The line $\frac{x - 2}{4} = \frac{y + 1}{- 1} = \frac{z - 1}{3}$ is perpendicular to the plane

2 x - y + z = 3

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4 x - y + 3 z = 1

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4 x + y + 3 z = 1

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x + 7 y + z = 3

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Solution

The correct option is B $4 x - y + 3 z = 1$
A line and a plane is perpendicular if the direction ratio of the line is parallel to the normal vector of the plane.
$\frac{x - 2}{4} = \frac{y + 1}{- 1} = \frac{z - 1}{3}$
Direction ratios of the line are $(4, - 1, 3)$
The plane in option (b) has the normal vector as $(4, - 1, 3)$
$∴$ Option (b) is correct.

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Similar questions

\frac{x - 2}{4} = \frac{y + 1}{- 1} = \frac{z - 1}{3}

(a) 2 x - y + z = 3

(b) 4 x - y + 3 z = 1

(c) 4 x + y + 3 z = 1

(d) x + 7 y + z = 3

x + 5 y + 4 z = 3

2 x + 2 y + 3 z = 4

4 x - 2 y + z = 9

x, y, z

Equation of the line passing through (1,1,1) and perpendicular to $2 x - 3 y + z = 0$ is

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

x + y + z = 1

(2, 3, 1)

x - 2 y - z + 5 = 0 and x + y + 3 z = 6

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