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  • Explain solution RD Sharma class 12 chapter The Plane exercise 28.3 question 16 maths

Explain solution RD Sharma class 12 chapter The Plane exercise 28.3  question 16 maths

Answers (1)

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Answer:

The answer of given question is -14x+6y+2z=178  or -7x+3y+z=89

Hint:

By using formula \vec{A} \cdot \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z}

Given:

A(4,-1,2) and B(-10, 5,4).

Solution:

It means that the plane is passing through B(-10, 5,4).

Therefore, the position vector of this point is,

\Rightarrow \vec{a}=-10 \hat{\imath}+5 \hat{\jmath}+4 \hat{k} \ldots(i)

And also given the line segment joining the points A(4,-1,2) and B(-10, 5,4). and is at right angle to it.

Then \vec{n}=\overrightarrow{A B}

? \vec{n} =position vector of \vec{B} -  Position vector of \vec{A}
\begin{aligned} &\vec{n}=(-10 \hat{\imath}+5 \hat{\jmath}+4 \hat{k})-(4 \hat{\imath}-\hat{\jmath}+2 \hat{k}) \\ &\vec{n}=-14 \hat{\imath}+6 \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}-(-10 \hat{\imath}+5 \hat{\jmath}+4 \hat{k})] \cdot(-14 \hat{\imath}+6 \hat{\jmath}+2 \hat{k})=0} \\ &\Rightarrow \vec{r} \cdot(-14 \hat{\imath}+6 \hat{\jmath}+2 \hat{k})-(-10 \hat{\imath}+5 \hat{\jmath}+4 \hat{k}) \cdot(-14 \hat{\imath}+6 \hat{\jmath}+2 \hat{k})=0 \\ &\Rightarrow \vec{r} \cdot(-14 \hat{\imath}+6 \hat{\jmath}+2 \hat{k})-[(-10)(-14)+(5)(6)+(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(-14 \hat{\imath}+6 \hat{\jmath}+2 \hat{k})-[140+30+8]=0 \\ &\Rightarrow \vec{r} \cdot(-14 \hat{\imath}+6 \hat{\jmath}+2 \hat{k})-178=0 \end{aligned}

\Rightarrow \vec{r} \cdot(-14 \hat{\imath}+6 \hat{\jmath}+2 \hat{k})=178 is the vector equation of a required plane.

\vec{r}=(x \hat{\imath}+y \hat{\jmath}+z \hat{k})

Then the above vector equation of a plane becomes,

\begin{aligned} &\Rightarrow(x)(-14)+(y)(6)+(z)(2)=178 \\ &\Rightarrow-14 x+6 y+2 z=178 \text { or }-7 x+3 y+z=89 \end{aligned}

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

Posted by

infoexpert26

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