Equation of a plane which passes through the point of intersection of lines x-13-y-21-z-32 and x-31-y-12-z-23 and at greatest distance from the origin is

The correct option is B 4x+3y+5z=50 L_1:x 13=y 21=z 32 L_2:x 31=y 12=z 23 Any point on L_1(3k_1+1,k_1+2,2k_1+3) Any point on L_2(k_2+3 ...

214

You visited us 214 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Introduction

Equation of a...

Question

Equation of a plane which passes through the point of intersection of lines
$\frac{x - 1}{3} = \frac{y - 2}{1} = \frac{z - 3}{2}$ and $\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ and at greatest distance from the origin is

7 x + 2 y + 4 z = 54

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

4 x + 3 y + 5 z = 50

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

3 x + 4 y + 5 z = 49

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

x + y + z = 12

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

Open in App

Solution

The correct option is B $4 x + 3 y + 5 z = 50$
$L_{1} : \frac{x - 1}{3} = \frac{y - 2}{1} = \frac{z - 3}{2}$

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

Any point on $L_{1} (3 k_{1} + 1, k_{1} + 2, 2 k_{1} + 3)$
Any point on $L_{2} (k_{2} + 3, 2 k_{2} + 1, 3 k_{2} + 2)$
$∴ 3 k_{1} + 1 = k_{2} + 3$
$\Rightarrow 3 k_{1} - k_{2} - 2 = 0$ ....(i)
$k_{1} + 2 = 2 k_{2} + 1$
$\Rightarrow k_{1} - 2 k_{2} + 1 = 0$ ......(ii)
$(i) \times (i i)$
$\frac{\begin{matrix} 6 k_{1} - 2 k_{2} - 4 = 0 k_{1} - 2 k_{2} + 1 = 0 - + - \end{matrix}}{5 k_{1} = 5}$
$∴ k_{1} = 1$
$∴$ point of intersection is $(4, 3, 5)$
$\to P O =^n$
Direction of $O P (4, 3, 5)$
Equation of point is
$4 (x - 4) + 3 (y - 3) + 5 (z - 5) = 0$
$\Rightarrow 4 x + 3 y + 5 z = 50$

Suggest Corrections

Similar questions

Q. The equation of the plane which passes through the point of intersection of lines

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

and

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

and at greatest distance from origin is

Q. Find the equation of the plane passing through the line of intersection of planes

2 x - y + z = 3

4 x - 3 y + 5 z + 9 = 0

and parallel to the line

\frac{x + 1}{2} = - \frac{y + 3}{4} = \frac{z - 3}{5}

The equation of a plane passing through the line of intersection of the planes $x + 2 y + 3 z = 2$ and $x - y + z = 3$ and at a distance $\frac{2}{\sqrt{3}}$ from the point (3,1,-1) is

Q. Find the coordinates of the point of intersection of the line

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

with the plane

3 x + 4 y + 5 z = 5

Join BYJU'S Learning Program

Equation of a plane which passes through the point of intersection of linesx−13=y−21=z−32 and x−31=y−12=z−23 and at greatest distance from the origin is

Equation of a plane which passes through the point of intersection of lines
$\frac{x - 1}{3} = \frac{y - 2}{1} = \frac{z - 3}{2}$ and $\frac{x - 3}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$ and at greatest distance from the origin is