# Vector Along the Bisector of Given Two Vectors

## Vector Along the Bisector of Given Two Vectors

We know that the diagonal in a parallelogram is not necessarily the bisector of the angle formed by two adjacent sides. However, the diagonal in a rhombus bisects the angle between the two adjacent sides.

Consider vectors $$\overrightarrow{AB}=\overrightarrow{a}$$ and $$\overrightarrow{AD}=\overrightarrow{b}$$ forming a parallelogram ABCD as shown in figure.

Consider the two unit vector along the given vectors, which form a rhombus AB’C’D’.

Now $$\overrightarrow{AB’}=\frac{\overrightarrow{a}}{\left| \overrightarrow{a} \right|}$$ and $$\overrightarrow{AD’}=\frac{\overrightarrow{b}}{\left| \overrightarrow{b} \right|}$$.

$$\overrightarrow{AC’}=\frac{\overrightarrow{a}}{\left| \overrightarrow{a} \right|}+\frac{\overrightarrow{b}}{\left| \overrightarrow{b} \right|}$$,

So, any vector along the bisector is $$\lambda \left( \frac{\overrightarrow{a}}{\left| \overrightarrow{a} \right|}+\frac{\overrightarrow{b}}{\left| \overrightarrow{b} \right|} \right)$$.

Similarly, any vector along the external bisector is $$\overrightarrow{AC’}=\lambda \left( \frac{\overrightarrow{a}}{\left| \overrightarrow{a} \right|}+\frac{\overrightarrow{b}}{\left| \overrightarrow{b} \right|} \right)$$.

Example: Find a unit vector $$\overrightarrow{c}$$ if -i + j – k bisects the angle between vector $$\overrightarrow{c}$$ and 3i + 4j.

Solution: Let $$\overrightarrow{c}=xi+yj+zk$$,

Where x² + y² + z² = 1 … (1)

Unit vector along 3i + 4j is $$\frac{3i+4j}{\sqrt{{{(3)}^{2}}+{{(4)}^{2}}}}=\frac{3i+4j}{5}$$,

The bisector of these two is -i + j – k (given)

Therefore, $$-i+j-k=\lambda \left( xi+yj+zk+\frac{3i+4j}{5} \right)$$,

$$-i+j-k=\frac{1}{5}\lambda \left( (5x+3)i+(5y+4)j+5zk \right)$$,

$$\frac{\lambda }{5}(5x+3)=-1$$,

$$\frac{\lambda }{5}(5y+4)=1$$,

$$\frac{\lambda }{5}(5z)=-1$$,

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

$$y=\frac{5-4\lambda }{5\lambda }$$,

$$z=-\frac{1}{\lambda }$$,

Putting these values in (1) i.e., we get

$${{(5+3\lambda )}^{2}}+{{(5-4\lambda )}^{2}}+25=25{{\lambda }^{2}}$$,

$$25{{\lambda }^{2}}-10\lambda +75=25{{\lambda }^{2}}$$,

$$\lambda =\frac{15}{2}$$.

$$\overrightarrow{c}=\frac{1}{15}(-11i+10j-2k)$$.