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
\({{(5+3\lambda )}^{2}}+{{(5-4\lambda )}^{2}}+25=25{{\lambda }^{2}}\),
\(25{{\lambda }^{2}}-10\lambda +75=25{{\lambda }^{2}}\),
\(\lambda =\frac{15}{2}\).
\(\overrightarrow{c}=\frac{1}{15}(-11i+10j-2k)\).