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  • Explain solution RD sharma class 12 chapter 28 The Plane exercise 28.8, question 16

Explain solution RD sharma class 12 chapter 28 The Plane exercise 28.8, question 16

Answers (1)

The answer of the given question is \vec{r}. \left (\hat{i}-\hat{k} \right )+2=0.

Hints:-  We know that, equation of a plane passing through the line of intersection of two planes

a_{1}x+b_{1}y+c_{1}z+d_{1}=0 and a_{2}x+b_{2}y+c_{2}z+d_{2}=0 is given by
\left (a_{1}x+b_{1}y+c_{1}z+d_{1} \right )+k\left ( a_{2}x+b_{2}y+c_{2}z+d_{2} \right )=0
Given:-

x+y+z=1 ,\\ 2x+3y+4z=5 \\ x-y+z=0

Solution:-  Equation of the plane through the intersection of plane is
(x+y+z-1)+\lambda (2x+3y+4z-5)=0 or

\left (1+2\lambda \right )x + \left (1+3\lambda \right ) y+\left (1+4\lambda \right )z-\left (1+5\lambda \right )=0                              …(i)

This plane is perpendicular to x-y+z=0

We know that, two planes are perpendicular if  a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0

1 \left (1+2\lambda \right ) - 1 \left (1+3\lambda \right ) + 1 \left (1+4\lambda \right )=0 or

\lambda = -\frac{1}{3}

Equation of plane is
\left (x+y+z-1 \right )-\frac{1}{3}\left (2x+3y+4z-5 \right )=0\\ \Rightarrow x-z+2=0

Vector form of plane is \vec{r}. \left (\hat{i}-\hat{k} \right )+2=0

Yes, line lies on the plane on (-2, 3, 0) satisfies \vec{r}. \left (\hat{i}-\hat{k} \right )+2=0

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