \dfrac{4}{3}$
• $z$: $2 - 9t = -1 \implies t = \dfrac{1}{3}$

The three components give three different values of $t$ ($-\tfrac16$, $\tfrac43$, $\tfrac13$), which is inconsistent. So the point $(2,1,-1)$ is not on $l_2$, confirming $l_1$ and $l_2$ are parallel but distinct lines, they never meet.

Why the other options are wrong

• A: would require every point of $l_1$ to also satisfy $l_2$‘s equation, but Step 3 shows even one point of $l_1$ fails to lie on $l_2$.
• C: parallel, non-identical lines cannot intersect at all, so “exactly one point of intersection” is impossible here.
• D: skew lines must have non-parallel direction vectors; Steps 1–2 show $\mathbf{d_1}$ and $\mathbf{d_2}$ are parallel, so “skew” is ruled out.

Final answer

$\boxed{\text{Parallel but distinct lines} \quad \text{(Option B)}}$