Consider the line \(L: \displaystyle\frac{x-1}{2}=\frac{y+3}{0}=\frac{z-2}{-1}\).
(A) Every point of L has y = 3
(B) L is perpendicular to the y-axis
(C) L passes through the point (5, −3, 1)
(D) L is parallel to the line \(\vec{r} = \hat{\imath}+\lambda(4\hat{\imath}-2\hat{k})\)
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (B) and (D) only
A zero denominator means that the matching numerator is zero for every point of the line, so \(y+3 = 0\). With the common value \(\lambda\), the general point is
\((1+2\lambda,\ -3,\ 2-\lambda)\)
The direction ratios are 2, 0, −1. With the y-axis (0, 1, 0):
\((2)(0)+(0)(1)+(-1)(0) = 0\)
For the point with x = 5: \(1+2\lambda = 5\) gives \(\lambda = 2\), and then \(z = 2-2 = 0\).
The line in (D) has direction ratios 4, 0, −2, which is 2 times 2, 0, −1.
So every point of L has y = −3, L is perpendicular to the y-axis, its point with x = 5 is (5, −3, 0), and its direction 2, 0, −1 is parallel to 4, 0, −2.
Checking the statements
(A) Every point of L has y = 3 — Incorrect. \(y+3 = 0\) gives y = −3.
(B) L is perpendicular to the y-axis — Correct. Its direction ratio along y is 0, so the dot product with (0, 1, 0) is 0.
(C) L passes through the point (5, −3, 1) — Incorrect. x = 5 gives \(\lambda = 2\) and z = 0, not 1.
(D) L is parallel to the line \(\vec{r} = \hat{\imath}+\lambda(4\hat{\imath}-2\hat{k})\) — Correct. 4, 0, −2 is twice 2, 0, −1.