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Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.3  question 18

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Answer:

The answer of given question is 2x+3y-z=20

Hint:

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

Given:

The plane is passing through P(5,2,-4) and perpendicular to the line having 2,3,-1 as the direction ratio.

Solution:

Let, the position vector of this point P be

\Rightarrow \vec{a}=5 \hat{\imath}+2 \hat{\jmath}-4 \hat{k} \ldots(i)

And also given the plane is normal having 2,3,-1

As the direction ratio.

\vec{n}=2 \hat{\imath}+3 \hat{\jmath}-\hat{k} \ldots(ii)

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}-(5 \hat{\imath}+2 \hat{\jmath}-4 \hat{k})] \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})=0} \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})-(5 \hat{\imath}+2 \hat{\jmath}-4 \hat{k}) \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})=0 \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})-[(2)(5)+(2)(3)+(-4)(-1)]=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}+3 \hat{\jmath}-\hat{k})-[10+6+4]=0 \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})-20=0 \end{aligned}

\Rightarrow \vec{r} \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})=20 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,

(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(2 \hat{\imath}+3 \hat{\jmath}-\hat{k})=20

Now multiplying the two vectors using the formula

\begin{aligned} &\vec{A} . \vec{B}=A_{x} B_{x}+A_{y} B_{y}+A_{z} B_{z} \\ &\Rightarrow(x)(2)+(y)(3)+(z)(-1)=20 \\ &\Rightarrow 2 x+3 y-z=20 \end{aligned}

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

 

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infoexpert26

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