If $\vec a, \vec b, \vec c$ are position vectors of three non-collinear points A, B, C respectively, the shortest distance of A from BC is
Answer & explanation
Correct answer: option 2
The vector equation of the line BC is
$\vec r=\vec b+λ(\vec c-\vec b)$
Clearly,
BM = Projection of $\vec{BA}$ on $\vec{BC}$
$⇒BM =\frac{\vec{BA}.\vec{BC}}{|\vec{BC}|}=\frac{(\vec a-\vec b).(\vec c-\vec b)}{|\vec c-\vec b|}$
∴ Required distance = $AM=\sqrt{AB^2-BM^2}$
⇒ Required distance = $\sqrt{|\vec b-\vec a|^2-\left(\frac{(\vec a-\vec b)(\vec c-\vec b)}{|\vec c-\vec b|}\right)^2}$