Question

The shortest distance between the straight lines through the points $A_1=(6,2,2)$ and
$A_2=(-4,0,-1)$, in the directions of $(1,-2,2)$ and $(3,-2,-2)$ is

• A. 6
• B. 8
• C. 12
• D. 9

Answer 1

Answer: (D)

Equation of first line,
$\frac{x-6}{1}=\frac{y-2}{2}=\frac{z-2}{2}=k\left(say\right)$
$\therefore x=k+6,y=-2k+2,z=2k+2$
Hence, general point on the first line,
$P\equiv \left(k+6,-2k+2,2k+2\right)$
Equation of second line,
$\frac{x+4}{3}=\frac{y}{-2}=\frac{Z+1}{-2}=l\left(say\right)$
$\therefore x=3l-4,y=-2l,z=-2l-1$
Hence, general point on the second line,
$Q\equiv \left(3l-4,-2l,-2l-1\right)$
Direction ratios of PQ are
$3l-4-k-6,-2l+2k-2,-2l-1-2k-2$
i.e. $3l-k-10,-2l+2k-2,-2l-2k-3$
Now |PQ| will be the shortest distance
between the two lines if PQ is perpendicular
to both the lines. Hence,
$1\left(3l-k-10\right)+\left(-2\right)$
$\left(-2l+2k-2\right)+2\left(-2l-2k-3\right)=0$
and $3\left(3l-k-10\right)+\left(-2\right)\left(-2l+2k-2\right)+\left(-2\right)\left(-21-2k-3\right)=0$
$i.e.3l-9k=12orl-3k=4...\left(i\right)$
and $17l-3k=20...\left(ii\right)$
Subtracting equation $\left(i\right)$ from$\left(ii\right)$, we get
$161=16\therefore l=1$
Putting this value of l in equation \left(i\right), we get
$-3k=3,\therefore k=-1$
$\therefore P\equiv \left(-1+6,-2\left(-1\right)+2,2\left(-1\right)+2\right)$
$\equiv \left(5,4,0\right)$
Similarly, $Q=\left(-1,-2,-3\right)$
Hence, shortest distance, PQ,
=$\sqrt{{\left(-1-5\right)}^2+{\left(-2-4\right)}^2+{\left(-3-0\right)}^2}$
$=\sqrt{{\left(-6\right)}^2+{\left(-6\right)}^2+{\left(-3\right)}^2}=\sqrt{36+36+9}$
=9 units