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  • Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.15 question 11

Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.15 question 11

Answers (1)

Answer:  (7,0,3)(-1,4,-1), \sqrt{6}

Hint:

\hat{N} \quad(3,2,1)

                                                    

If M is lying on the plane and the normal vector as well \hat{N}

(a x+b y+c z+d), \hat{N}=\frac{(a)}{\sqrt{a^{2}+b^{2}+i^{2}}} \hat{\imath}+\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}} \hat{\jmath}+\frac{c}{\sqrt{a^{2}+b^{2}} t^{2}} \hat{k}

Given:

Find the coordinates of the foot of the perpendicular and the perpendicular distance of the point P(3,2,1) from the plane 2x-y+z+1=0. Find also the image of the point in the plane

Solution:

\begin{aligned} &\hat{N}=\frac{2}{\sqrt{6}} \hat{\imath}-\frac{\hat{\jmath}}{\sqrt{6}}+\frac{\hat{k}}{\sqrt{6}}\\ &\hat{N} \end{aligned}

                                                    

\begin{aligned} &\therefore \frac{x-3}{\frac{2}{\sqrt{6}}}=\frac{y-2}{\left(-\frac{1}{\sqrt{6}}\right)}=\frac{3-1}{\left(\frac{1}{\sqrt{6}}\right)}=\lambda \\\\ &x=\frac{3+2 \lambda}{\sqrt{6}}, y=\frac{2-\lambda}{\sqrt{6}}, z=\frac{1+\lambda}{\sqrt{6}} \end{aligned}

\begin{aligned} &=2\left(3+\frac{2 \lambda}{\sqrt{6}}\right)-\left(2-\frac{\lambda}{\sqrt{6}}\right)+\left(1+\frac{\lambda}{\sqrt{6}}\right)+1=0 \\\\ &=6+\frac{4 \lambda}{\sqrt{6}}-2+\frac{\lambda}{\sqrt{6}}+2+\frac{\lambda}{\sqrt{6}}=0 \end{aligned}

\begin{aligned} &6+\frac{6}{\sqrt{6}}=0 \\\\ &\Rightarrow \lambda=-\frac{6 \sqrt{6}}{6}=-\sqrt{6} \\\\ &M[3-2,2+1,1-1] \end{aligned}

\begin{aligned} &\therefore M(1,3,0) \\ &\rightarrow \quad Q=\sqrt{2^{2}+1+1}=\sqrt{6} \\ &P(3,2,1) \end{aligned}

                                    

\begin{aligned} &\Rightarrow \frac{(3-a)}{(2 / \sqrt{6})}=\frac{(2-b)}{(-1 / \sqrt{6})}=\frac{(1-c)}{1 / \sqrt{6}}=2 \sqrt{6} \\\\ &\Rightarrow(3-a)=4 \text { or }-4 \\\\ &\Rightarrow-10 r+7 \end{aligned}

\begin{aligned} &\Rightarrow(2-b)=-2 \text { or } 2 \\\\ &\Rightarrow b=-4 \text { or } 0 \\\\ &\Rightarrow(1-c)=2 \text { or }-2 \end{aligned}

\begin{aligned} &\Rightarrow c=-1 \\\\ &p !(-1,4,-1) \text { or } \\\\ &p=7,0,3 \end{aligned}

 

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