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  • Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.15 question 14

Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.15 question 14

Answers (1)

best_answer

Answer: \left(0, \frac{5}{2}, 0\right) ; \sqrt{6}

Hint:

Direction ratio on (2,-2,4)

Equation of line =\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}

                            

                                2 x-2 y+4 z-6=0

Given:

Find the length and foot of the perpendicular from the point (1,3/2,2) to the plane

2x-2y+4z-6=0

Solution:

2x-2y+4z+5=0

\begin{aligned} &D R^{\prime} \Delta=(2,-2,4) \\\\ &A B D R^{\prime} S=(2,-2,4) \end{aligned}

\begin{aligned} &\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{3-3}{c} \\\\ &\frac{x-1}{2}=\frac{y-3 / 2}{-2}=\frac{2-2}{4}=1 \end{aligned}

\begin{aligned} &x=2 \lambda+1 \\\\ &y=-2 \lambda+\frac{3}{2} \\\\ &z=4 \lambda+2 \end{aligned}

\begin{aligned} &\Rightarrow 2(2 \lambda+1)-2\left(-2 \lambda+\frac{3}{2}\right)+4(4 \lambda+2)+5=0 \\\\ &\Rightarrow 4 \lambda+2+4 \lambda-3+16 \lambda+8+5=0 \end{aligned}

\begin{aligned} &24 \lambda=-12 \\\\ &\lambda=-\frac{12}{24} \\\\ &=-\frac{1}{2} \end{aligned}

Foot of the perpendicular,

\begin{aligned} &x=2 \lambda+1 \\\\ &=2\left(\frac{-1}{2}\right)+1=0 \\\\ &y=-2\left(\frac{-1}{2}\right)+\frac{3}{2}=\frac{5}{2} \\\\ &z=4\left(\frac{-1}{2}\right)+2=0 \end{aligned}

Foot of the perpendicular \left(0, \frac{5}{2}, 0\right)

 

A B=\sqrt{1+1+4}=\sqrt{6}

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