Question

The vector $-\hat{i}+\hat{j}+\hat{k}$ bisects angle between the vectors $\overrightarrow{c}$ and $3\hat{i}+4\hat{j}.$ Find unit vector along $\overrightarrow{c}$

Answer 1

Let $x\hat{i}+y\hat{j}+z\hat{k}$ be the unit vectors Along vector c.

since $-\hat{i}+\hat{j}+\hat{k}$ bisects the angle between vector

c & $3\hat{i}+4\hat{j}$

therefore $(-\hat{i}+\hat{j}-\hat{k})=(x\hat{i}+yhatj+z\hat{k})+\frac{3\hat{i}+4\hat{j}}{5}$

$x+\frac{3}{5}=-\lambda$

$y+\frac{4}{5}=-\lambda$ [ on comparing the co. effo of $\hat{i},\hat{j},\hat{k}$]

$z=-\lambda$

Now, $x^2+y^2+z^2=1[$ $\because x\hat{i}+y\hat{j}+z\hat{k}$ is a unit vector]

Or $(-\lambda -\frac{3}{5}{)}^2+(\lambda -\frac{4}{5}{)}^2+{\lambda }^2=1$ putting value of x, y, z.

or $3{\lambda }^2\frac{2}{5}\lambda =0$

$\lambda =0$ & $\lambda =\frac{2}{15}$ [ but $\lambda \ne 0\hat{i}f\lambda =0$ it implies that the given vector are parall]

now $x+\frac{3}{5}=\frac{-2}{15}\Rightarrow x=\frac{-2}{15}-\frac{3}{5}$

$x=\frac{-2-9}{15}=\frac{-11}{15}$

$y+\frac{4}{5}=\frac{-2}{15}$

$\Rightarrow y=\frac{-2}{15}-\frac{4}{5}=\frac{-2-12}{15}=\frac{-14}{15}$

$z=\frac{-2}{15}$

Hence $x\hat{i}+y\hat{j}+z\hat{k}=\frac{-1}{15}(11\hat{i}+10\hat{j}+2\hat{k})$