JEE MATHEMATICS MCQ: 3D PLANES

1 Easy

1 of 1

Analyzing +4 | -1

Statement I The point

A(3,1,6)

is the mirror image of the point

B(1,3,4)

in the plane

x-y+z=5

.

Statement II The plane

x-y+z=5

bisects the line segment joining

A(3,1,6)

and

B(1,3,4)

View answer and explanation

Correct option: A

Statement

\mathrm{I}

is true, Statement

\mathrm{II}

is true;
Statement II is the correct explanation of Statement I

Option A (Statement I is true, Statement II is true; Statement II is the correct explanation of Statement I) is correct because the point A is indeed the mirror image of B in the given plane, and the plane bisects the line segment joining A and B.

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