Introduction to 3DHard
Question
The equation of a plane passing through the line of intersection of the planes x + 2y + 3z = 2 and x - y + z = 3 and at a distance 
from the point (3, 1, - 1) is
from the point (3, 1, - 1) isOptions
A.5x - 11y + z = 17
B.√2x + y = 3√2 - 1
C.x + y + z = √3
D.x - #8730;2y = 1 - √2
Solution
Equation of required plane is
P ≡ (x + 2y + 3z - 2) + λ(x - y + z - 3) = 0
⇒ (1 + λ)x + (2 - λ)y + (3 + λ)z - (2 + 3λ) = 0
Its distance from (3, 1, - 1) is
⇒ 3λ2 + 4λ+ 14 = 3λ2
-5x + 11y - z + 17 = 0.
P ≡ (x + 2y + 3z - 2) + λ(x - y + z - 3) = 0
⇒ (1 + λ)x + (2 - λ)y + (3 + λ)z - (2 + 3λ) = 0
Its distance from (3, 1, - 1) is
⇒ 3λ2 + 4λ+ 14 = 3λ2-5x + 11y - z + 17 = 0.
Create a free account to view solution
View Solution FreeMore Introduction to 3D Questions
The point of intersection of the line and the plane 2x + 3y + z = 0 is -...The shortest distance between the lines and is -...The distance of the point (1, 2, 3) from x-axis is -...The equation of the plane through intersection of planes x + 2y + 3z = 4 and 2x + y − z = − 5 and perpendicu...A point moves in such a way that sum of square of its distnaces from the co-ordinate axis are 36, then distance of these...