Question

Find out the vector perpendicular to both vectors $\mathop i\limits^ \wedge - \mathop j\limits^ \wedge + \mathop k\limits^ \wedge $ and $\mathop i\limits^ \wedge + \mathop j\limits^ \wedge + \mathop k\limits^ \wedge $ ?
A. $\mathop i\limits^ \wedge + \mathop j\limits^ \wedge $
B. $\dfrac{{ - \mathop i\limits^ \wedge + \mathop k\limits^ \wedge }}{{\sqrt 2 }}$
C. $\mathop j\limits^ \wedge + \mathop k\limits^ \wedge $
D. $\left( {\mathop j\limits^ \wedge + \mathop k\limits^ \wedge } \right)\sqrt 2 $

Answer 1

Hint: We can solve the problem with the concept of a vector. Vector is an object which has both the magnitude and direction. It can be presented by drawing a pointed arrow, length shows the magnitude of the vector and the arrow shows the direction of vectors.

We can also apply mathematical laws on vectors but it is not as simple as with scalar quantities. Vectors have triangle law for addition and subtraction. Two vectors can be multiplied by the cross product and dot product.

Complete answer:
Cross product of the two vectors is perpendicular to each vector. For example : we have two vectors \[\overrightarrow a \] and $\overrightarrow b $. $\theta $ is the angle between these two vectors. Then the cross product is given as below:

 $a \times b = \left| a \right|\left| b \right|\sin \left( \theta \right)n$ , where $\left| a \right|$ is the magnitude of a vector $a$ , $\left| b \right|$ is the magnitude of vector $b$, $n$ is the unit vector perpendicular at both vectors. The product is maximum when both vectors are at right angles and minimum when they are at same or opposite directions.

For given question, we use cross product because it is the only law in which the product is the vector which is perpendicular to both given vectors , therefore let $\mathop n\limits^ \wedge $ be a unit vector perpendicular to both $\overrightarrow a $ and $\overrightarrow b $ , so $\mathop n\limits^ \wedge = \dfrac{{\overrightarrow a \times \overrightarrow b }}{{\left| {\overrightarrow a \times \overrightarrow b } \right|}}$ , where $
  \overrightarrow a = \mathop i\limits^ \wedge - \mathop j\limits^ \wedge + \mathop k\limits^ \wedge \\
  \overrightarrow b = \mathop i\limits^ \wedge + \mathop j\limits^ \wedge + \mathop k\limits^ \wedge \\
 $
$\left| {\overrightarrow a \times \overrightarrow b } \right| = \left( {\mathop i\limits^ \wedge - \mathop j\limits^ \wedge + \mathop k\limits^ \wedge } \right) \times \left( {\mathop i\limits^ \wedge + \mathop j\limits^ \wedge + \mathop k\limits^ \wedge } \right) = \mathop k\limits^ \wedge - \mathop j\limits^ \wedge + \mathop k\limits^ \wedge - \mathop i\limits^ \wedge + \mathop j\limits^ \wedge - \mathop i\limits^ \wedge $$ \Rightarrow 2\left( { - \mathop i\limits^ \wedge + \mathop k\limits^ \wedge } \right)$
$\mathop n\limits^ \wedge = \dfrac{{2\left( { - \mathop i\limits^ \wedge + \mathop k\limits^ \wedge } \right)}}{{2\sqrt 2 }} = \dfrac{{\left( { - \mathop i\limits^ \wedge + \mathop k\limits^ \wedge } \right)}}{{\sqrt 2 }}$

The vector perpendicular to both vectors $\mathop i\limits^ \wedge - \mathop j\limits^ \wedge + \mathop k\limits^ \wedge $ and $\mathop i\limits^ \wedge + \mathop j\limits^ \wedge + \mathop k\limits^ \wedge $ is $\dfrac{{ - \mathop i\limits^ \wedge + \mathop k\limits^ \wedge }}{{\sqrt 2 }}$ .

So, the correct answer is “Option B”.

Note:
If two vectors are multiplied by the cross product then the product is also a vector quantity but in dot product, the output is a scalar quantity. So the cross product and the dot product are also known as vector products and the scalar product respectively.