Find the coordinates of the foot of the perpe | vedclass

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Question

(A) Let the coordinates of the foot of the perpendicular $P$ from the origin to the plane be $(x_1, y_1, z_1)$.The equation of the plane is $x+y+z=1$

Vedclass · Accepted Answer

$\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)$

Vedclass · Answer

(A) Let the coordinates of the foot of the perpendicular $P$ from the origin to the plane be $(x_1, y_1, z_1)$.The equation of the plane is $x+y+z=1$  $(1)$.The direction ratios of the normal to the plane are $1, 1, 1$.The magnitude of the normal vector is $\sqrt{1^2+1^2+1^2} = \sqrt{3}$.Dividing the equation of the plane by $\sqrt{3}$,we get the normal form $lx+my+nz=d$:$\frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}}y + \frac{1}{\sqrt{3}}z = \frac{1}{\sqrt{3}}$.Here,the direction cosines are $l = \frac{1}{\sqrt{3}}, m = \frac{1}{\sqrt{3}}, n = \frac{1}{\sqrt{3}}$ and the distance $d = \frac{1}{\sqrt{3}}$.The coordinates of the foot of the perpendicular from the origin are given by $(ld, md, nd)$.Substituting the values,we get:$x_1 = \left(\frac{1}{\sqrt{3}}\right) \left(\frac{1}{\sqrt{3}}\right) = \frac{1}{3}$,$y_1 = \left(\frac{1}{\sqrt{3}}\right) \left(\frac{1}{\sqrt{3}}\right) = \frac{1}{3}$,$z_1 = \left(\frac{1}{\sqrt{3}}\right) \left(\frac{1}{\sqrt{3}}\right) = \frac{1}{3}$.Thus,the coordinates of the foot of the perpendicular are $\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right)$.

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane $x+y+z=1$.

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