Question

If the vectors \[\overrightarrow {\text{k}} \] and \[\overrightarrow {\text{A}} \] are parallel to each other ,then what is \[{\text{k}}\overrightarrow {\text{k}} \times \overrightarrow {\text{A}} \] equal to?
A) \[{k^2}\overrightarrow A \]
B) \[\overrightarrow 0 \]
C) \[ - {k^2}\overrightarrow A \]
D) \[\overrightarrow A \]

Answer 1

Hint: Here since the vectors \[\overrightarrow {\text{k}} \] and \[\overrightarrow {\text{A}} \] are parallel to each other therefore the angle between them is . Hence apply the formula for \[|\overrightarrow X \times \overrightarrow {Y|} \] to get the answer.
Formula of magnitude is given by:
\[|\overrightarrow {\text{X}} \times \overrightarrow {{\text{Y}}|} = {\text{|}}\overrightarrow {\text{X}} {\text{| |}}\overrightarrow {{\text{Y|}}} \sin \theta \]

Complete step by step solution:
We are given two vectors \[\overrightarrow {\text{k}} \] and \[\overrightarrow {\text{A}} \] such that they are parallel to each other.
Hence the angle between these two vectors is \[{0^ \circ }\].
According to the following formula of vector product of two vectors:
\[|\overrightarrow {\text{X}} \times \overrightarrow {{\text{Y}}|} = {\text{|}}\overrightarrow {\text{X}} {\text{| |}}\overrightarrow {{\text{Y|}}} \sin \theta \]
Applying this formula for given vectors we get:
\[\overrightarrow {{\text{|k}}} \times \overrightarrow {\text{A}} | = \overrightarrow {{\text{|k}}} |\overrightarrow {{\text{|A|}}} \sin {0^ \circ }\]
Since , \[\sin {0^ \circ } = 0\]
Therefore , \[\overrightarrow {{\text{|k}}} \times \overrightarrow {\text{A}} | = 0\]
Hence,
\[
  {\text{k}}\overrightarrow {\text{k}} \times \overrightarrow {\text{A}} = {\text{k}}\left( 0 \right) \\
  {\text{k}}\overrightarrow {\text{k}} \times \overrightarrow {\text{A}} = \overrightarrow 0 \\
 \]

Therefore option B is the correct answer.

Note:
This question can be directly solved in one step as the vector product of all the vectors which are parallel to each other is zero.
Therefore,
 \[
  {\text{k}}\overrightarrow {\text{k}} \times \overrightarrow {\text{A}} = {\text{k}}\left( 0 \right) \\
  {\text{k}}\overrightarrow {\text{k}} \times \overrightarrow {\text{A}} = \overrightarrow 0 \\
 \]