Back to 3D PLANES

JEE MATHEMATICS MCQ: 3D PLANES

1 Easy

1 of 1

Analyzing +4 | -1

The shortest distance between the

\mathrm{Z}

-axis and the line

x+y+2 z-3=0,2 x+3 y+4 z-4=0

is

View answer and explanation

Correct option: B

2

The correct answer is B. The shortest distance between the Z-axis and the given line is found by finding a plane containing the line that is parallel to the Z-axis, and then calculating the distance from a point on the Z-axis to that plane, which results in 2.

More Questions from 3D PLANES

Practice All Chapter MCQs →

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

lie on the plane

p x-q y+

z=5

p, q \in \mathbb{R}

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

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

b x+a y-z=0

pass through a straight line, then the value of

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

is: Let the plane

2 x-3 y+9 z=0

P

, the equation of line passing through

(0,1,-1)

and lying-in plane

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

P

is: A plane which passes through the point

(3,2,0)

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

a x+

=0

is rotated through an angle

\alpha

about its line of intersection with the plane

z=0

. If the equation of plane in new position is

\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)

x, y

z

A, B

C

respectively. A plane is drawn parallel to

y z

A

, a second plane is drawn parallel

z x

B

and a third plane is drawn parallel to

x y

C

. Then the locus of the point of intersection of these three planes, is