How would you find a unit vector parallel to the resultant of the vectors \[A = 2i - 6j - 3k\] and \[B = 4i + 3j - k\]?

Hint: Here in this question, we have to find the unit vector which is parallel to the resultant of two vectors. The vectors A and B are given. By using the formula \[\hat x = \dfrac{{\overrightarrow A + \overrightarrow B }}{{||\overrightarrow A + \overrightarrow B ||}}\] the unit vector is determined. Where A, and B are points which are already in the question then substituting the values we obtain the required result for the question.

Complete step-by-step solution:
A vector that has a magnitude of 1 is a unit vector. It is also known as Direction Vector. The given vectors are \[A = 2i - 6j - 3k\] and \[B = 4i + 3j - k\]. Therefore, the resultant vector of A and B is the sum of vectors A and B.
Then the unit vector is determined by using the formula \[\dfrac{{\overrightarrow A + \overrightarrow B }}{{||\overrightarrow A \times \overrightarrow B ||}}\]----- (1)
The vector \[\overrightarrow A + \overrightarrow B \] is determined by adding two vectors, substituting the values of A and B we get
\[ \Rightarrow \overrightarrow A + \overrightarrow B = (2i - 6j - 3k) + (4i + 3j - k)\]
Add the terms which are having the same unit vectors that I, j and k
\[ \Rightarrow \overrightarrow A + \overrightarrow B = (2 + 4)i + ( - 6 + 3)j + ( - 3 - 1)k\]
On simplifying we get
\[ \Rightarrow \overrightarrow A + \overrightarrow B = 6i - 3j - 4k\]
Hence we have determined the resultant vectors of A and B
Now the \[\left\| {\overrightarrow A + \overrightarrow B } \right\|\] is determined by
\[
   \Rightarrow \left\| {\overrightarrow A + \overrightarrow B } \right\| = \sqrt {{{(6)}^2} + {{( - 3)}^2} + {{( - 4)}^2}} \\
   \Rightarrow \left\| {\overrightarrow A + \overrightarrow B } \right\| = \sqrt {36 + 9 + 16} \\
   \Rightarrow \left\| {\overrightarrow A + \overrightarrow B } \right\| = \sqrt {61} \\
 \]
Therefore the unit vector is given by
\[\dfrac{{\overrightarrow A + \overrightarrow B }}{{||\overrightarrow A + \overrightarrow B ||}} = \left( {\dfrac{6}{{\sqrt {61} }},\dfrac{{ - 3}}{{\sqrt {61} }},\dfrac{{ - 4}}{{\sqrt {61} }}} \right)\]
Hence this is the unit vector parallel to the resultant vector AB.

Note: To the vectors the arithmetic operations are applicable. The vectors are multiplied by the two kinds one is dot product and the other one is cross product. The dot product is like multiplication itself. But in case of cross product while multiplying the terms we consider the determinant for the points or vector.