If the perpendicular distance from the point $(- 1, 2, - 2)$ to the plane $3 x - 4 y + 2 z + k = 0$ is $\sqrt{29}$ , then $k =$ ____
( $k < 0$ ).

44

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44

and

- 14

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- 44

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- 14

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Solution

The correct option is D $- 14$
given the perpendicular distance

from the formula of perpendicular distance of a point to a plane we get

$| \frac{3 (- 1) - 4 (2) + 2 (- 2) + k}{\sqrt{9 + 16 + 4}} | = \sqrt{29}$

$| k - 15 | = 29$

$\Rightarrow$ k=44 or -14

but $k < 0$

therefore

$k = - 14$

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. Find the equation of this perpendicular and the perpendicular distance between the point and the plane.

\frac{18}{29} \times \frac{44}{35} = \frac{- 44}{35} \times \frac{18}{29}

If true then enter

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and if false then enter

0

Image of a point P(2, -3, 1 ) with respect to line L is I. Find the coordinates of I, if foot of the perpendicular of P with respect to L is $(\frac{- 22}{7}, \frac{- 3}{14}, \frac{- 13}{14})$

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If the perpendicular distance from the point (−1,2,−2) to the plane 3x−4y+2z+k=0 is √29, then k= ____(k<0).

The correct option is D −14given the perpendicular distancefrom the formula of perpendicular distance of a point to a plane we get|3(−1)−4(2)+2(−2)+k√9+16+4|=√29|k−15|=29⇒k=44 or -14but k<0therefore k=−14

If the perpendicular distance from the point $(- 1, 2, - 2)$ to the plane $3 x - 4 y + 2 z + k = 0$ is $\sqrt{29}$ , then $k =$ ____
( $k < 0$ ).

The correct option is D $- 14$
given the perpendicular distance

from the formula of perpendicular distance of a point to a plane we get

$| \frac{3 (- 1) - 4 (2) + 2 (- 2) + k}{\sqrt{9 + 16 + 4}} | = \sqrt{29}$

$| k - 15 | = 29$

$\Rightarrow$ k=44 or -14

but $k < 0$

therefore

$k = - 14$