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  • Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.5 question 2

Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.5 question 2

Answers (1)

Answer:

 The vector equation of required plane is

\vec{r}.(2\hat{i}+3\hat{j}+4\hat{k})=7

Hint:

 Using formula

\vec{r}.\vec{n}=\vec{a}.\vec{n}

Given:

 Vector equation of plane passing through the points P(2, 5, -3), Q(-2, -3, 5) and R(5, 3,-3)

Solution:

The required plane passes through the point (2, 5, - 3) whose position vector is

\vec{a}=(2\hat{i}+5\hat{j}-3\hat{k})

and is normal to the vector \vec{n} given by

\begin{aligned} &\vec{n}=\vec{PQ}\times \vec{PR}\\ &\vec{PQ}=\vec{OQ}- \vec{OP}=(-2\hat{i}-3\hat{j}+5\hat{k})-(2\hat{i}+5\hat{j}-3\hat{k})=-4\hat{i}-8\hat{j}+8\hat{k}\\ &\vec{PR}=\vec{OR}- \vec{OP}=(5\hat{i}+3\hat{j}-3\hat{k})-(2\hat{i}+5\hat{j}-3\hat{k})=3\hat{i}-2\hat{j}-0\hat{k}\\ &\vec{n}=\vec{PQ}\times \vec{PR}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ -4 &-8 &8 \\ 3 &-2 &0 \end{vmatrix}=16\hat{i}+24\hat{j}+32\hat{k} \end{aligned}

The vector equation of the required plane is

\begin{aligned} &\vec{r}.\vec{n}=\vec{a}.\vec{n}\\ &\vec{r}.(16\hat{i}+24\hat{j}+32\hat{k})=(2\hat{i}+5\hat{j}-3\hat{k}).(16\hat{i}+24\hat{j}+32\hat{k})\\ &\vec{r}.[8(2\hat{i}+3\hat{j}+4\hat{k})]=32+120-96\\ &\vec{r}.[8(2\hat{i}+3\hat{j}+4\hat{k})]=56\\ &\vec{r}.(2\hat{i}+3\hat{j}+4\hat{k})=7 \end{aligned}

Posted by

Gurleen Kaur

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