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  • Provide Solution For R.D. Sharma Maths Class 12 Chapter 28 The Plane Exercise 28.12 Question 3 Maths Textbook Solution.

Provide Solution For  R.D. Sharma Maths Class 12 Chapter 28 The Plane Exercise 28.12 Question 3 Maths Textbook Solution.

Answers (1)

Answer:Therefore ,the distance is 13 Units .

Hint : Use Distance formula \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-_y{1} \right )^{2}+\left ( z_{2}-z_{1} \right )^{2}}

Given:

\begin{aligned} &\vec{r}=2 \hat{\imath}-\hat{\jmath}+2 \hat{k}+\lambda(3 \hat{\imath}+4 \hat{\jmath}+2 \hat{k}) \text { and } \\ &\vec{r} \cdot(\hat{\imath}-\hat{\jmath}+\hat{k})=5 \text { and point }(-1,-5,-10) \end{aligned}

Solution :

General\; point \: on \: the \: line \: is

(2+3 \lambda,-1+4 \lambda, 2+2 \lambda)

As\: point \: lies \: on\: the\: plane

\therefore 2+3 \lambda+1-4 \lambda+2+2 \lambda=5

\Rightarrow \lambda=0

    \therefore \: So\: the \: \: coordinate \: \: of \: \: point \: \: is (2,-1,2)

Distance =\sqrt{(2-(-1))^{2}+(-1-(-5))^{2}+(2-(-10))^{2}}

=\sqrt{(2+1)^{2}+(-1+5)^{2}+(2+10)^{2}}

=\sqrt{(3)^{2}+(4)^{2}+(12)^{2}}

=\sqrt{9+16+144}

=\sqrt{169}=13 unit

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