Equation of a line in the plane π=2x−y+z−4=0 which is perpendicular to the line l whose equation is 1x−2=−1y−2=−2z−3 and which passes through the point of intersection of l and π is (A) 1x−2=5y−1=−1z−1 (B) 3x−1=5y−3=−1z−5 (C) 2x+2=−1y+1=1z+1 (D) 2x−2=−1y−1=1z−1

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Equation of a line in the plane ={2}{x}-{y}+{z}-{4}={0} which

Question

Equation of a line in the plane $π = 2 x - y + z - 4 = 0$ which is perpendicular to the line $l$ whose equation is $\frac{x - 2}{1} = \frac{y - 2}{- 1} = \frac{z - 3}{- 2}$ and which passes through the point of intersection of $l$ and $π$ is (A) $\frac{x - 2}{1} = \frac{y - 1}{5} = \frac{z - 1}{- 1}$ (B) $\frac{x - 1}{3} = \frac{y - 3}{5} = \frac{z - 5}{- 1}$ (C) $\frac{x + 2}{2} = \frac{y + 1}{- 1} = \frac{z + 1}{1}$ (D) $\frac{x - 2}{2} = \frac{y - 1}{- 1} = \frac{z - 1}{1}$

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Elizabeth Discussed

Equation of a line in the plane

π = 2 x - y + z - 4 = 0

which is perpendicular to the line

l

whose equation is

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

and which passes through the point of intersection of

l

and

π

is (A)

\frac{x - 2}{1} = \frac{y - 1}{5} = \frac{z - 1}{- 1}

(B)

\frac{x - 1}{3} = \frac{y - 3}{5} = \frac{z - 5}{- 1}

(C)

\frac{x + 2}{2} = \frac{y + 1}{- 1} = \frac{z + 1}{1}

(D)

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

11 mins ago

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