Q202 / 270 · NDA 2025 (Phase 2) · Mathematics
For the following two (02) items:
A plane $P$ is parallel to the line having direction ratios $\langle 1, 3, 2 \rangle$ and contains the line of intersection of the planes $6x + 4y - 5z = 2$ and $x - 2y + 3z = 0$.
What is the equation of the plane $P$?
A plane $P$ is parallel to the line having direction ratios $\langle 1, 3, 2 \rangle$ and contains the line of intersection of the planes $6x + 4y - 5z = 2$ and $x - 2y + 3z = 0$.
What is the equation of the plane $P$?
- Correct answerA$2x - 20y + 29z + 2 = 0$
- B$2x - 20y + 29z - 2 = 0$
- C$2x + 3y + 2z - 4 = 0$
- D$x - 3y + 2z + 5 = 0$
Solution
The equation of the plane passing through the line of intersection of the given planes is
$(6x + 4y - 5z - 2) + \lambda(x - 2y + 3z) = 0$
$(6 + \lambda)x + (4 - 2\lambda)y + (-5 + 3\lambda)z - 2 = 0$
Since this plane is parallel to the line with direction ratios $\langle 1, 3, 2 \rangle$, the normal to the plane is perpendicular to the line. Thus,
$1(6 + \lambda) + 3(4 - 2\lambda) + 2(-5 + 3\lambda) = 0$
$6 + \lambda + 12 - 6\lambda - 10 + 6\lambda = 0$
$\lambda + 8 = 0 \Rightarrow \lambda = -8$
Substituting $\lambda = -8$ in the equation of the plane, we get
$(6 - 8)x + (4 - 2(-8))y + (-5 + 3(-8))z - 2 = 0$
$-2x + 20y - 29z - 2 = 0$
$2x - 20y + 29z + 2 = 0$
Answer: $2x - 20y + 29z + 2 = 0$
$(6x + 4y - 5z - 2) + \lambda(x - 2y + 3z) = 0$
$(6 + \lambda)x + (4 - 2\lambda)y + (-5 + 3\lambda)z - 2 = 0$
Since this plane is parallel to the line with direction ratios $\langle 1, 3, 2 \rangle$, the normal to the plane is perpendicular to the line. Thus,
$1(6 + \lambda) + 3(4 - 2\lambda) + 2(-5 + 3\lambda) = 0$
$6 + \lambda + 12 - 6\lambda - 10 + 6\lambda = 0$
$\lambda + 8 = 0 \Rightarrow \lambda = -8$
Substituting $\lambda = -8$ in the equation of the plane, we get
$(6 - 8)x + (4 - 2(-8))y + (-5 + 3(-8))z - 2 = 0$
$-2x + 20y - 29z - 2 = 0$
$2x - 20y + 29z + 2 = 0$
Answer: $2x - 20y + 29z + 2 = 0$
Hence, the correct answer is option: 'A'.