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  • Please solve RD Sharma class 12 chapter The Plane exercise 28.15 question 9 maths textbook solution

Please solve RD Sharma class 12 chapter The Plane exercise 28.15 question 9 maths textbook solution

Answers (1)

Answer:  13

Hint:

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

The coordinate of the point corresponding to the position vector

Given:

Find the distance of the point with position vector -\hat{\imath}-5 \hat{\jmath}-10 \hat{k}  from the point of intersection of the line \vec{r}=(2 \hat{\imath}-\hat{\jmath}+2 \hat{k})+\lambda(3 i+4 \hat{\jmath}+12 \hat{k}) with the plane\vec{r} \cdot(\hat{\imath}-j+k)=5

Solution:

The given equation of the line is

\begin{aligned} &\vec{r}=(2 \hat{\imath}-\hat{\jmath}+2 \hat{k})+\lambda(3 \hat{\imath}+4 \hat{\jmath}+12 \hat{k}) \\\\ &\Rightarrow \vec{r}=(2+3 \lambda) \hat{\imath}+(-1+4 \lambda) \hat{\jmath}+(2+2 \lambda) \hat{k} \end{aligned}

The coordinate of any point in this are of the form

(2+3 \lambda) i+(-1+4 \lambda) j+(2+2 \lambda) \hat{k}

Since this point lies on the plane  \vec{r} \cdot(i-\hat{\jmath}+\hat{k})=5

\begin{aligned} &{[(2+3 \lambda) \hat{\imath}+(-1+4 \lambda) \hat{\jmath}+(2+2 \lambda) \hat{k}] \hat{1}-\hat{\jmath}+\hat{k}=5} \\\\ &\Rightarrow 2+3 \lambda+1-4 \lambda+2+2 \lambda-5=0 \\\\ &\Rightarrow \lambda=0 \end{aligned}

So the coordinate of the point are

\begin{aligned} &(2+3 \lambda,-1+4 \lambda, 2+2 \lambda) \\\\ &=(2+0,-1+0,2+0) \\\\ &=(2,-1,2) \end{aligned}.............(1)

The coordinate of the point corresponding to the position vector  -i-5 \hat{\jmath}-10 \hat{k}  are (-1,5,-10)    … (2)

Distance between (1) and (2)

\begin{aligned} &\therefore \sqrt{(-1-2)^{2}+(-5+1)^{2}+(-10-2)^{2}} \\\\ &=\sqrt{9+16+144} \\\\ &=13 \end{aligned}

Posted by

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