JEE MATHEMATICS MCQ: 3D PLANES

1 Medium

1 of 1

Analyzing +4 | -1

If a plane passes through the points

(-1, k, 0)

(2, k,-1),(1,1,2)

and is parallel to the line

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

, then the value of

\frac{k^2+1}{(k-1)(k-2)}

View answer and explanation

Correct option: A

\frac{13}{6}

Option A (

\frac{{13}}{{6}}

) is correct because it is obtained by finding the normal vector to the plane, using the parallel line condition to solve for k, and then substituting k into the expression.

More Questions from 3D PLANES

Practice All Chapter MCQs →

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 : If the planes

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

and

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

& parallel to

P

is: A plane which passes through the point

(3,2,0)

and the line

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

is The plane

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)

and meets

x, y

and

z

axes at

A, B

and

C

respectively. A plane is drawn parallel to

y z

-plane through

A

, a second plane is drawn parallel

z x

-plane through

B

and a third plane is drawn parallel to

x y

-plane through

C

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