Question

If the vector $-\hat{i}+\hat{j}-\hat{k}$ bisects the angle between the vector c and the vector$3\hat{i}+4\hat{j}$,then the unit vector in the direction of c,is

Answer 1

$\begin{array}{c}\overrightarrow{a}=3\hat{i}+4\hat{j}\\ \overrightarrow{c}bisect\\ \overrightarrow{b}=-\hat{i}+\hat{j}-\hat{k}\end{array}$

So, we can write

$\begin{array}{cc}\overrightarrow{a}+\overrightarrow{c}=\lambda \overrightarrow{b}& \\ \hat{a}+\hat{c}=\lambda \hat{b}& \left[\hat{a}=\frac{3\hat{i}+4\hat{j}}{\sqrt{9+16}}=\frac{3\hat{i}+4\hat{j}}{\sqrt{15}}=\frac{3\hat{i}+4\hat{j}}{5}\right]\\ \frac{3i}{5}+\frac{4j}{5}+\hat{c}=\lambda \left(\frac{-\hat{i}+\hat{j}-\hat{k}}{\sqrt{3}}\right)\to \left(i\right)& \end{array}$

Let the angle between a & b is $\theta$

$\begin{array}{c}\overrightarrow{a}.\overrightarrow{b}=\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \theta \Rightarrow \cos \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\frac{-3+4}{5\sqrt{3}}=\frac{1}{5\sqrt{3}}\\ \overrightarrow{a}+\overrightarrow{c}=\lambda \overrightarrow{b}\\ \overrightarrow{a}.\overrightarrow{b}+\overrightarrow{c}.\overrightarrow{b}=\lambda \left({\left|\overrightarrow{b}\right|}^2\right)\\ \left|\sigma \right|\left|\overrightarrow{a}\right|\cos \theta +\left|\overrightarrow{c}\right|\left|\overrightarrow{b}\right|\cos \theta =3\lambda \\ 1\times \sqrt{3}\times \frac{1}{5\sqrt{3}}+1.\sqrt{3}\times \frac{1}{5\sqrt{3}}=3\lambda \\ \frac{1}{5}+\frac{1}{5}=3\lambda \Rightarrow \frac{2}{5}=3\lambda \Rightarrow \frac{2}{15}=\lambda \\ \frac{3\hat{i}}{5}+\frac{4\hat{j}}{5}+\hat{c}=\frac{2}{15}\left(-\hat{i}+\hat{j}-\hat{k}\right)\\ \overrightarrow{c}=\frac{-2}{15}\hat{i}+\frac{2}{15}\hat{j}-\frac{2}{15}\hat{k}-\frac{3\hat{i}}{5}-\frac{4\hat{j}}{5}\\ \overrightarrow{c}=\frac{-11}{15}\hat{i}-\frac{10}{15}\hat{j}-\frac{2}{15}\hat{k}\\ \overrightarrow{c}=\frac{-1}{15}\left(11\hat{i}+10\hat{j}+2\hat{k}\right)\end{array}$