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  • Need solution for RD sharma maths class 12 chapter 28 The Plane exercise 28.8 question 13

Need solution for RD sharma maths class 12 chapter 28 The Plane exercise 28.8 question 13

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The answer of the given question is \vec{r}.\left ( 20\hat{i}+23\hat{j}+26\hat{k} \right )=69.

Hint:- 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:-\vec{r}.(\hat{i}+\hat{j}+\hat{k})=6           and

\vec{r}.(2\hat{i}+3\hat{j}+4\hat{k})=-5  and Point (1,1,1)

Solution:-  The Cartesian equation of the given planes are x+y+z-6=0 and  2x+3y+4z+5=0

The family of planes is x+y+z-6+\lambda (2x+3y+4z+5)=0                      …(i)

Since it pass through (1,1,1)

1+1+1-6+\lambda(2+3+4+5)=0\\ -3+\lambda(14)=0\\ \lambda=\frac{3}{14}

x+y+z-6+\frac{3}{14} \left (2x+3y+4z+5 \right )=0\\ 14x+14y+14z-84+6x+9y+12z+15=0\\ 20x+23y+26z-69=0
Vector equation is  \vec{r}. (20\hat{i}+23\hat{j}+26\hat{k})=69

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