JEE MATHEMATICS MCQ: 3D PLANES

1 Easy

1 of 1

Application +4 | -1

Let

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

lie on the plane

p x-q y+

z=5

, for some

p, q \in \mathbb{R}

. The shortest distance of the plane from the origin is :

View answer and explanation

Correct option: B

\sqrt{\frac{5}{142}}

Option B (

\sqrt{\frac{5}{142}}

) is correct because after finding the equation of the plane, the shortest distance from the origin is calculated correctly.

More Questions from 3D PLANES

Practice All Chapter MCQs →

x-c y-b z=0, c x-y+a z=0

b x+a y-z=0

a^2+b^2+c^2+2 a b c

2 x-3 y+9 z=0

P

(0,1,-1)

4 x-2 y+7 z+9=0

P

(3,2,0)

\frac{x-4}{1}=\frac{y-7}{5}=\frac{z-4}{4}

a x+

=0

\alpha

z=0

\mathrm{ax}+\mathrm{by} \pm \mathrm{z} \sqrt{\mathrm{a}^2+\mathrm{b}^2} \tan \alpha+\mathrm{c}=0

(3,2,1)

x, y

z

A, B

C

y z

A

z x

B

x y

C

(a, b, c)

(-a,-c,-b)

x y