The shortest distance between the lines (x/2) = (y/2) = (z/1) and (x +2/-1) = (y -4/8) = (z -5/4) lies in the interval :

Question

The shortest distance between the lines (x/2) = (y/2) = (z/1) and (x +2/-1) = (y -4/8) = (z -5/4) lies in the interval : (A) [0, 1) (B) [1, 2) (C) (2, 3] (D) (3, 4]

Tardigrade Admin · Accepted Answer

Shortest distance ( beginvmatrixx2-x1&y2-y1&z2-z1 a1&b1&c1 a2&b2&c2 endvmatrix/ beginVmatrixi&j&k a1&b1&c1 a2&b2&c2 endVmatrix) =( beginvmatrix-2&4&5 2&2&1 -1&3&4 endvmatrix/ beginvmatrixi&j&k 2&2&1 -1 &8&4 endvmatrix)=(254/20.1)=2.7

Q. 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 :

$[0, 1)$

$[1, 2)$

$(2, 3]$

$(3, 4]$

Q. The shortest distance between the lines 2x​=2y​=1z​ and −1x+2​=8y−4​=4z−5​ lies in the interval :

[0,1)

[1,2)

(2,3]

(3,4]

Shortest distance ∥∥​ia1​a2​​jb1​b2​​kc1​c2​​∥∥​∣∣​x2​−x1​a1​a2​​y2​−y1​b1​b2​​z2​−z1​c1​c2​​∣∣​​ =∣∣​i2−1​j28​k14​∣∣​∣∣​−22−1​423​514​∣∣​​=20.1254​=2.7

Q. 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 :

$[0, 1)$

$[1, 2)$

$(2, 3]$

$(3, 4]$