Let \(\vec{a} = \hat{\imath}-2\hat{\jmath}+3\hat{k}\) and \(\vec{b} = 5\hat{\imath}+2\hat{\jmath}-\hat{k}\) be the position vectors of the points A and B, and let L be the line \(\vec{r} = \vec{a}+\lambda(\vec{b}-\vec{a})\).
(A) \(\lambda = \frac{1}{2}\) gives the point (3, 0, 1)
(B) \(\lambda = 2\) gives the point (9, 6, −3)
(C) \(\lambda = -1\) gives a point lying between A and B
(D) The distance between the points given by \(\lambda = 0\) and \(\lambda = 3\) is \(12\sqrt{3}\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) and (D) only
The direction of L is
\(\vec{b}-\vec{a} = 4\hat{\imath}+4\hat{\jmath}-4\hat{k}\)
so the general point of L is
\((1+4\lambda,\ -2+4\lambda,\ 3-4\lambda)\)
In this two-point form, \(\lambda = 0\) gives A, \(\lambda = 1\) gives B, and only \(0 \lt \lambda \lt 1\) gives points between A and B.
\( \displaystyle \lambda = \frac{1}{2}:\ (3, 0, 1) \)
\(\lambda = 2:\ (9, 6, -5)\)
\(\lambda = -1:\ (-3, -6, 7)\)
The points for \(\lambda = 0\) and \(\lambda = 3\) differ by \(3(\vec{b}-\vec{a})\):
\(3\sqrt{16+16+16} = 3\times 4\sqrt{3} = 12\sqrt{3}\)
So \(\lambda = \frac{1}{2}\) gives (3, 0, 1), \(\lambda = 2\) gives (9, 6, −5), \(\lambda = -1\) gives (−3, −6, 7) beyond A, and the distance for \(\lambda = 0\) and 3 is \(12\sqrt{3}\).
Checking the statements
(A) \(\lambda = \frac{1}{2}\) gives the point (3, 0, 1) — Correct. \((1+2, -2+2, 3-2) = (3, 0, 1)\), the midpoint of AB.
(B) \(\lambda = 2\) gives the point (9, 6, −3) — Incorrect. The z-coordinate is \(3-8 = -5\).
(C) \(\lambda = -1\) gives a point lying between A and B — Incorrect. Points between A and B need \(0 \lt \lambda \lt 1\); \(\lambda = -1\) gives (−3, −6, 7), beyond A.
(D) The distance between the points given by \(\lambda = 0\) and \(\lambda = 3\) is \(12\sqrt{3}\) — Correct. It is \(3|\vec{b}-\vec{a}| = 3\times 4\sqrt{3}\).