JEE MATHEMATICS MCQ: 3D GEOMETRY SYSTEM

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Application +4 | -1

The shortest distance (S.D) between lines

\frac{x-3}{3}=\frac{y-8}{-1}=\frac{z-3}{1}

and

\frac{x+3}{-3}-\frac{y+7}{2}=\frac{z-6}{4}

View answer and explanation

Correct option: A

3 \sqrt{30}

Option A (

3 \sqrt{30}

) is correct because it represents the shortest distance between the two given lines, calculated by finding a common perpendicular.

More Questions from 3D GEOMETRY SYSTEM

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\left( { - 10,\;8} \right)

\left( { - 6,\;12} \right)

\left( {4,\; - 2} \right)

\left( { - 2,\;4} \right)

y z

(1,-2,4)

(3,-4

5)

p

2 x+y-3=0

3 x-y-2=0

x-p y+5=0

O\left( {0,0} \right),A\left( {a,0} \right)

B\left( {0,a} \right).

A(1,0,3)

B(4,7,1)

C(3,5,3)

\frac{x}{1}=\frac{y}{2}=\frac{z}{3}, \frac{x}{2}=\frac{y}{1}=\frac{z}{3}

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