The vector equation of the line through the point P(2, 2, −1) which meets the line \(\displaystyle\frac{x+1}{2}=\frac{y-3}{-1}=\frac{z+2}{2}\) at right angles is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\vec{r} = (2\hat{\imath}+2\hat{\jmath}-\hat{k})+\mu(\hat{\imath}-\hat{k})\)
The required line joins P to the foot of the perpendicular F on the given line. A general point of the given line is
\(F(-1+2\lambda,\ 3-\lambda,\ -2+2\lambda)\)
\(\overrightarrow{PF} = (-3+2\lambda)\hat{\imath}+(1-\lambda)\hat{\jmath}+(-1+2\lambda)\hat{k}\)
PF must be perpendicular to the direction 2, −1, 2:
\(2(-3+2\lambda)-(1-\lambda)+2(-1+2\lambda) = 0\)
\(9\lambda-9 = 0\)
\(\lambda = 1\)
So F is (1, 2, 0) and
\(\overrightarrow{PF} = -\hat{\imath}+0\hat{\jmath}+\hat{k}\)
The line through P along PF (or along \(\hat{\imath}-\hat{k}\), the opposite direction) is
\(\vec{r} = (2\hat{\imath}+2\hat{\jmath}-\hat{k})+\mu(\hat{\imath}-\hat{k})\)
Why the other options are incorrect
Option (1) \(\vec{r} = (2\hat{\imath}+2\hat{\jmath}-\hat{k})+\mu(2\hat{\imath}-\hat{\jmath}+2\hat{k})\) — Incorrect. This is parallel to the given line, so it never meets it.
Option (2) \(\vec{r} = (2\hat{\imath}+2\hat{\jmath}-\hat{k})+\mu(3\hat{\imath}-\hat{\jmath}+\hat{k})\) — Incorrect. This joins P to the point (−1, 3, −2) of the given line, but \(2(3)-1(-1)+2(1) = 9 \ne 0\), so it does not meet the line at right angles.
Option (4) \(\vec{r} = (2\hat{\imath}+2\hat{\jmath}-\hat{k})+\mu(\hat{\imath}+2\hat{\jmath})\) — Incorrect. Its direction is perpendicular to the given line (\(2-2+0 = 0\)), but every point of it has z = −1, and the given line has z = −1 only at \(\left(0, \frac{5}{2}, -1\right)\); there x = 0 needs \(\mu = -2\), giving y = −2, so the two lines never meet.