Find the distance of the plane 2x - 3y + 4z - | vedclass

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Question

(A) The equation of the plane is $2x - 3y + 4z - 6 = 0$,which can be written as $2x - 3y + 4z = 6$. The distance $d$ of a plane $Ax + By + Cz = D$ fro

Vedclass · Accepted Answer

$\frac{6}{\sqrt{29}}$

Vedclass · Answer

(A) The equation of the plane is $2x - 3y + 4z - 6 = 0$,which can be written as $2x - 3y + 4z = 6$. The distance $d$ of a plane $Ax + By + Cz = D$ from the origin $(0, 0, 0)$ is given by the formula $d = \frac{|D|}{\sqrt{A^2 + B^2 + C^2}}$. Here,$A = 2$,$B = -3$,$C = 4$,and $D = 6$. Substituting these values into the formula: $d = \frac{|6|}{\sqrt{2^2 + (-3)^2 + 4^2}}$ $d = \frac{6}{\sqrt{4 + 9 + 16}}$ $d = \frac{6}{\sqrt{29}}$. Thus,the distance of the plane from the origin is $\frac{6}{\sqrt{29}}$.

Find the distance of the plane $2x - 3y + 4z - 6 = 0$ from the origin.

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