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  • Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.7 question 2 subquestion (ii)

Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.7 question 2 subquestion (ii)

Answers (1)

Answer:

 Required Cartesian equation is 2y - z = 1

Hint:

 Using the equation of the plane

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

Given:

 \vec{r}=(1+s+t)\hat{i}+(2-s+t)\hat{j}+(3-2s+2t)\hat{k}

Solution:

Since from the given equation

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

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

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

Substituting the value of \begin{aligned} &\vec{a} \end{aligned} and \begin{aligned} &\vec{n} \end{aligned} in equation (1)

We get

\begin{aligned} &\vec{r}.(-4\hat{j}+2\hat{k})=(\hat{i}+2\hat{j}+3\hat{k}).(-4\hat{j}+2\hat{k})\\ &=(1)(0)+(2)(-4)+(3)(2)\\ &\vec{r}.2(-2\hat{j}+\hat{k})=-2\\ &\vec{r}.(2\hat{j}-\hat{k})=1 \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})(-4\hat{j}+2\hat{k})=-2\\ &(x)(0)+(y)(-4)+(z)(2)=-2\\ &-4y+2z=-2\\ &2y-z=1 \end{aligned}

Hence the required equation of the plane is

\begin{aligned} &2y-z=1 \end{aligned}

Posted by

Gurleen Kaur

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