The two adjacent faces of a cuboid are represented by planes $P_{1} : 2 x - 3 {sin}^{2} θ \cdot y + z = - 10$ and $P_{2} : 3 x + 4 y + 12 {cos}^{2} θ \cdot z = - 10$ . If there is a line L which is perpendicular to the line of intersection of planes $P_{1}$ and $P_{2}$ and whose one point $A (x_{1}, y_{1}, \frac{10}{3})$ lies on $P_{1}$ and two points $B (x_{2}, y_{2}, z_{2})$ and $C (x_{3}, y_{3}, z_{3})$ lies on plane $P_{2}$ . If equation of L is $\frac{x - x_{1}}{l} = \frac{y - y_{1}}{- m} = \frac{z - \frac{10}{3}}{n}; (l, m, n \in N)$ , then which of the following is /are TRUE

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

${(l + m + n)}_{min} = 21$

$3 x_{1} + y_{1} = - 20$

$| \frac{x_{3} - x_{2}}{z_{3} - z_{2}} | = 2$

line of intersection of $P_{1}$ and $P_{2}$ is $\frac{x + \frac{250}{59}}{43} = \frac{y - \frac{40}{59}}{12} = \frac{z - 0}{- 59}$

answer is A, B, C, D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$P_{1} : 8 x - 9 y + 4 z = - 40$
$P_{2} : 3 x + 4 y + 3 z = - 10$
Line L must lie in $P_{2}$ and perpendicular to $P_{1}$
$∴ A (x_{1}, y_{1}, \frac{10}{3})$ must lie in both the plane
Question Image
$∴ x_{1} = - \frac{20}{3}, y_{1} = 0, z_{1} = \frac{10}{3}$