JEE MATHEMATICS MCQ: 3D PLANES

1 Easy

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Analyzing +4 | -1

If the lines

\vec r = \left( {\hat i - \hat j + \hat k} \right) + \lambda \left( {3\hat j - \hat k} \right)

and

\vec r = \left( {\alpha \hat i - \hat j} \right) + \mu \left( {2\hat j - 3\hat k} \right)

are co-planar, then the distance of the plane containing these two lines from the point

\left( {\alpha ,\,0,\,0} \right)

is:

View answer and explanation

Correct option: B

\frac{2}{{11}}

Option B (

\frac{2}{{11}}

) is correct because it represents the distance of the plane containing the coplanar lines from the point

(\alpha, 0, 0)

after finding the value of

\alpha

and the equation of the plane.

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