The coordinates of the foot of the perpendicu | vedclass

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(A) The equation of the plane is $2x + y - 2z = 18$. The normal vector to the plane is $\vec{n} = 2\hat{i} + \hat{j} - 2\hat{k}$. The length of the no

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$(4, 2, -4)$

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(A) The equation of the plane is $2x + y - 2z = 18$. The normal vector to the plane is $\vec{n} = 2\hat{i} + \hat{j} - 2\hat{k}$. The length of the normal vector is $|\vec{n}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$. Dividing the equation of the plane by $3$,we get the normal form $\frac{2}{3}x + \frac{1}{3}y - \frac{2}{3}z = 6$. The distance of the plane from the origin is $d = 6$. The coordinates of the foot of the perpendicular from the origin to the plane $ax + by + cz = d$ are given by $(\frac{ad}{a^2+b^2+c^2}, \frac{bd}{a^2+b^2+c^2}, \frac{cd}{a^2+b^2+c^2})$. Here,$a=2, b=1, c=-2$ and $d=18$. The denominator $a^2+b^2+c^2 = 9$. Thus,the coordinates are $(\frac{2 	imes 18}{9}, \frac{1 	imes 18}{9}, \frac{-2 	imes 18}{9}) = (4, 2, -4)$.

The coordinates of the foot of the perpendicular drawn from the origin to the plane $2x + y - 2z = 18$ are

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