Question

The equation of the line of shortest distance between the lines $\frac{x+4}{4}=\frac{y-2}{-2}=\frac{z-3}{0}$ and $\frac{x-5}{5}=\frac{y-3}{3}=\frac{z}{0}$, is

• A. $\frac{x+4}{0}=\frac{y-2}{0}=\frac{z-3}{1}$
• B. $\frac{x-5}{0}=\frac{y-3}{0}=\frac{z}{1}$
• C. $\frac{x}{0}=\frac{y}{0}=\frac{z-3}{1}$
• D. None of the above

Answer 1

Answer: (C)

$\frac{x}{0}=\frac{y}{0}=\frac{z-3}{1}$

Since, line of shortest distance is perpendicular to both the lines, its direction ratios can be obtained by cross-product of direction ratios of the two lines.

$(4\hat{i}-2\hat{j})\times (5\hat{i}+3\hat{j})=22\hat{k}$

Direction of line of shortest distance$=\hat{k}$

Let $\frac{x+4}{4}=\frac{y-2}{-2}=\frac{z-3}{0}=a$

and $\frac{x-5}{5}=\frac{y-3}{3}=\frac{z}{0}=b$

Point of contact of first line and line of shortest distance $=(4a-4,-2a+2,3)$

Point of contact of second line and line of shortest distance $=(5b+5,3b+3,0)$

Since, line of shortest distance is perpendicular to both the lines,

$4(4a-5b-9)-2(-2a-3b-1)=0\Rightarrow 10a-7b=17$

$5(4a-5b-9)+3(-2a-3b-1)=0\Rightarrow 7a-17b=24$

On solving, we get $a=1,b=-1$

Substituting $a=1,$ we get a point $(0,0,3)$ that lies on the line of shortest distance

So, equation of line of shortest distance $:\frac{x}{0}=\frac{y}{0}=\frac{z-3}{1}$

Hence, Option (C)