Distance of the point $P(\vec{c})$ from the line $\vec{r} = \vec{a}+λ \vec{b}$, is
Answer & explanation
Correct answer: option 1
We have,
QM = Projection of $\vec{QP}$ on $\vec{b}$
$⇒ QM = |\vec{QP}.\vec{b}|= |(\vec{c}-\vec{a}).\vec{b}| = \frac{|(\vec{c}-\vec{a}).\vec{b}|}{|\vec{b}|}$

In right angled triangle PMQ, we have
$PM^2 = PQ^2 - QM^2$
$⇒ PM = \sqrt{|\vec{c}-\vec{a}|^2=\frac{|(\vec{c}-\vec{a}).\vec{b}|^2}{|\vec{b}|^2}}$
$⇒ PM = \sqrt{\frac{|\vec{c} - \vec{a}|^2|\vec{b}|^2-|(\vec{c}-\vec{a}).\vec{b}|^2}{|\vec{b}|^2}}$
$⇒ PM = \frac{|(\vec{c}-\vec{a}).\vec{b}|}{|\vec{b}|}$ $[∵ |\vec{α}×\vec{β}|^2 = |\vec{α}|^2|\vec{β}|^2 - (\vec{α}.\vec{β})^2 ]$