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  • Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.7 question 3 subquestion (i)

Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.7 question 3 subquestion (i)

Answers (1)

Answer:

 The required equation is \vec{r}.(2\hat{i}-5\hat{j}-\hat{k})+15=0

Hint:

 The plane is perpendicular to \vec{b} X \vec{c} so we find \vec{n}=\vec{b}\times \vec{c}

Given:

 \vec{r}=(\lambda -2\mu )\hat{i}+(3-\mu )\hat{j}+(2\lambda +\mu )\hat{k}

Solution:

Here

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

We know that the equation

\vec{r}=\vec{a}+\lambda \vec{b}+\mu \vec{c}

represents a plane passing through a point having position \vec{a} and parallel to the vectors \vec{b} and \vec{c}

Clearly

\begin{aligned} &\vec{a}=3\hat{j}\\ &\vec{b}=\hat{i}+2\hat{k}\\ &\vec{c}=-2\hat{i}-\hat{j}+\hat{k} \end{aligned}

Now the plane is perpendicular to \begin{aligned} &\vec{b}\times \vec{c} \end{aligned}

Hence

\begin{aligned} &\vec{n}=\vec{b}\times \vec{c}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 1 &0 &2 \\ -2 &-1 &1 \end{vmatrix}\\ &=(0+2)\hat{i}-(1+4)\hat{j}+(-1+0)\hat{k}\\ &=2\hat{i}-5\hat{j}-\hat{k} \end{aligned}

We know that the vector equation of plane in scalar product form is given by

\begin{aligned} &\vec{r}.\vec{n}=\vec{a}.\vec{n}\\ &\vec{r}.(2\hat{i}-5\hat{j}-\hat{k})=(3\hat{j}).(2\hat{i}-5\hat{j}-\hat{k})\\ &\vec{r}.(2\hat{i}-5\hat{j}-\hat{k})=-15 \end{aligned}

Hence the required equation is

\begin{aligned} &\vec{r}.(2\hat{i}-5\hat{j}-\hat{k})+15=0 \end{aligned}

Posted by

Gurleen Kaur

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