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  • Provide Solution For R.D. Sharma Maths Class 12 Chapter 28 The Plane Exercise 28.12 Question 9 Maths Textbook Solution.

Provide Solution For  R.D. Sharma Maths Class 12 Chapter 28 The Plane Exercise 28.12 Question 9 Maths Textbook Solution.

Answers (1)

Answer: Therefore, required distance is \vec{r}=-2 \hat{\imath}-4 \hat{\jmath}+7 \hat{k}+\mu(3 \hat{\imath}+3 \hat{\jmath}-3 \hat{k}) & distance is 3\sqrt{3}units

Hint: Use properties of vector

Given:

\begin{aligned} &\vec{r}=3 \hat{\imath}-2 \hat{\jmath}+6 \hat{k}+\lambda(2 \hat{\imath}-\hat{\jmath}+2 \hat{k}) \text { and }\\ &\vec{r} \cdot(\hat{\imath}-\hat{\jmath}+\hat{k})=6\\ \end{aligned}

Solution:

\begin{aligned} &\vec{r}=3 \hat{\imath}-2 \hat{\jmath}+6 \hat{k}+\lambda(2 \hat{\imath}-\hat{\jmath}+2 \hat{k})\\ &\vec{r}=(3+2 \lambda) \hat{\imath}+(-2-\lambda) \hat{\jmath}+(6+2 \lambda) \hat{k}\\ &\text { Plane is } \vec{r} \cdot(\hat{\imath}-\hat{\jmath}+\hat{k})=6\\ &\text { To find the point of intersection, }\\ &\therefore 3+2 \lambda+2 \lambda+6+2 \lambda=6\\ &5 \lambda=-5 \Rightarrow \lambda=-1\\ &\therefore \text { Point } \mathrm{Q}(1,-1,4)\\ &P(-2,-4,7)\\ &\text { Line } \mathrm{PQ} \text { is } \vec{r}=-2 \hat{\imath}-4 \hat{\jmath}+7 \hat{k}+\mu(3 \hat{\imath}+3 \hat{\jmath}-3 \hat{k})\\ &\text { Required distance of } \mathrm{PQ}=\sqrt{(1-(-2))^{2}+((-4)+1)^{2}+(7-4)^{2}}=\sqrt{3^{2}+3^{2}+3^{2}}=\sqrt{27} \text { units } \end{aligned}

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