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Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.15 question 7

Answers (1)

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Answer: $(5.2,6) ; \sqrt{11}$

Hint:

Let $M$ be the foot of the perpendicular of the point $P(2,3,7)$ in the plane $3x-y-z=7$

Given:

Find the coordinate of foot of the perpendicular from the point $(2,3,7)$ to the plane

$3x-y-z=7$ . Also find the length of the perpendicular.

Solution:

Let $M$ be the foot of the perpendicular of the point $P (2,3,7)$ in the plane $3x-y-z=7$ then $PM$ is normal to the plane so the direction ratio of $PM$ are proportional to $3,-1,-1$

Let the coordination of $M$ be $(3r+2,-r+3,-r+7)$

Since $M$ lies in the plane $3x-y-z=7,$

$9r+6+r-3+r-7=7$

$11r=11$

$r=1$

Substituting this in coordinates of $M$ ,

$M=(3 r+4,-r+3,-r+\lambda)=(5,2,6)$

Now length of the perpendicular from $P$ onto the plane

$\begin{aligned} &=\left|\frac{3(2)-3-7-7}{\sqrt{9+1+1}}\right| \\\\ &=2 \frac{11}{\sqrt{11}} \end{aligned}$

$\frac{\sqrt{11} \times \sqrt{11}}{\sqrt{11}}=\sqrt{11} \; \mathrm{Ans}$

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