Distance of point P(−2,3,4) from the line 3x+2=42y+3=53z+4 measured parallel to the plane 8x+6y−9z+1=0 is

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Distance of point {P}{(-{2},{3},{4})} from the line {{{x}+{2}}

Question

Distance of point $P (- 2, 3, 4)$ from the line $\frac{x + 2}{3} = \frac{2 y + 3}{4} = \frac{3 z + 4}{5}$ measured parallel to the plane $8 x + 6 y - 9 z + 1 = 0$ is

\frac{301}{2}

\frac{401}{2}

\frac{307}{2}

\frac{207}{2}

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Noah Discussed

Distance of point

P (- 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

8 x + 6 y - 9 z + 1 = 0

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Question Text	Distance of point $P (- 2, 3, 4)$ from the line $\frac{x + 2}{3} = \frac{2 y + 3}{4} = \frac{3 z + 4}{5}$ measured parallel to the plane $8 x + 6 y - 9 z + 1 = 0$ is
Answer Type	Video solution: 1
Upvotes	150