The distance of the plane 6x - 3y + 2z - 14 = | vedclass

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Question

(A) The distance of a plane $ax + by + cz + d = 0$ from the origin $(0, 0, 0)$ is given by the formula: Distance $= \frac{|d|}{\sqrt{a^2 + b^2 + c^2}}

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$2$

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(A) The distance of a plane $ax + by + cz + d = 0$ from the origin $(0, 0, 0)$ is given by the formula: Distance $= \frac{|d|}{\sqrt{a^2 + b^2 + c^2}}$ Here,the equation of the plane is $6x - 3y + 2z - 14 = 0$. Comparing this with the standard form,we have $a = 6$,$b = -3$,$c = 2$,and $d = -14$. Substituting these values into the formula: Distance $= \frac{|-14|}{\sqrt{6^2 + (-3)^2 + 2^2}}$ $= \frac{14}{\sqrt{36 + 9 + 4}}$ $= \frac{14}{\sqrt{49}}$ $= \frac{14}{7} = 2$. Thus,the distance of the plane from the origin is $2$ units.

The distance of the plane $6x - 3y + 2z - 14 = 0$ from the origin is

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