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  • Provide solution for RD Sharma maths class 12 chapter 28 The Plane exercise multiple choice question 10

Provide solution for RD Sharma maths class 12 chapter 28 The Plane exercise multiple choice question 10

Answers (1)

Answer:

 Option (a)

Hint:

 Put  \lambda = 1

Given:

 \overrightarrow{r}=(-2\widehat{i}+3\widehat{j}+4\widehat{k})+\lambda (3\widehat{i}-2\widehat{j}-\widehat{k})

Solution:

The plane contains the line

\overrightarrow{r}=(-2\widehat{i}+3\widehat{j}+4\widehat{k})+\lambda (3\widehat{i}-2\widehat{j}-\widehat{k})

so, the plane contains the point

(\widehat{i}+2\widehat{j}+3\widehat{k})

 Put  \lambda = 1

-2\widehat{i}+3\widehat{j}+4\widehat{k}+3\widehat{i}-2\widehat{j}-\widehat{k} \text { i.e } (\widehat{i}-5\widehat{j}+3\widehat{k})

so, we got the plane, they are:

(\widehat{i}+2\widehat{j}+3\widehat{k}); (-2\widehat{i}-3\widehat{j}+4\widehat{k}); \text { and } (\widehat{i}-5\widehat{j}+3\widehat{k})

Let,

\begin{aligned} &\overrightarrow{a}=(\widehat{i}-5\widehat{j}+3\widehat{k})-(\widehat{i}+2\widehat{j}+3\widehat{k}) \text { and }\\ &\overrightarrow{b}=(\widehat{i}-5\widehat{j}+3\widehat{k})-(-2\widehat{i}-3\widehat{j}+4\widehat{k})\\ &\text { So, }\overrightarrow{a}=-7\widehat{j}\text { and }\overrightarrow{b}=3\widehat{i}-2\widehat{j}-\widehat{k} \end{aligned}

The normal of two vectors \begin{aligned} &\overrightarrow{a} \end{aligned} and \begin{aligned} &\overrightarrow{b} \end{aligned}

\begin{aligned} &\overrightarrow{a}=\begin{vmatrix} \widehat{i} &\widehat{j} &\widehat{k} \\ 0 &-7 &0 \\ 3 &-2 &1 \end{vmatrix}\\ &=\left [ \left \{ (-7)\times (-1)-((-2)\times 0) \right \}\widehat{i}+\left \{ (0\times (-1))-(0\times (-2)) \right \}\widehat{j}+\left \{ (0\times (-2))-((-7)\times 3) \right \}\widehat{k} \right ]\\ &=7\widehat{i}+21\widehat{k} \end{aligned}

The general equation of plane is,

\begin{aligned} &\left [ (x-1)\widehat{i}+(y-2)\widehat{j}+(z-3)\widehat{k} \right ].(7\widehat{i}+21\widehat{k})=0\\ &7(x-1)+21(z-3)=0\\ &7x-7+21z-63=0\\ &7x+21z=70\\ &x+3z=10\\ &\text { or }\\ &\overrightarrow{r}.(\widehat{i}+3\widehat{k})=10, \qquad \left [ \overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k} \right ] \end{aligned}

Hence, the vector equation of the plane containing the line

\overrightarrow{r}=(-2\widehat{i}+3\widehat{j}+4\widehat{k})+\lambda (3\widehat{i}-2\widehat{j}-\widehat{k})

and of the point

(\widehat{i}+2\widehat{j}+3\widehat{k})

is

\begin{aligned}&\overrightarrow{r}.(\widehat{i}+3\widehat{k})=10 \end{aligned}

Posted by

Gurleen Kaur

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