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  • Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.3 question 7

Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.3  question 7

Answers (1)

Answer:

12x-4y+3z=169

Hint:

(\vec{r}-\vec{a}) \cdot \vec{n}=0

Given:

The co-ordinates of the foot of the perpendicular drawn from the origin to a plane are

(12,-4, 3)

Solution:

Let O (0,0,0)  and P(12, -4, 3)

\begin{aligned} &d . r \text { of } O P=12-0, \quad 4-0, \quad 3-0 \\ &=12,-4,3 \end{aligned}

\begin{aligned} &\vec{n}=12 \hat{\imath}-4 \hat{\jmath}+3 \hat{k} \\ &\vec{a}=12 \hat{\imath}-4 \hat{\jmath}+3 \hat{k} \\ &(\vec{r}-\vec{a}) \cdot \vec{n}=0 \end{aligned}

\begin{aligned} &\Rightarrow \vec{r} \cdot \vec{n}=\vec{a} \cdot \vec{n} \\ &\Rightarrow(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(12 \hat{\imath}-4 \hat{\jmath}+3 \hat{k})=(12 \hat{\imath}-4 \hat{\jmath}+3 \hat{k}) \cdot(12 \hat{\imath}-4 \hat{\jmath}+3 \hat{k}) \\ &\Rightarrow \vec{r} \cdot(12 \hat{\imath}-4 \hat{\jmath}+3 \hat{k})=169 \end{aligned}..is the required vector equation

\begin{aligned} &\Rightarrow 12 x-4 y+3 z=144+16+9 \\ &\Rightarrow 12 x-4 y+3 z=169 \end{aligned}..is the required equation of plane

 

Posted by

infoexpert26

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