JEE MATHEMATICS MCQ: DCS AND DRS

1 Easy

1 of 1

Application +4 | -1

The shortest distance between the lines

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

and

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

lies in the interval :

View answer and explanation

Correct option: B

(2,3]

Option B ((2,3]) is correct because the shortest distance between the given lines is approximately 2.68, which lies in the interval (2,3].

More Questions from DCS AND DRS

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\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

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

a, b, c

\frac{1}{b c}, \frac{1}{c a}, \frac{1}{a b}