Line of intersection of the planes x-2 y-0 and y-3 z-3-0 isA- x -6- y -3- z -1B- x-2-y-3-1-z-1C- x-6-2-y-31-1-z-2-10D- x-6-6-y-3-3-z-2-1

The correct option is D x+6 6=y 33=z 21y= x/2, y=3z 3 x/ 2=y/1=z 1/1/3 ⇒x/ 6=y/3=z 1/1 nbsp; The above equation represents a line passing through (0, 0, 1) ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Condition for a Line to Lie on a Plane

Line of inter...

Question

Line of intersection of the planes $x + 2 y = 0$ and $y - 3 z + 3 = 0$ is

\frac{x}{- 6} = \frac{y}{3} = \frac{z}{1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{x + 6}{- 6} = \frac{y - 3}{3} = \frac{z - 2}{1}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{x + 6}{- 2} = \frac{y - 3}{1} = \frac{- z - 2}{1}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{x + 6}{- 6} = \frac{y - 3}{3} = \frac{z - 2}{1}$
$y = - \frac{x}{2}$ , $y = 3 z - 3$

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

$\Rightarrow \frac{x}{- 6} = \frac{y}{3} = \frac{z - 1}{1}$
The above equation represents a line passing through $(0, 0, 1)$ with DR's $- 6, 3, 1$ .

$\frac{x + 6}{- 6} = \frac{y - 3}{3} = \frac{z - 2}{1}$ have same DR's and passes through $(0, 0, 1)$

Suggest Corrections

Similar questions

Q. Equation of the line passing through the point

(2, 3, 1)

and parallel to the line of the intersection of the plane

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

Q. The point of intersection of line

\frac{x - 6}{- 1} = \frac{y + 1}{0} = \frac{z + 3}{4}

and plane

x + y - z = 3

Q. The equation of plane containing intersecting lines

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

and

\frac{x - 3}{4} = \frac{y - 2}{2} = \frac{z - 6}{3}

is _______

Q. The equation of the line passing through

(- 4, 3, 1)

, parallel to the plane

x + 2 y - z - 5 = 0

and intersecting the line

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

is:

Q. The equation of the plane containing the two lines of intersection of the two pairs of planes x + 2y – z – 3 = 0 and 3x – y + 2z – 1 = 0, 2x – 2y + 3z = 0 and x – y + z + 1 =0 is :

Join BYJU'S Learning Program

Line of intersection of the planes x+2y=0 and y−3z+3=0 is

The correct option is B x+6−6=y−33=z−21y=−x2, y=3z−3 x−2=y1=z−113 ⇒x−6=y3=z−11 The above equation represents a line passing through (0,0,1) with DR's −6,3,1. x+6−6=y−33=z−21 have same DR's and passes through (0,0,1)

Line of intersection of the planes $x + 2 y = 0$ and $y - 3 z + 3 = 0$ is