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  • Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.15 question 10

Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.15 question 10

Answers (1)

Answer:   \frac{13}{12} \sqrt{6},\left(-\frac{1}{12}, \frac{25}{12},-\frac{2}{12}\right)

Hint:

For P as distance put

\begin{aligned} &\vec{r}=i+\hat{\jmath}+2 \hat{k} \text { in } \\\\ &\vec{r} \cdot(\hat{\imath}-2 \hat{\jmath}+4 \hat{k})+5=0 P(1,1,2) \end{aligned}

                                        

Given:

Find the length and foot of the perpendicular from the point (1,1,2) to the plane \vec{r} \cdot(\hat{\imath}-2 \hat{\jmath}+4 \hat{k})+5=0

Solution:

Distance  =\frac{(\hat{\imath}+\hat{\jmath}+2 \hat{k}) \cdot(\hat{\imath}-2 \hat{\jmath}+4 \hat{i})+5}{\sqrt{1^{2}+(-2)^{2}+4^{2}}}

                \begin{aligned} &=\frac{1-2+8+5}{\sqrt{1+4+16}} \\\\ &=\frac{12}{\sqrt{21}} \end{aligned}

Direction of  \vec{n}=(1,-2,4)
 

equation of line \frac{x-1}{1}=\frac{y-1}{-2}=\frac{3-2}{4}=\lambda(\text { say })

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

P u t\; \vec{r}=(\lambda+1) i+(-2 \lambda+1) \hat{\jmath}+(4 \lambda+2) \hat{k} \text { in } \vec{r} \cdot(\hat{\imath}-\hat{y}+4 \hat{k})+5=0

\begin{aligned} &{[(\lambda+1) i+(-2 \lambda+1) \hat{\jmath}+(4 \lambda+2) \hat{k}] \cdot[\hat{l}-\hat{y}+4 \hat{k}]+5=0} \\\\ &\lambda+1+4 \lambda-2+16 \lambda+8+5=0 \\\\ &21 \lambda+12=0 \end{aligned}

\begin{aligned} &\lambda=-\frac{12}{21} \\\\ &x=\frac{-4}{7}+1 \\\\ &=\frac{3}{7} \end{aligned}

\begin{aligned} &y=\frac{8}{7}+1 \\\\ &=\frac{15}{7} \\\\ &z=\frac{-16}{7}+2 \end{aligned}

\begin{aligned} &=-\frac{2}{7} \\\\ &\left(\frac{3}{7}, \frac{15}{7}, \frac{-2}{7}\right) \text { Ans. } \end{aligned}

Posted by

infoexpert26

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