Consider the line \(L: \vec{r} = (\hat{\imath}+2\hat{\jmath}-\hat{k})+\lambda(2\hat{\imath}-\hat{\jmath}+2\hat{k})\).
(A) L passes through the point (5, 0, 3)
(B) \(\displaystyle\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}\) are direction cosines of L
(C) The Cartesian equation of L is \(\displaystyle\frac{x+1}{2}=\frac{y+2}{-1}=\frac{z-1}{2}\)
(D) L is perpendicular to every line with direction ratios 1, 4, 1
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A), (B) and (D) only
L passes through (1, 2, −1) and has direction ratios 2, −1, 2. Its general point is
\((1+2\lambda,\ 2-\lambda,\ -1+2\lambda)\)
With \(\lambda = 2\) this is (5, 0, 3).
\(\sqrt{2^2+(-1)^2+2^2} = 3\)
so the direction cosines are \(\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}\). The Cartesian form through (1, 2, −1) is
\( \displaystyle \frac{x-1}{2}=\frac{y-2}{-1}=\frac{z+1}{2} \)
For direction ratios 1, 4, 1:
\((2)(1)+(-1)(4)+(2)(1) = 0\)
So L contains (5, 0, 3), has direction cosines \(\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}\), passes through (1, 2, −1), and is perpendicular to direction 1, 4, 1.
Checking the statements
(A) L passes through (5, 0, 3) — Correct. \(\lambda = 2\) gives (5, 0, 3).
(B) \(\frac{2}{3}, -\frac{1}{3}, \frac{2}{3}\) are direction cosines of L — Correct. They are 2, −1, 2 divided by 3.
(C) The Cartesian equation of L is \(\frac{x+1}{2}=\frac{y+2}{-1}=\frac{z-1}{2}\) — Incorrect. That line passes through (−1, −2, 1), not (1, 2, −1).
(D) L is perpendicular to every line with direction ratios 1, 4, 1 — Correct. \(2-4+2 = 0\).