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  • Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.3 question 19

Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.3  question 19

Answers (1)

Answer:

The answer of given question is x+2y-3z=14

Hint:

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

Given:

The plane is passing through P(1,2,-3) and perpendicular to OP

Solution:

The required plane is passing through P(1,2,-3)  and perpendicular to OP. let the position vector of this point P be

\Rightarrow \vec{a}=\hat{\imath}+2 \hat{\jmath}-3 \hat{k} \ldots(i)

And it is also given the planes is normal to the line joining the points O(0,0,0) and position vector of point P(1,2,-3).

Then

\vec{n}=\overrightarrow{O P}

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

\begin{aligned} &\vec{n}=(\hat{\imath}+2 \hat{\jmath}-3 \hat{k})-(0 \hat{\imath}+0 \hat{\jmath}+0 \hat{k}) \\ &\vec{n}=\hat{\imath}+2 \hat{\jmath}-3 \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}-(\hat{\imath}+2 \hat{\jmath}-3 \hat{k})] \cdot(\hat{\imath}+2 \hat{\jmath}-3 \hat{k})=0} \\ &\Rightarrow \vec{r} \cdot(\hat{\imath}+2 \hat{\jmath}-3 \hat{k})-(\hat{\imath}+2 \hat{\jmath}-3 \hat{k}) \cdot(\hat{\imath}+2 \hat{\jmath}-3 \hat{k})=0 \\ &\Rightarrow \vec{r}-(\hat{\imath}+2 \hat{\jmath}-3 \hat{k})-[(1)(1)+(2)(2)+(-3)(-3)]=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(\hat{\imath}+2 \hat{\jmath}-3 \hat{k})-[1+4+9]=0 \\ &\Rightarrow \vec{r} \cdot(\hat{\imath}+2 \hat{\jmath}-3 \hat{k})-14=0 \end{aligned}is the vector equation of the required plane

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

Then, the above vector equation of the plane becomes

(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(\hat{\imath}+2 \hat{\jmath}-3 \hat{k})=14

Now 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(x)(1)+(y)(2)+(z)(-3)=14 \\ &\Rightarrow x+2 y-3 z=14 \end{aligned}

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

Posted by

infoexpert26

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