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Please solve RD Sharma class 12 chapter The Plane exercise 28.4 question 9 maths textbook solution

Answers (1)

Answer:

 x-y+3z-2=0\: ;\: \frac{2}{\sqrt{11}}

Hint:

  You must know the rules of solving vector functions

Given:

 Find the equation of plane passing through (1,2,1) and perpendicular to the line joining the points (1,4,2) and (2,3,5). Find perpendicular distance.

Solution:

We have

The normal is passing through the points A(1,4,2) and B(2,3,5)
\begin{aligned} &\vec{n}=\vec{AB}=\vec{OB}-\vec{OA}\\ &=(2\hat{i}+3\hat{j}+5\hat{k})-(\hat{i}+4\hat{j}+2\hat{k})\\ &=(\hat{i}-\hat{j}+3\hat{k}) \end{aligned}

We know that the vector equation of the plane passing through a point (1,2,1)(\begin{aligned} &\vec{a} \end{aligned})
and normal to \begin{aligned} &\vec{n} \end{aligned} is
\begin{aligned} &\vec{r}.\vec{n}=\vec{a}.\vec{n} \end{aligned}

Putting

\begin{aligned} &\vec{a}=\hat{i}+2\hat{j}+\hat{k}\\ &\vec{n}=\hat{i}-\hat{j}+3\hat{k}\\ &\vec{r}.(\hat{i}-\hat{j}+3\hat{k})=(\hat{i}+2\hat{j}+\hat{k})(\hat{i}-\hat{j}+3\hat{k})\\ &\vec{r}.(\hat{i}-\hat{j}+3\hat{k})=(1-2+3)\\ &\vec{r}.(\hat{i}-\hat{j}+3\hat{k})=2 \qquad \qquad \qquad \rightarrow (1) \end{aligned}

To find the perpendicular distance of this plane from this origin,
we have
\begin{aligned} &\vec{n}=\hat{i}-\hat{j}+3\hat{k}\\ &\left | \vec{n} \right |=\sqrt{(1)^2+(-1)^2+(3)^2}\\ &=\sqrt{1+1+9}\\ &=\sqrt{11} \end{aligned}

Divide equation (1) by

\begin{aligned} &\sqrt{11} \end{aligned}

\begin{aligned} &\vec{r}.\left ( \frac{1}{\sqrt{11}}\hat{i}-\frac{1}{\sqrt{11}}\hat{j}+\frac{3}{\sqrt{11}}\hat{k} \right )=\frac{2}{\sqrt{11}} \qquad \qquad \rightarrow (2) \end{aligned}

The perpendicular distance of this plane from this origin

\begin{aligned} &=\frac{2}{\sqrt{11}} \end{aligned}

from equation (1)

\begin{aligned} &=\frac{1}{\sqrt{11}}x-\frac{1}{\sqrt{11}}y+\frac{3}{\sqrt{11}}z=\frac{2}{\sqrt{11}}\\ &x-y+3z=2\\ &x-y+3z-2=0 \end{aligned}

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Gurleen Kaur

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