The equation of plane containing intersecting lines x -3-3- y -1- z -2-2 and x -3-4- y -2-2- z -6-3A- x-2 y-2 z-9-0B- 2 x-y-z-9-0C- x-y-z-5-0D- x-y-2 z-7-0

The correct option is D x+y 2z+7=0Equation of plane containing both lines is nbsp; x+3 amp;y amp;z 2 3 amp;1 amp;2 4 amp;2 amp;3=0 ⇒ (x+3) y+2( ...

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

Byju's Answer

Standard XII

Mathematics

Equation of a Plane : Point Normal Form

The equation ...

Question

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 _______

x + 2 y - 2 z + 9 = 0

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

2 x - y + z + 9 = 0

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

x + y + z + 5 = 0

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

x + y - 2 z + 7 = 0

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

Open in App

Solution

The correct option is D $x + y - 2 z + 7 = 0$
Equation of plane containing both lines is
$∣ ∣ ∣ ∣ \begin{matrix} x + 3 & y & z - 2 3 & 1 & 2 4 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣ = 0$

$\Rightarrow - (x + 3) - y + 2 (z - 2) = 0 \Rightarrow x + y - 2 z + 7 = 0$

Suggest Corrections

Similar questions

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

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 :

Q. The equation of the plane which contains the lines

L_{1} : x + y + z - 6 = 0 = x - z + 6

and

L_{2} : \frac{x}{2} = \frac{y}{1} = \frac{z - 6}{2}

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 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

The equation of plane containing intersecting lines x+33=y1=z−22 and x−34=y−22=z−63 is _______

The correct option is D x+y−2z+7=0Equation of plane containing both lines is ∣∣ ∣∣x+3yz−2312423∣∣ ∣∣=0 ⇒−(x+3)−y+2(z−2)=0⇒x+y−2z+7=0

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 _______

The correct option is D $x + y - 2 z + 7 = 0$
Equation of plane containing both lines is
$∣ ∣ ∣ ∣ \begin{matrix} x + 3 & y & z - 2 3 & 1 & 2 4 & 2 & 3 \end{matrix} ∣ ∣ ∣ ∣ = 0$

$\Rightarrow - (x + 3) - y + 2 (z - 2) = 0 \Rightarrow x + y - 2 z + 7 = 0$