JEE MATHEMATICS MCQ: 3D PLANES

1 Easy

1 of 1

Application +4 | -1

The vector equation of the plane through the point

(2,1,-1)

and passing through the line of intersection of the plane

\mathbf{r} \cdot(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}})=0

and

\mathbf{r} \cdot(\hat{\mathbf{j}}+2 \hat{\mathbf{k}})=0

, is

View answer and explanation

Correct option: A

\mathbf{r} \cdot(\hat{\mathbf{i}}+9 \hat{\mathbf{j}}+11 \hat{\mathbf{k}})=0

Option A is correct because the vector equation of the plane is found by using the line of intersection of two planes and the given point to solve for the parameter and obtain the final equation.

More Questions from 3D PLANES

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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