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  4. A plane P passes through the point (1,1 , 1) and is parallel to the vectors overset arrow a=- hati+ hatj and overset arrow b= hati- hatk . The distance of the point ((3 √3/2) , 3 √3 , 3) from the plane is equal to

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Q. A plane $P$ passes through the point $\left(1,1 , 1\right)$ and is parallel to the vectors $\overset{ \rightarrow }{a}=-\hat{i}+\hat{j}$ and $\overset{ \rightarrow }{b}=\hat{i}-\hat{k}$ . The distance of the point $\left(\frac{3 \sqrt{3}}{2} , 3 \sqrt{3} , 3\right)$ from the plane is equal to

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Solution:

Equation of the plane is $\begin{vmatrix} x-1 & y-1 & z-1 \\ -1 & 1 & 0 \\ 1 & 0 & -1 \end{vmatrix}=0$
$\Rightarrow \left(x - 1\right)\left(- 1\right)-\left(y - 1\right)\left(1\right)+\left(z - 1\right)\left(- 1\right)=0$
$\Rightarrow x+y+z-3=0$
The distance of point $\left(\frac{3 \sqrt{3}}{2} , 3 \sqrt{3} , 3\right)$ from $x+y+z-3=0$ is
$\left|\frac{\frac{3 \sqrt{3}}{2} + 3 \sqrt{3} + 3 - 3}{\sqrt{3}}\right|=\frac{9}{2}$ units