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  • Please solve RD Sharma class 12 chapter The Plane exercise 28.7 question 1 subquestion (i) maths textbook solution

Please solve RD Sharma class 12 chapter The Plane exercise 28.7 question 1 subquestion (i) maths textbook solution

Answers (1)

Answer:

 Required equation is

\vec{r}.(\hat{j}-2\hat{k})=2

Hint:

 Using scalar form

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

Given:

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

Solution:

We know that

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

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

Clearly,

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

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

Hence

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

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

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

Put \begin{aligned} &\vec{a} \end{aligned} and \begin{aligned} &\vec{n} \end{aligned} in equation (a) we get

\begin{aligned} &\vec{r}.(\hat{j}-2\hat{k})=(2\hat{i}-\hat{k})(\hat{j}-2\hat{k})\\ &\Rightarrow \vec{r}.(\hat{j}-2\hat{k})=2(0)+(0)(1)+(-1)(-2)\\ &\Rightarrow \vec{r}.(\hat{j}-2\hat{k})=2 \end{aligned}

Hence the require equation is

\begin{aligned} &\vec{r}.(\hat{j}-2\hat{k})=2 \end{aligned}

Posted by

Gurleen Kaur

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