Question

Given the vectors $\overrightarrow{AB}=b$ and $\overrightarrow{AC}=c$ coincident with two sides of a triangle ABC. Find resolution (w.r.t. the basis b, c) of the vector drawn from the vertex B of the $\Delta ABC$ and coinciding with the altitude BD.

• A. $BD=\frac{b.c}{{\left|b\right|}^2}c+b.$
• B. $BD=\frac{b.c}{{\left|b\right|}^2}c-b.$
• C. $BD=\frac{b.c}{{\left|c\right|}^2}c+b.$
• D. $BD=\frac{b.c}{{\left|c\right|}^2}c-b.$

Answer 1

The correct option is D $BD=\frac{b.c}{{\left|c\right|}^2}c-b.$

We are given: $\overrightarrow{AB}=b,\overrightarrow{AC}=c$ and $\overrightarrow{BD}\perp \overrightarrow{AC}.$

To find the resolution of $\overrightarrow{BD}$ w. r. t. b and c.

$\overrightarrow{BD}=\overrightarrow{AD}-AB$ ...(1)

But $\overrightarrow{AD}$ =projection of b on c$=\frac{b.c}{\left|c\right|}.$

Also the unit vector in the direction of AC is $\frac{c}{\left|c\right|}.$

Hence $\overrightarrow{AD}=\left(\frac{b.c}{\left|c\right|}\right)\left(\frac{c}{\left|c\right|}\right)=\frac{b.c}{{\left|c\right|}^2}c.$

Hence from (1), we get. $BD=\frac{b.c}{{\left|c\right|}^2}c-b.$