You visited us 3 times! Enjoying our articles? Unlock Full Access!

Definition of Vector

Let two plane...

Question

Let two planes $p_{1} : 2 x - y + z = 2$ , and $p_{2} : x + 2 y - z = 3$ are given. The equation of the bisector of angle of the planes $P_{1}$ and $P_{2}$ which does not contains origin, is

x - 3 y + 2 z + 1 = 0

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

x + 3 y = 5

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

x + 3 y + 2 z + 2 = 0

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

3 x + y = 5

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

Open in App

Solution

The correct option is A $3 x + y = 5$
Given planes are $p_{1} : 2 x - y + z = 2$ and $p_{2} : x + 2 y - z = 3$

Normals to the planes
$N_{1} : \frac{1}{\sqrt{6}} (2, - 1, 1)$
$N_{2} : \frac{1}{\sqrt{6}} (1, 2, - 1)$
Let $N$ be the normal vector of angle bisector
$N = N_{1} + N_{2}$ or $N_{1} - N_{2}$
$N = (3, 1, 0)$ or $(1, - 3, 2)$
The equation of plane is
$P = P_{1} + λ P_{2}$
$P = 2 x - y + z - 2 + λ (x + 2 y - z - 3)$
If $N = (3, 1, 0)$ , then $λ = 1$ ,
Equation of Plane $= P = 3 x + y - 5$
It does not pass through origin.

Hence, option D is correct.

Suggest Corrections

Similar questions

p_{1} : 2 x - y + z = 2

p_{2} : x + 2 y - z = 3

P_{1}

P_{2}

\frac{x - 2}{4} = \frac{y + 2}{8} = \frac{z - 3}{6}

P_{1} : 2 x + 3 y - 4 z + 6 = 0

P_{1}

P_{2}

P_{1}

P_{2}

P_{2}

p_{1} : 2 x - y + z = 2

p_{2} : x + 2 y - z = 3

P_{1}

P_{2}

P_{1} : 4 x - 2 y + 6 z = 3

P_{2} : 4 x - 2 y - 2 z = 6

P_{3} : - 6 x + 3 y - 9 z = 5

P_{4} : 2 x - y - z = 3

p_{1} : 2 x - y + z = 2

p_{2} : x + 2 y - z = 3

P_{1}

P_{2}

(3, 2, 1)

Join BYJU'S Learning Program