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  • Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.15 question 15

Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.15 question 15

Answers (1)

best_answer

Answer: \left(3, \frac{7}{2}, \frac{11}{2}\right),(4,4,7)

Hint:

Distance \rightarrow \sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}+\left(z_{1}-z_{2}\right)^{2}}

\vec{d}=\vec{r} \cdot \vec{n}

                            

Given:

Find the position vector of the foot of the perpendicular and the perpendicular distance from the point P with position vector 2 \tilde{i}+3 \hat{\jmath}+4 \hat{k}  to the plane  \vec{r} \cdot(\hat{\imath}+\hat{\jmath}+3 \hat{k})-26=0 . Also find the image of P in the plane

Solution:

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

\begin{aligned} &\Rightarrow[(2+2 t) \hat{i}+(3+t) \hat{j}+(4+3 t) \hat{k}][2 \hat{i}+\hat{j}+3 \hat{k}]=26 \\\\ &\Rightarrow 19+14 t=26 \\\\ &t=\frac{1}{2} \end{aligned}

Putting the value of t in equation 1
i.e,

\begin{aligned} & x=3, \\\\ &y=\frac{7}{2}, \\\\ &z=\frac{11}{2} \end{aligned}

\mathrm{PQ} \Rightarrow(3-2)^{2}+\left(\frac{7}{2}-3\right)^{2}+\left(\frac{11}{2}-2\right)^{2}

        \begin{aligned} &=\sqrt{1+\left(\frac{1}{2}\right)^{2}+\left(\frac{11}{2}\right)^{2}}, \\\\ &=\sqrt{1+\frac{1}{4}+\frac{9}{4}} \end{aligned}

        \begin{aligned} &=\sqrt{\frac{14}{4}} \\\\ &=\frac{\sqrt{14}}{2} \end{aligned}

Image ,

x=2+2(1)=4

\begin{aligned} &y=3+1=4 \\\\ &z=4+3(1)=7 \end{aligned}

Image (4,4,7) Ans
Foot \left(3, \frac{7}{2}, \frac{11}{2}\right) Ans


 

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infoexpert26

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