JEE MATHEMATICS MCQ: DCS AND DRS

1 Easy

1 of 1

Analyzing +4 | -1

Consider the following 3 lines in space

{L_1}:r = 3i - j + 2k + \lambda (2i + 4j - k)

{L_2}:r = i + j - 3k + \mu (4i + 2j + 4k)

{L_3}:r = 3i + 2j - 2k + t(2i + j + 2k)

Then, which one of the following pair(s) is/are in the same plane?

View answer and explanation

Correct option: D

L_1 L_2

and

L_2 L_3

Option D (

L_1 L_2

and

L_2 L_3

) is correct because

L_1

and

L_2

intersect, and

L_2

and

L_3

are parallel, implying they are coplanar.

More Questions from DCS AND DRS

Practice All Chapter MCQs →

\frac{2-x}{3}=\frac{3 y-2}{4 \lambda+1}=4-z

\frac{x+3}{3 \mu}=\frac{1-2 y}{6}=\frac{5-z}{7},

4 \lambda+9 \mu

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

\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}

4 {~m}

2 {~kg}

10 {~kg}

45^{\circ}

60^{\circ}

X

Y

Z

\mathrm{AB}

-2 / \sqrt{17}

3 / \sqrt{17},-2 / \sqrt{17}

\mathrm{AB}=\sqrt{17}

\mathrm{A}

(3,-6,10)

\mathrm{B}

A B

45^{\circ}

120^{\circ}

X

Y

A B

\theta

Z

\theta