The distance of the point (-1,9,-16) from the plane 2 x+3 y-z=5 measured parallel to the line (x+4/3)=(2-y/4)=(z-3/12) is

Question

The distance of the point (-1,9,-16) from the plane 2 x+3 y-z=5 measured parallel to the line (x+4/3)=(2-y/4)=(z-3/12) is (A) 20 √2 (B) 31 (C) 13 √2 (D) 26

Tardigrade Admin · Accepted Answer

Equation of line (x+1/3)=(y-9/-4)=(z+16/12) G.P on line (3 λ-1,-4 λ+9,12 λ-16) point of intersection of line plane 6 λ-2-12 λ+27-12 λ+16=5 λ=2 Point (5,1,8) Distance =√36+64+576=26

Q. The distance of the point $(- 1, 9, - 16)$ from the plane $2 x + 3 y - z = 5$ measured parallel to the line $\frac{x + 4}{3} = \frac{2 - y}{4} = \frac{z - 3}{12}$ is

$202$

31

$132$

26

Equation of line
$\frac{x + 1}{3} = \frac{y - 9}{- 4} = \frac{z + 16}{12}$
G.P on line $(3 λ - 1, - 4 λ + 9, 12 λ - 16)$
point of intersection of line \& plane
$6 λ - 2 - 12 λ + 27 - 12 λ + 16 = 5$
$λ = 2$
Point $(5, 1, 8)$
Distance $= 36 + 64 + 576 = 26$

Q. The distance of the point (−1,9,−16) from the plane 2x+3y−z=5 measured parallel to the line 3x+4​=42−y​=12z−3​ is

202​

31

132​

26

Equation of line 3x+1​=−4y−9​=12z+16​ G.P on line (3λ−1,−4λ+9,12λ−16) point of intersection of line \& plane 6λ−2−12λ+27−12λ+16=5 λ=2 Point (5,1,8) Distance =36+64+576​=26

Q. The distance of the point $(- 1, 9, - 16)$ from the plane $2 x + 3 y - z = 5$ measured parallel to the line $\frac{x + 4}{3} = \frac{2 - y}{4} = \frac{z - 3}{12}$ is

$202$

$132$

Equation of line
$\frac{x + 1}{3} = \frac{y - 9}{- 4} = \frac{z + 16}{12}$
G.P on line $(3 λ - 1, - 4 λ + 9, 12 λ - 16)$
point of intersection of line \& plane
$6 λ - 2 - 12 λ + 27 - 12 λ + 16 = 5$
$λ = 2$
Point $(5, 1, 8)$
Distance $= 36 + 64 + 576 = 26$