The perpendicular distance from original to the plane 3x-2y-2x=2 is

1 / \sqrt{17}

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2 / \sqrt{17}

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3 / \sqrt{17}

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4 / \sqrt{17}

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Solution

The correct option is B $2 / \sqrt{17}$
Plane : $3 x - 2 y - 2 z = 2$
$∴$ distance from origin $= \frac{| 3 \times 0 - 2 \times 0 - 2 \times 0 - 2 |}{\sqrt{3^{2} + 2^{2} + 2^{2}}} = \frac{2}{\sqrt{17}}$ units.

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P (3, 8, 2)

\frac{x - 1}{2} = \frac{y - 3}{4} = \frac{z - 2}{3}

3 x + 2 y - 2 z + 17 = 0

Solve for x and y:

$\frac{2}{(3 x + 2 y)} + \frac{3}{(3 x - 2 y)} = \frac{17}{5}, \frac{5}{(3 x + 2 y)} + \frac{1}{(3 x - 2 y)} = 2.$

\sqrt{17} units

2^i - 3^j - 2^k,

\frac{2}{3 x + 2 y} + \frac{3}{3 x - 2 y} = \frac{17}{5}

\frac{5}{3 x + 2 y} + \frac{1}{3 x - 2 y} = 2

Solve the following systems of equations:

$\frac{2}{3 x + 2 y} + \frac{3}{3 x - 2 y} = \frac{17}{5}$

$\frac{5}{3 x + 2 y} + \frac{1}{3 x - 2 y} = 2$

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