Q206 / 270 · NDA 2026 (Phase 1) · Mathematics
Passage: A line $L$ passing through the point $(-1,2,-3)$ is perpendicular to the plane $P$ given by $2x+3y+z+5=0$.
Question: What are the direction ratios of a line $M$ parallel to the plane $P$ ?
Question: What are the direction ratios of a line $M$ parallel to the plane $P$ ?
- A$\langle-3,2,1\rangle$
- B$\langle3,2,-6\rangle$
- C$\langle1,3,2\rangle$
- Correct answerD$\langle2,2,-10\rangle$
Solution
The plane $P$ is given by $2x + 3y + z + 5 = 0$.
The direction ratios of the normal to the plane $P$ are $\langle 2, 3, 1 \rangle$.
Since the line $M$ is parallel to the plane $P$, its direction vector must be perpendicular to the normal vector of the plane.
Let the direction ratios of the line $M$ be $\langle a, b, c \rangle$.
Then, $2a + 3b + c = 0$.
Checking the given options:
For $\langle -3, 2, 1 \rangle$, $2(-3) + 3(2) + 1 = 1 \neq 0$
For $\langle 3, 2, -6 \rangle$, $2(3) + 3(2) - 6 = 6 \neq 0$
For $\langle 1, 3, 2 \rangle$, $2(1) + 3(3) + 2 = 13 \neq 0$
For $\langle 2, 2, -10 \rangle$, $2(2) + 3(2) - 10 = 0$
Thus, the direction ratios of the line $M$ can be $\langle 2, 2, -10 \rangle$.
Answer: $\langle2,2,-10\rangle$
The direction ratios of the normal to the plane $P$ are $\langle 2, 3, 1 \rangle$.
Since the line $M$ is parallel to the plane $P$, its direction vector must be perpendicular to the normal vector of the plane.
Let the direction ratios of the line $M$ be $\langle a, b, c \rangle$.
Then, $2a + 3b + c = 0$.
Checking the given options:
For $\langle -3, 2, 1 \rangle$, $2(-3) + 3(2) + 1 = 1 \neq 0$
For $\langle 3, 2, -6 \rangle$, $2(3) + 3(2) - 6 = 6 \neq 0$
For $\langle 1, 3, 2 \rangle$, $2(1) + 3(3) + 2 = 13 \neq 0$
For $\langle 2, 2, -10 \rangle$, $2(2) + 3(2) - 10 = 0$
Thus, the direction ratios of the line $M$ can be $\langle 2, 2, -10 \rangle$.
Answer: $\langle2,2,-10\rangle$
Hence, the correct answer is option: 'D'.