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  • Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.7 question 1 subquestion (ii)

Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.7 question 1 subquestion (ii)

Answers (1)

Answer:

 Required vector equation is

\vec{r}.(-2\hat{i}-\hat{k})+5=0

Hint:

 Using scalar form

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

Given:

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

Solution:

We know that the equation

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

represent a plane passing through a point whose position vector \vec{a}

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

Here

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

Normal vector

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

The vector equation of plane is in scalar product form in

\begin{aligned} &\Rightarrow \vec{r}(-2\hat{i}-\hat{k})=(\hat{i}+2\hat{j}+3\hat{k}).(-2\hat{i}-\hat{k})\\ &\Rightarrow \vec{r}(-2\hat{i}-\hat{k})=-2+0-3=-5\\ &\therefore \vec{r}(-2\hat{i}-\hat{k})+5=0 \end{aligned}

Is the required vector equation

Posted by

Gurleen Kaur

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