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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)

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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