question_answer

The distance of the point (4, 2, 5) from the line\[\frac{x-1}{6}=\frac{y+2}{3}=\frac{z+1}{1}\] is

A) 3         

B) 4     

C) 5                     

D) 2

Correct Answer: C

Solution :

[c] : The distance of the point C from the line \[\vec{r}=\vec{a}+s\vec{b}\] is \[\frac{|(\vec{c}-\vec{a})\times \vec{b}|}{|\vec{b}|}\] Here, \[\vec{a}=\hat{i}-2\hat{j}-\hat{k},\vec{c}=4\hat{i}+2\hat{j}+5\hat{k}\]and \[\vec{b}=6\hat{i}+3\hat{j}+2\hat{k}\] \[\therefore \] \[=-10\hat{i}+30\hat{j}-15\hat{k}=5(-2\hat{i}+6\hat{j}-3\hat{k})\] Hence,\[d=\frac{|(\vec{c}-\vec{a})\times \vec{b}|}{|\vec{b}|}=\frac{5\times 7}{7}=5\].