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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

Answers (1)

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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