If $4 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+8 \hat{\mathbf{k}}, 2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, 2 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}$ are respectively the position vectors of the vertices $A, B, C$ of $\triangle A B C$, then the position vector of the point where the bisector of angle $A$ meet $\mathbf{B C}$ is

Question

If $4 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+8 \hat{\mathbf{k}}, 2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}, 2 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}$ are respectively the position vectors of the vertices $A, B, C$ of $\triangle A B C$, then the position vector of the point where the bisector of angle $A$ meet $\mathbf{B C}$ is, $2 \hat{\mathbf{i}}+\frac{13}{3} \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$, $2 \hat{\mathbf{i}}-\frac{13}{3} \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$, $2 \hat{\mathbf{i}}+13 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$, $2 \hat{\mathbf{i}}+\frac{13}{3} \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$

MARKS App · Accepted Answer

As we know bisector of angle $A$ divides the $B C$ in ratio $c: b$. where $c$ is length of side $A B=\sqrt{4+16+16}=6$ and $b$ is length of side $A C=\sqrt{4+4+1}=3$ $\therefore$ Position vector of the point where the bisector of angle $A$ meet $\mathbf{B C}$ is $\begin{aligned} & \frac{6(2 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+7 \hat{\mathbf{k}})+3(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})}{6+3} \ & =\frac{18 \hat{\mathbf{i}}+39 \hat{\mathbf{j}}+54 \hat{\mathbf{k}}}{9}=2 \hat{\mathbf{i}}+\frac{13}{3} \hat{\mathbf{j}}+6 \hat{\mathbf{k}} \end{aligned}$ Hence, option (4) is correct.

Search any question & find its solution

Looking for more such questions to practice?