Shortest distance between the lines ( x -1/2)=( y +8/-7)=( z -4/5) and ( x -1/2)=( y -2/1)=( z -6/-3) is

Question

Shortest distance between the lines ( x -1/2)=( y +8/-7)=( z -4/5) and ( x -1/2)=( y -2/1)=( z -6/-3) is (A) 2 √3 (B) 4 √3 (C) 3 √3 (D) 5 √3

Tardigrade Admin · Accepted Answer

(x-1/2)=(y+8/-7)=(z-4/5) vec a = hat i -8 hat j +4 hat k ( x -1/2)=( y -2/1)=( z -6/-3) vec b = hat i +2 hat j +6 hat k vec p =2 hat i -7 hat j +5 hat k , vec q =2 hat i + hat j -3 hat k vec p × vec q =| hat i hat j hat k 2 -7 5 2 1 -3| = hat i (16)- hat j (-16)+ hat k (16) =16( hat i + hat j + hat k ) d =|(( a - b ) ⋅( vec p × vec q )/| vec p × vec q |)|=|((-10 hat j -2 hat k ) ⋅ 16( hat i + hat j + hat k ) mid/16 √3)| =|(-12/√3)|=4 √3

Q. Shortest distance between the lines $\frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}$ and $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}$ is

$23$

$43$

$33$

$53$

Q. Shortest distance between the lines 2x−1​=−7y+8​=5z−4​ and 2x−1​=1y−2​=−3z−6​ is

23​

43​

33​

53​

Q. Shortest distance between the lines $\frac{x - 1}{2} = \frac{y + 8}{- 7} = \frac{z - 4}{5}$ and $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z - 6}{- 3}$ is

$23$

$43$

$33$

$53$