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  • Explain solution RD Sharma class 12 chapter The Plane exercise 28.7 question 3 subquestion (ii) maths

Explain solution RD Sharma class 12 chapter The Plane exercise 28.7 question 3 subquestion (ii) maths

Answers (1)

Answer:

 The required equation is \vec{r}.(5\hat{i}+2\hat{j}-3\hat{k})=17

Hint:

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

Given:

\vec{r}=(2\hat{i}+2\hat{j}-\hat{k})+\lambda (\hat{i}+2\hat{j}+3\hat{k})+\mu (5\hat{i}-2\hat{j}+7\hat{k}) 

Solution:

We know that \vec{r}=\vec{a}+\lambda \vec{b}+\mu \vec{c} represents

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

Clearly

\begin{aligned} &\vec{a}=2\hat{i}+2\hat{j}-\hat{k}\\ &\vec{b}=\hat{i}+2\hat{j}+3\hat{k}\\ &\vec{c}=5\hat{i}-2\hat{j}+7\hat{k} \end{aligned}

Hence

\begin{aligned} &\vec{n}=\vec{b}\times \vec{c}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ 1 &2 &3 \\ 5 &-2 &7 \end{vmatrix}\\ &=(14+6)\hat{i}-(7-15)\hat{j}+(-2-10)\hat{k}\\ &=20\hat{i}+8\hat{j}-12\hat{k} \end{aligned}

We know that vector equation of a plane in scalar product is given as

\begin{aligned} &\vec{r}.\vec{n}=\vec{a}.\vec{n}\\ &\vec{r}.(20\hat{i}+8\hat{j}-12\hat{k})=(2\hat{i}+2\hat{j}-\hat{k}).(20\hat{i}+8\hat{j}-12\hat{k})\\ &\vec{r}.(20\hat{i}+8\hat{j}-12\hat{k})=40+16+12\\ &\vec{r}.(20\hat{i}+8\hat{j}-12\hat{k})=68\\ &\vec{r}.(5\hat{i}+2\hat{j}-3\hat{k})=17 \end{aligned}

Hence the require equation is

\begin{aligned} &\vec{r}.(5\hat{i}+2\hat{j}-3\hat{k})=17 \end{aligned}

Posted by

Gurleen Kaur

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