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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)

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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