Let \(\vec{a}\) and \(\vec{b}\) be non-zero vectors that are not parallel. The lines \(\vec{r} = \vec{a}+\lambda\vec{b}\) and \(\vec{r} = \vec{b}+\mu\vec{a}\) intersect at the point whose position vector is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\vec{a}+\vec{b}\)
At a common point the two expressions are equal:
\(\vec{a}+\lambda\vec{b} = \vec{b}+\mu\vec{a}\)
\((1-\mu)\vec{a}+(\lambda-1)\vec{b} = \vec{0}\)
Since \(\vec{a}\) and \(\vec{b}\) are not parallel, a combination of them is zero only when both coefficients are zero:
\(\mu = 1,\quad \lambda = 1\)
So the lines meet at \(\vec{a}+\vec{b}\), which is \(\lambda = 1\) on the first line and \(\mu = 1\) on the second.
Why the other options are incorrect
Option (1) \(\vec{a}-\vec{b}\) — Incorrect. It is on the first line (\(\lambda = -1\)), but \(\vec{b}+\mu\vec{a} = \vec{a}-\vec{b}\) would need \(2\vec{b} = (1-\mu)\vec{a}\), which is impossible.
Option (3) \(\vec{0}\) — Incorrect. \(\vec{a}+\lambda\vec{b} = \vec{0}\) would make \(\vec{a}\) parallel to \(\vec{b}\), so the origin is not on the first line.
Option (4) \(\vec{a}+2\vec{b}\) — Incorrect. It is on the first line (\(\lambda = 2\)), but on the second line the coefficient of \(\vec{b}\) is always 1, not 2.