Question

The distance of the point (-2, 4, -5) from the line $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$ is

• A. $\frac{37}{\sqrt{10}}$
• B. $\frac{37}{10}$
• C. $\sqrt{\frac{37}{10}}$
• D. $\frac{\sqrt{37}}{10}$

Answer 1

The correct option is C $\sqrt{\frac{37}{10}}$

$\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}=t$
$\therefore x=3t-3$
$y=5t+4$ $z=6t-8$
Any point on the given line is $Q(3t-3,5t+4,6t-8)P(-2,4,-5)$
Direction ratios of PQ are $3t-3+2,5t+4-4,6t-8+5$ i.e. $3t-1,5t,6t-3$
PQ is perpendicular to given line
$\therefore 3(3t-1)+5\left(5t\right)+6(6t-3)=0$
$9t-3+25t+36t-18=0$
$70t=21$
$t=\frac{21}{70}=\frac{3}{10}$
$Q[\frac{9}{10}-3,\frac{15}{10}+4,\frac{18}{10}-8]$
$Q[\frac{-21}{10},\frac{55}{10},\frac{-62}{10}]$
$PQ=\sqrt{{(\frac{-21}{10}+2)}^2+{(\frac{55}{10}-4)}^2+{(\frac{-62}{10}+5)}^2}$
$=\sqrt{{\left(\frac{1}{10}\right)}^2+{\left(\frac{15}{10}\right)}^2+{\left(\frac{-12}{10}\right)}^2}$
$=\sqrt{\frac{1+225+144}{100}}=\sqrt{\frac{370}{100}}$
$PQ=\sqrt{\frac{37}{10}}$