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  • Please solve RD Sharma class 12 chapter The Plane exercise 28.7 question 2 subquestion (i) maths textbook solution

Please solve RD Sharma class 12 chapter The Plane exercise 28.7 question 2 subquestion (i) maths textbook solution

Answers (1)

Answer:

 Required Cartesian equation is

x-y+z=2

Hint:

 The plane is perpendicular to \vec{n}

Given:

 \vec{r}=(\hat{i}-\hat{j})+s(-\hat{i}+\hat{j}+2\hat{k})+t(\hat{i}+2\hat{j}+\hat{k})

Solution:

\vec{r}=(\hat{i}-\hat{j})+s(-\hat{i}+\hat{j}+2\hat{k})+t(\hat{i}+2\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} parallel to the vectors \vec{b} and \vec{c}

Here

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

The given plane is perpendicular to \vec{b} X \vec{c}

\begin{aligned} &\vec{b}\times \vec{c}=\begin{vmatrix} \hat{i} &\hat{j} &\hat{k} \\ -1 &1 &2 \\ 1 &2 &1 \end{vmatrix}\\ &=\hat{i}(1-4)-\hat{j}(-1-2)+\hat{k}(-2-1)\\ &=-3\hat{i}+3\hat{j}-3\hat{k} \end{aligned}

Vector equation of the plane in scalar product form is

\begin{aligned} &\vec{r}.\vec{n}=\vec{a}.\vec{n} \qquad \qquad \dots (1) \end{aligned}

\begin{aligned} &\vec{r}.(-3\hat{i}+3\hat{j}-3\hat{k})=(\hat{i}-\hat{j}).(-3\hat{i}+3\hat{j}-3\hat{k})\\ &\vec{r}.(-3\hat{i}+3\hat{j}-3\hat{k})=1(-3)+(-1)(3)+0(3)\\ &\vec{r}.(-3\hat{i}+3\hat{j}-3\hat{k})=-6\\ &\vec{r}.(-\hat{i}+\hat{j}-\hat{k})=-2\\ &\vec{r}.(\hat{i}-\hat{j}+\hat{k})=2 \end{aligned}

Now put

\begin{aligned} &\vec{r}=x\hat{i}+y\hat{j}+z\hat{k} \end{aligned}

So, we have

\begin{aligned} &(x\hat{i}+y\hat{j}+z\hat{k}).(-3\hat{i}+3\hat{j}-3\hat{k})=-6\\ &x(-3)+y(3)+2(-3)=-6\\ &-3x+3y-3z=-6 \end{aligned}

Divide by -3, we get

\begin{aligned} &x-y+z=2 \end{aligned}

Hence the required equation is

\begin{aligned} &x-y+z=2 \end{aligned}

Posted by

Gurleen Kaur

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