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  • Explain solution RD Sharma class 12 chapter The Plane exercise 28.15 question 4 maths

Explain solution RD Sharma class 12 chapter The Plane exercise 28.15 question 4 maths

Answers (1)

Answer:   (1,2,1), 2 i+\frac{3}{2} \hat{\jmath}+\frac{3}{2} \hat{k}

  \vec{r}=3 i+\hat{\jmath}+2 \hat{k}+\lambda(2 \hat{\imath}-\hat{\jmath}+\hat{k})

Hint:

Point (3,1,2)
Suppose \vec{r}=x \hat{\imath}+y \hat{\jmath}+z \hat{k}

Given:

Find the image of the point with position vector 3 i+2 \hat{\jmath}+\hat{k} in the plane \vec{r} \cdot(2 \hat{\imath}+\hat{j}+\hat{k})=\mathrm{Y} .. Also find the position vectors of the foot of the perpendicular and the equation of the perpendicular line through  3 i+2 \hat{\jmath}+\hat{k} P(3,1,2)

                                                

Q(1,2,1)

Solution:

\begin{aligned} &\vec{r}=x \hat{\imath}+y \hat{\jmath}+z \hat{k} \\\\ &2 x-y+z=4 \\\\ &\frac{x-3}{2}=\frac{y-1}{-1}=\frac{z-2}{1} \end{aligned}

\begin{aligned} &=\frac{-2(6-1+2-4)}{(4+1+1)} \\\\ &=\frac{-2(3)}{6} \\\\ &=-1 \end{aligned}

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

\begin{aligned} &x=1 \\ &y=2 \\ &z=1 \end{aligned}

Point(1,2,1)

Passing (3,1,2), the point.


direction \rightarrow 2,-1,1

\begin{aligned} &\frac{x-3}{2}=\frac{y-1}{-1}=\frac{z-2}{1} \\\\ &\Rightarrow \vec{r}=3 \hat{\imath}+\hat{\jmath}+2 \hat{k} \\\\ &(2 \hat{\imath}+-\hat{\jmath}+\hat{k}) \end{aligned}

point (3,1,2)
point q=(1,2,1)
So the middle point =2+\frac{3}{2}+\frac{3}{2}
 

i.e, 2 i+\frac{3}{2} \hat{\jmath}+\frac{3}{2} k


 

 

Posted by

infoexpert26

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