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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

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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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