Let $P$ be the plane containing the straight line $\frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5}$ and perpendicular to the plane containing the straight lines $\frac{ x }{2}=\frac{ y }{3}=\frac{ z }{5}$ and $\frac{ x }{3}=\frac{ y }{7}=\frac{ z }{8}$ If $d$ is the distance of $P$ from the point $(2,-5,11)$, then $d ^2$ is equal to :
- $\frac{147}{2}$
- 96
- $\frac{32}{3}$
- 54
The Correct Option is C
Solution and Explanation
The correct answer is option (C): \(\frac{32}{3}\)
a(x - 3)+b(y + 4)+c(z - 7) = 0
P : 9a - b -5c = 0
-11a - b + 5c = 0
After solving DR's ∝ (1, -1 , 2)
Equation of plane
x - y + 2z = 21
\(d=\frac{8}{\sqrt6}\)
\(d^2=\frac{32}{2}\)
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Concepts Used:
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When a plane intersects a cone in multiple sections, several types of curves are obtained. These curves can be a circle, an ellipse, a parabola, and a hyperbola. When a plane cuts the cone other than the vertex then the following situations may occur:
Let ‘β’ is the angle made by the plane with the vertical axis of the cone
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