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  • Explain solution RD Sharma class 12 chapter The Plane exercise 28.7 question 1 subquestion (iv) maths

Explain solution RD Sharma class 12 chapter The Plane exercise 28.7 question 1 subquestion (iv) maths

Answers (1)

Answer:

 Required vector equation is

\vec{r}.(5\hat{i}+\hat{j}-6\hat{k})=4

Hint:

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

Given:

 \vec{r}=(\hat{i}-\hat{j})+\lambda (\hat{i}+\hat{j}+\hat{k})+\mu (4\hat{i}-2\hat{j}+3\hat{k})

Solution:

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} parallel to the vectors \vec{b} and \vec{c}

Clearly

\begin{aligned} &\vec{a}=\hat{i}-\hat{j}\\ &\vec{b}=\hat{i}+\hat{j}+\hat{k}\\ &\vec{c}=4\hat{i}-2\hat{j}+3\hat{k} \end{aligned}

Now the plane is perpendicular to \vec{b} X \vec{c}

Hence

\begin{aligned} &\vec{n}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ \hat{1} &\hat{1} &\hat{1} \\ \hat{4} &\hat{-2} &\hat{3} \end{vmatrix} \end{aligned}

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

\vec{r}. \vec{n}= \vec{a}.\vec{n} \qquad \qquad \dots (a)

Put \vec{a} and \vec{n} in equation (a) we get

\begin{aligned} &\vec{r}.(5\hat{i}+\hat{j}-6\hat{k})=(\hat{i}-\hat{j})(5\hat{i}+\hat{j}-6\hat{k})\\ &\Rightarrow \vec{r}.(5\hat{i}+\hat{j}-6\hat{k})=(1)(5)+(-1)(1)+(0)(-6)\\ &\Rightarrow \vec{r}.(5\hat{i}+\hat{j}-6\hat{k})=4 \end{aligned}

Hence the required equation is

\begin{aligned} &\vec{r}.(5\hat{i}+\hat{j}-6\hat{k})=4 \end{aligned}

Posted by

Gurleen Kaur

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