The point on the line \(\vec{r} = (\hat{\imath}+2\hat{k})+\lambda(\hat{\imath}+\hat{\jmath}-\hat{k})\) which is equidistant from the points A(3, 0, 2) and B(1, 2, 4) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (0, −1, 3)
Let P be the general point of the line:
\(P(1+\lambda,\ \lambda,\ 2-\lambda)\)
The squares of the distances from A and B are
\(PA^2 = (\lambda-2)^2+\lambda^2+\lambda^2 = 3\lambda^2-4\lambda+4\)
\(PB^2 = \lambda^2+(\lambda-2)^2+(\lambda+2)^2 = 3\lambda^2+8\)
Setting \(PA^2 = PB^2\):
\(-4\lambda+4 = 8\)
\(\lambda = -1\)
So P is \((0, -1, 3)\). Check: \(PA^2 = 9+1+1 = 11\) and \(PB^2 = 1+9+1 = 11\).
Why the other options are incorrect
Option (1) (2, 1, 3) — Incorrect. This is the midpoint of AB, so it is equidistant from A and B, but it is not on the line: x = 2 needs \(\lambda = 1\), which gives z = 1, not 3.
Option (3) (2, 1, 1) — Incorrect. It is on the line (\(\lambda = 1\)), but \(PA^2 = 3\) and \(PB^2 = 11\).
Option (4) (1, 0, 2) — Incorrect. It is on the line (\(\lambda = 0\)), but \(PA^2 = 4\) and \(PB^2 = 8\).