Skip to main content
Back to 3D PLANES

JEE MATHEMATICS MCQ: 3D PLANES

1 Easy
1 of 1
Analyzing +4 | -1
If the three planes x=5,2x−5ay+3z−2=0x=5,2 x-5 a y+3 z-2=0 and 3bx+y−3z=03 b x+y-3 z=0 contain a common line, then (a,b)(a, b) is equal to
View answer and explanation
Correct option: B
(15,−815)\left(\frac{1}{5},-\frac{8}{15}\right)
Option B ((15,−815)\left(\frac{1}{5},-\frac{8}{15}\right)) is correct because it satisfies the condition for the three planes to contain a common line.

More Questions from 3D PLANES

Practice All Chapter MCQs →
Let x−23=y+1−2=z+3−1\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z+3}{-1} lie on the plane px−qy+p x-q y+ z=5z=5, for some p,q∈Rp, q \in \mathbb{R}. The shortest distance of the plane from the origin is : If the planes x−cy−bz=0,cx−y+az=0x-c y-b z=0, c x-y+a z=0 and bx+ay−z=0b x+a y-z=0 pass through a straight line, then the value of a2+b2+c2+2abca^2+b^2+c^2+2 a b c is: Let the plane 2x−3y+9z=02 x-3 y+9 z=0 is PP, the equation of line passing through (0,1,−1)(0,1,-1) and lying-in plane 4x−2y+7z+9=04 x-2 y+7 z+9=0 & parallel to PP is: A plane which passes through the point (3,2,0)(3,2,0) and the line x−41=y−75=z−44\frac{x-4}{1}=\frac{y-7}{5}=\frac{z-4}{4} is The plane ax+a x+ by =0=0 is rotated through an angle α\alpha about its line of intersection with the plane z=0z=0. If the equation of plane in new position is ax+by±za2+b2tan⁡α+c=0\mathrm{ax}+\mathrm{by} \pm \mathrm{z} \sqrt{\mathrm{a}^2+\mathrm{b}^2} \tan \alpha+\mathrm{c}=0. Find c. A variable plane passes through a fixed point (3,2,1)(3,2,1) and meets x,yx, y and zz axes at A,BA, B and CC respectively. A plane is drawn parallel to yzy z-plane through AA, a second plane is drawn parallel zxz x-plane through BB and a third plane is drawn parallel to xyx y-plane through CC. Then the locus of the point of intersection of these three planes, is
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.