Question

The vector $\overrightarrow C $ is directed along the internal bisector of the angle between vectors $\overrightarrow A = \hat i - 4\hat j - 4\hat k$ and $\overrightarrow B = - 2\hat i - \hat j + 2\hat k$. Then $\overrightarrow C $=?

Answer 1

Hint: We can find a vector that bisects the angle between two given vectors by adding their unit vectors. When we add 2 vectors with the same magnitude, the resultant vectors will be in the direction of its internal angle bisector.

Complete step by step solution: We have $\overrightarrow A = \hat i - 4\hat j - 4\hat k$ . We can get its unit vector as follows
$\hat A = \dfrac{{\vec A}}{{\left| A \right|}}$
As, $\left| A \right| = \sqrt {{x^2} + {y^2} + {z^2}} $, on substituting values we get,
$\hat A = \dfrac{{\hat i - 4\hat j - 4\hat k}}{{\sqrt {{1^2} + {4^2} + {4^2}} }}$
On simplification we get,
$
  \hat A = \dfrac{{\hat i - 4\hat j - 4\hat k}}{{\sqrt {33} }} \\
  \hat A = \dfrac{1}{{\sqrt {33} }}\hat i - \dfrac{4}{{\sqrt {33} }}\hat j - \dfrac{4}{{\sqrt {33} }}\hat k \\
 $
Similarly, we can also find the unit vector of $\overrightarrow B = - 2\hat i - \hat j + 2\hat k$ as follows
$\hat B = \dfrac{{\vec B}}{{\left| B \right|}}$
As, $\left| A \right| = \sqrt {{x^2} + {y^2} + {z^2}} $, on substituting values we get,
$\hat B = \dfrac{{ - 2\hat i - \hat j + 2\hat k}}{{\sqrt {{2^2} + {1^2} + {2^2}} }}$
On simplification we get,
$
  \hat B = \dfrac{{ - 2\hat i - \hat j + 2\hat k}}{{\sqrt 9 }} \\
  \hat B = \dfrac{{ - 2\hat i - \hat j + 2\hat k}}{3} \\
  \hat B = \dfrac{{ - 2}}{3}\hat i - \dfrac{1}{3}\hat j + \dfrac{2}{3}\hat k \\
 $
Now we have 2 vectors with magnitude 1. By adding these vectors, we will get a vector in the direction on its internal angle bisector.
Thus $\overrightarrow C = {\text{ }}\hat A + \hat B$
$
  \overrightarrow C = {\text{ }}\left( {\dfrac{1}{{\sqrt {33} }}\hat i - \dfrac{4}{{\sqrt {33} }}\hat j - \dfrac{4}{{\sqrt {33} }}\hat k} \right) + \left( {\dfrac{{ - 2}}{3}\hat i - \dfrac{1}{3}\hat j + \dfrac{2}{3}\hat k} \right) \\
   = \dfrac{{3 - 2\sqrt {33} }}{{3\sqrt {33} }}\hat i - {\text{ }}\dfrac{{12 + \sqrt {33} }}{{3\sqrt {33} }}\hat j - \dfrac{{12 - 2\sqrt {33} }}{{3\sqrt {33} }}\hat k \\
 $
Thus, the required vector is $\vec C = \dfrac{{3 - 2\sqrt {33} }}{{3\sqrt {33} }}\hat i - {\text{ }}\dfrac{{12 + \sqrt {33} }}{{3\sqrt {33} }}\hat j - \dfrac{{12 - 2\sqrt {33} }}{{3\sqrt {33} }}\hat k$

Note: The method used here for finding the vector in the direction of the internal angle bisector can be derived from the parallelogram law of vector addition. If the two vectors to be added are equal in magnitude, the parallelogram will become a rhombus. The diagonal of a rhombus is the angle bisector. Thus, we get the required vector.