Find the distance of the point (2, 1, 0) from | vedclass

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(B) The distance $d$ of a point $(x_1, y_1, z_1)$ from a plane $ax + by + cz + d = 0$ is given by the formula: $d = \frac{|ax_1 + by_1 + cz_1 + d|}{\s

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$10/3$

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(B) The distance $d$ of a point $(x_1, y_1, z_1)$ from a plane $ax + by + cz + d = 0$ is given by the formula: $d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$ Given point is $(2, 1, 0)$ and the plane is $2x + y + 2z + 5 = 0$. Here,$a = 2, b = 1, c = 2, d = 5$ and $x_1 = 2, y_1 = 1, z_1 = 0$. Substituting these values into the formula: $d = \frac{|2(2) + 1(1) + 2(0) + 5|}{\sqrt{2^2 + 1^2 + 2^2}}$ $d = \frac{|4 + 1 + 0 + 5|}{\sqrt{4 + 1 + 4}}$ $d = \frac{|10|}{\sqrt{9}}$ $d = \frac{10}{3}$ Thus,the distance is $10/3$ units.

Find the distance of the point $(2, 1, 0)$ from the plane $2x + y + 2z + 5 = 0$.

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