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Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.15 question 6

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Answer:   1

Hint:

Let equation of line passing through P(1,-2,3)

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

Given:

Find the distance of the point (1,-2,3)  from the plane x-y+z=5  measured along a line parallel to \frac{x}{2}=\frac{y}{3}=\frac{z}{-6}

Solution:

Let the equation of line passing through P(1,-2,3) is

\begin{aligned} &\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{3-31}{c} \\\\ &\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-3}{-6}=\lambda \text { (say) } \\\\ &Q=(2 x+1,3 \lambda-2,-6 \lambda+3) \end{aligned}

Q Lies in the plane x-y+z=5

\begin{aligned} &2 \lambda+1-3 \lambda+2-6 \lambda+3=5 \\\\ &-7 \lambda=-1 \\\\ &\lambda=\frac{1}{7} \end{aligned}

\begin{aligned} Q &=\left(\frac{2}{7}+1, \frac{3}{7} \cdot-2,-\frac{6}{7}+3\right) \\\\ &=\left(\frac{9}{7}, \frac{-11}{7}, \frac{15}{7}\right) \end{aligned}

\begin{aligned} &P=(1,-2,3) \\\\ &P \theta=\sqrt{\left(\frac{9}{7}-1\right)^{2}+\left(-\frac{11}{7}+2\right)^{2}+\left(\frac{15}{7}-3\right)^{2}} \end{aligned}

       =\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{4.9}}

       \begin{aligned} &=\sqrt{\frac{4+9+36}{49}} \\\\ &=\sqrt{\frac{49}{49}} \\\\ &=1 \text { unit } A n s \end{aligned}


 

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