Question

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

Answer 1

Let $\hat{c}=x\hat{i}+y\hat{j}+z\hat{k}$. where $x^2+y^2+z^2=1......\left(2\right)$

And unit vector along $3\hat{i}+4\hat{j}=\frac{3\hat{i}+4\hat{j}}{\sqrt{3^2+4^4}}=\frac{3\hat{i}+4\hat{j}}{5}$

The bisectors of these two is given by

$r=t(\hat{a}+\hat{b})$

$\Rightarrow r=t(x\hat{i}+y\hat{j}+z\hat{k}+\frac{3\hat{i}+4\hat{j}}{5})$

$\Rightarrow r=\frac{t}{5}\left(\right(5x+3)\hat{i}+(5y+4)\hat{j}+5\hat{k})$ ..... $\left(2\right)$

But the bisector is given by $-\hat{i}+\hat{j}-\hat{k}$ ........ $\left(3\right)$

Comparing $\left(2\right)$ and $\left(3\right)$, we get,

$\frac{t}{5}(5x+3)=-1\Rightarrow x=-\frac{5+3t}{5t}$

$\frac{t}{5}(5y+4)=1\Rightarrow y=\frac{5-4t}{5t}$

$\frac{t}{5}=-1\Rightarrow z=-\frac{1}{t}$

Put all the value in equation $\left(1\right)$, we have,

${(-\frac{5+3t}{5t})}^2+{\left(\frac{5-4t}{5t}\right)}^2+{(-\frac{1}{t})}^2=1$

$\Rightarrow \frac{25+9t^2+30t+25+16t^2-40t+25}{25t^2}=1$

$\Rightarrow 25t^2-10t+75=25t^2$

$\Rightarrow t=7.5$

Thus,

$x=-\frac{5+3t}{5t}\Rightarrow x=-\frac{5+3\times 7.5}{5\times 7.5}=-\frac{5+22.5}{37.5}=-\frac{11}{15}$

$y=\frac{5-4t}{5t}\Rightarrow y=\frac{5-4\times 7.5}{5\times 7.5}=\frac{5-30}{37.5}=-\frac{10}{15}$

$z=-\frac{1}{t}\Rightarrow z=-\frac{1}{7.5}=-\frac{2}{15}$

hence, $\hat{c}=-\frac{11}{15}\hat{i}-\frac{10}{15}\hat{j}-\frac{2}{15}\hat{k}$