JEE MATHEMATICS MCQ: 3D PLANES

1 Easy

1 of 1

Application +4 | -1

A vector

\vec{n}

is inclined to

x

-axis at

45^{\circ}

, to

y

-axis at

60^{\circ}

and at an acute angle to

z

-axis. If

\vec{n}

is a normal to a plane passing through the point

(\sqrt{2},-1,1)

then the equation of the plane is:

View answer and explanation

Correct option: B

2 x+y+2 z=2 \sqrt{2}+1

Option B (

2 x+y+2 z=2 \sqrt{{2}}+1

) is correct because it represents the equation of the plane with the given normal vector and passing through the specified point.

More Questions from 3D PLANES

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

p x-q y+

z=5

p, q \in \mathbb{R}

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