Equation of the plane perpendicular to the line x -1-3- y -5-2- z -3-4 and passing through the point 1-2-3 is A- 3 x-2 y-4 z-13-0B- 3 x-2 y-4 z-13-0C- x-5 y-3 z-18-0D- x-5 y-3 z-18-0

The correct option is A 3x+2y 4z+13=0 nbsp; nbsp; nbsp;Direction ratios of the normal vector to the plane will be (3,2, 4). Also, the plane passes throu ...

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Standard XII

Mathematics

Angle between a Plane and a Line

Equation of t...

Question

Equation of the plane perpendicular to the line $\frac{x - 1}{3} = \frac{y - 5}{2} = \frac{z + 3}{- 4}$ and passing through the point $(1, - 2, 3)$ is

3 x + 2 y - 4 z + 13 = 0

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3 x + 2 y - 4 z - 13 = 0

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x + 5 y - 3 z + 18 = 0

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x + 5 y - 3 z - 18 = 0

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Solution

The correct option is A $3 x + 2 y - 4 z + 13 = 0$

Direction ratios of the normal vector to the plane will be $(3, 2, - 4) .$
Also, the plane passes through $(1, - 2, 3) .$
Therefore, equation of the plane is
$3 (x - 1) + 2 (y + 2) - 4 (z - 3) = 0$
$\Rightarrow 3 x + 2 y - 4 z + 13 = 0$

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Similar questions

Q. The equation of the plane containing the straight line

\frac{x - 1}{2} = \frac{y + 2}{- 3} = \frac{z}{5}

and perpendicular to the plane

x - y + z + 2 = 0

Q. The equation of the plane containing the straight line

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

and perpendicular to the plane containing the straight lines

\frac{x}{3} = \frac{y}{4} = \frac{z}{2} and \frac{x}{4} = \frac{y}{2} = \frac{z}{3}

is :

Equation of the plane containing the straight line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the staight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is

In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

(a)

(b)

(c)

(d)

(e)

The equation of the plane passing through the origin and perpendicular to the line x=2 y=3 z is

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Equation of the plane perpendicular to the line x−13=y−52=z+3−4 and passing through the point (1,−2,3) is

The correct option is A 3x+2y−4z+13=0 Direction ratios of the normal vector to the plane will be (3,2,−4). Also, the plane passes through (1,−2,3). Therefore, equation of the plane is 3(x−1)+2(y+2)−4(z−3)=0 ⇒3x+2y−4z+13=0

Equation of the plane perpendicular to the line $\frac{x - 1}{3} = \frac{y - 5}{2} = \frac{z + 3}{- 4}$ and passing through the point $(1, - 2, 3)$ is

The correct option is A $3 x + 2 y - 4 z + 13 = 0$

Direction ratios of the normal vector to the plane will be $(3, 2, - 4) .$
Also, the plane passes through $(1, - 2, 3) .$
Therefore, equation of the plane is
$3 (x - 1) + 2 (y + 2) - 4 (z - 3) = 0$
$\Rightarrow 3 x + 2 y - 4 z + 13 = 0$