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  • Explain solution RD Sharma class 12 chapter The Plane exercise 28.3 question 12 maths

Explain solution RD Sharma class 12 chapter The Plane exercise 28.3  question 12 maths

Answers (1)

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Answer:

The answer of given question is 2x+2y+2z=18

Hint:

\vec{a}=\frac{\text { Position vector of } A+\text { Position vector of } B}{2}

Given:

The given plane bisects the line segment joining pointsA(1,2,3) and B(3,4,5)  and is at right angle to it..

Solution:

As the given plane bisects the line segment

Therefore,

\vec{a}=\frac{\text { Position vector of } A+\text { Position vector of } B}{2}

\begin{aligned} &\Rightarrow \vec{a}=\frac{(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})+(3 \hat{\imath}+4 \hat{\jmath}+5 \hat{k})}{2} \\ &\Rightarrow \vec{a}=\frac{4 \hat{\imath}+6 \hat{\jmath}+8 \hat{k}}{2} \\ &\Rightarrow \vec{a}=2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k} \ldots(i) \end{aligned}

And it is also given the plane B normal to the line joining the points A(1,2,3) and B(3,4,5)

Then  \vec{n}=\overrightarrow{A B}

? \vec{n} = position vector of\vec{B} -  Position vector of \vec{A}

\begin{aligned} &\vec{n}=(3 \hat{\imath}+4 \hat{\jmath}+5 \hat{k})-(\hat{\imath}+2 \hat{\jmath}+3 \hat{k}) \\ &\vec{n}=2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k} \ldots(i i) \end{aligned}

We know that

(\vec{r}-\vec{a}) \cdot \vec{n}=0

Substituting the values from equation (i) and equation (ii) in the above equation we get

\begin{aligned} &{[\vec{r}-(2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k})] \cdot(2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k})=0} \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k})-(2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k}) \cdot(2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k})=0 \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k})-[(2)(2)+(3)(2)+(4)(2)]=0 \end{aligned}

By multiplying the two vectors using the formula

\begin{aligned} &\vec{A} \cdot \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z} \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k})-[4+6+8]=0 \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k})-18=0 \\ &\Rightarrow 2 x+2 y+2 z=18 \end{aligned}

This is the cartesian form of equation of the required plane.

 

Posted by

infoexpert26

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