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Please solve RD Sharma class 12 chapter The Plane exercise 28.15 question 5 maths textbook solution

Answers (1)

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Answer: $\left(-\frac{1}{12}, \frac{25}{12},-\frac{1}{6}\right), \frac{13}{\sqrt{24}}$

Hint:

Let point

directional ratio are perpendicular to the line

Given:

Find the coordinate of the foot of the perpendicular from the point $(1,1,2)$ to the plane $2x-2y+4z+5=0$ . Also find the length of the perpendicular

Solution:

Direction ratio of plane $\vec{n}=2,-2,4$

equation of line $\frac{x-1}{2}=\frac{y-1}{-2}=\frac{z-2}{4}=\lambda(\text { say })$

$\begin{aligned} &x=2 \lambda+1 \\\\ &y=-2 \lambda+1, \\\\ &z=4 \lambda+2 \end{aligned}$

Put in $2 x-2 y+4 z+5=0$

$2(2 \lambda+1)-2(-2 \lambda+1)+4(4 \lambda+2)+5=0$

$\begin{aligned} &4 \lambda+2+4 \lambda-2+16 \lambda+8+5=0 \\\\ &24 \lambda+13=0 \\\\ &\lambda=\frac{-13}{24} \end{aligned}$

Putting the value of λ iner

$\begin{aligned} &x=2 \lambda+1 \\\\ &y=-2 \lambda+1 \\\\ &z=4 \lambda+2 \end{aligned}$

$\begin{aligned} &x=2\left(-\frac{13}{24}\right)+1 \\\\ &y=-2\left(\frac{-13}{24}\right)+1 \\\\ &z=4\left(\frac{-13}{24}\right)+2 \end{aligned}$

$\begin{aligned} &x=\frac{-13+12}{12} \\\\ &y=\frac{13+12}{12} \\\\ &z=\frac{-13+12}{6} \end{aligned}$

Distance $=\left(-\frac{1}{12}, \frac{25}{12},-\frac{1}{6}\right)$
Given point $(1,1,2)$

Distance $=\sqrt{\left(1+\frac{1}{12}\right)^{2}+\left(1-\frac{25}{12}\right)^{2}+\left(2+\frac{1}{6}\right)^{2}}$

$\begin{aligned} &=\sqrt{\frac{169}{144}+\frac{169}{144}+\frac{169}{36}} \\\\ &=13 \sqrt{\frac{1}{144}+\frac{1}{144}+\frac{1}{36}} \end{aligned}$

$\begin{aligned} &=\frac{13}{12} \sqrt{1+1+4} \\\\ &=\frac{13 \sqrt{6}}{12} \end{aligned}$

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