Find the distance of the point (−2,3, −4)from the line 3x+2=42y+3=53z+4measured parallel to the plane 4x+12y−3z+1=0.

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Find the distance of the point {(-{2},{3},{ } -{4})}from the l

Question

Find the distance of the point $(- 2, 3, - 4)$ from the line $\frac{x + 2}{3} = \frac{2 y + 3}{4} = \frac{3 z + 4}{5}$ measured parallel to the plane $4 x + 12 y - 3 z + 1 = 0 .$

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Liam Discussed

Find the distance of the point

(- 2, 3, - 4)

from the line

\frac{x + 2}{3} = \frac{2 y + 3}{4} = \frac{3 z + 4}{5}

measured parallel to the plane

4 x + 12 y - 3 z + 1 = 0 .

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Question Text

Find the distance of the point

(- 2, 3, - 4)

from the line

\frac{x + 2}{3} = \frac{2 y + 3}{4} = \frac{3 z + 4}{5}

measured parallel to the plane

4 x + 12 y - 3 z + 1 = 0 .