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  • Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.3 question 11

Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.3  question 11

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Answer:

3x+y-z=-4 is the cartesian form of equation of the required plane  &

\vec{r} \cdot(3 \hat{\imath}+\hat{\jmath}-\hat{k})+4=0 is the vector equation of a required plane

Hint:

Let the position vector of this point Q be\vec{a}=\hat{\imath}-2 \hat{\jmath}+5 \hat{k} \ldots(i)

 

Given:

plane is passing through (1,-2,5) and is perpendicular to OP, where point O is the origin and position vector of point P is   3 \hat{\imath}+\hat{\jmath}-\hat{k}

Solution:

As per the criteria required plane is passing through Q(1,-2,5) and is perpendicular to OP, where point O is the origin and position vector of point P is 3 \hat{\imath}+\hat{\jmath}-\hat{k}

Let the position vector of the point Q be.

\vec{a}=\hat{\imath}-2 \hat{\jmath}+5 \hat{k}

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 is 3 \hat{\imath}+\hat{\jmath}-\hat{k}

Then

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

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

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

\Rightarrow \vec{r} \cdot(3 \hat{\imath}+\hat{\jmath}-\hat{k})+4=0   is the vector equation of a required plane

Then, the above vector equation of the plane becomes

(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(3 \hat{\imath}+\hat{\jmath}-\hat{k})+4=0

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(x)(3)+(y)(1)+(z)(-1)=-4 \end{aligned}

\Rightarrow 3 x+y-z=-4  is the cartesian form of equation of the required plane

 

Posted by

infoexpert26

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