Question

Let \[\overrightarrow x \], \[\overrightarrow y \] and \[\overrightarrow z \] be three vectors each of magnitude \[\sqrt 2 \] and the angle between each pair of them is \[\dfrac{\pi }{3}\]. If \[\overrightarrow a \] be a non-zero vector perpendicular to \[\overrightarrow x \] and \[\left( {\overrightarrow y \times \overrightarrow z } \right)\] and \[\overrightarrow b \] is another non-zero vector perpendicular to \[\overrightarrow y \] and \[\left( {\overrightarrow z \times \overrightarrow x } \right)\], then
(a) \[\overrightarrow b = \left( {\overrightarrow b .\overrightarrow z } \right)\left( {\overrightarrow z -\overrightarrow x } \right)\]
(b) \[\overrightarrow a = \left( {\overrightarrow a .\overrightarrow y } \right)\left( {\overrightarrow y -\overrightarrow z } \right)\]
(c) \[\left( {\overrightarrow a .\overrightarrow b } \right) = - \left( {\overrightarrow a .\overrightarrow y } \right)\left( {\overrightarrow b .\overrightarrow z } \right)\;\]
(d) \[\overrightarrow a = \left( {\overrightarrow a .\overrightarrow y } \right)\left( {\overrightarrow z -\overrightarrow y } \right)\]

Answer 1

Hint: Here we will use the fact that when a vector is perpendicular to two other vectors then it is in the direction of cross product of those vectors. The formula for vector triple product or three vectors i.e.
\[\overrightarrow x \times \left( {\overrightarrow y \times \overrightarrow z } \right) = \left( {\overrightarrow x .\overrightarrow z } \right)\overrightarrow y - \left( {\overrightarrow x .\overrightarrow y } \right)\overrightarrow z \]
Also, the dot product of two vectors having angle \[\theta \] between them is given by:-
\[\overrightarrow c .\overrightarrow d = \left| {\overrightarrow c } \right|\left| {\overrightarrow d } \right|\cos \theta \]

Complete step-by-step answer:
It is given that:-
\[\overrightarrow x \], \[\overrightarrow y \] and \[\overrightarrow z \] be three vectors each of magnitude \[\sqrt 2 \]
This implies, \[\left| {\overrightarrow x } \right| = \left| {\overrightarrow y } \right| = \left| {\overrightarrow z } \right| = \sqrt 2 \]
Also, the angle between each pair of them is \[\dfrac{\pi }{3}\].
Also, it is given that:
\[\overrightarrow a \] is a non-zero vector perpendicular to \[\overrightarrow x \] and \[\left( {\overrightarrow y \times \overrightarrow z } \right)\]
This implies \[\overrightarrow a \] is in the direction of \[\overrightarrow x \times \left( {\overrightarrow y \times \overrightarrow z } \right)\]
Hence we can write \[\overrightarrow a \] as:-
\[\overrightarrow a = {k_1}\left[ {\overrightarrow x \times \left( {\overrightarrow y \times \overrightarrow z } \right)} \right]\]
Now we know that:
\[\overrightarrow x \times \left( {\overrightarrow y \times \overrightarrow z } \right) = \left( {\overrightarrow x .\overrightarrow z } \right)\overrightarrow y - \left( {\overrightarrow x .\overrightarrow y } \right)\overrightarrow z \]
Hence putting the value we get:-
\[\overrightarrow a = {k_1}\left[ {\left( {\overrightarrow x .\overrightarrow z } \right)\overrightarrow y - \left( {\overrightarrow x .\overrightarrow y } \right)\overrightarrow z } \right]\]…………………………………(1)
Now we will find the value of \[\overrightarrow x .\overrightarrow z \]
We know that:
The dot product of two vectors having angle \[\theta \] between them is given by:-
\[\overrightarrow c .\overrightarrow d = \left| {\overrightarrow c } \right|\left| {\overrightarrow d } \right|\cos \theta \]
Hence,
\[\overrightarrow x .\overrightarrow z = \left| {\overrightarrow x } \right|\left| {\overrightarrow z } \right|\cos \left( {\dfrac{\pi }{3}} \right)\]
We know that:
\[\cos \left( {\dfrac{\pi }{3}} \right) = \dfrac{1}{2}\]
Hence putting in the respective values we get:-
\[\overrightarrow x .\overrightarrow z = \left( {\sqrt 2 } \right)\left( {\sqrt 2 } \right)\left( {\dfrac{1}{2}} \right)\]
Solving it further we get:-
\[
  \overrightarrow x .\overrightarrow z = 2\left( {\dfrac{1}{2}} \right) \\
   \Rightarrow \overrightarrow x .\overrightarrow z = 1...................\left( 2 \right) \\
 \]
Now we will find the value of \[\left( {\overrightarrow x .\overrightarrow y } \right)\]
The dot product of two vectors having angle \[\theta \] between them is given by:-
\[\overrightarrow c .\overrightarrow d = \left| {\overrightarrow c } \right|\left| {\overrightarrow d } \right|\cos \theta \]
Hence,
\[\overrightarrow x .\overrightarrow y = \left| {\overrightarrow x } \right|\left| {\overrightarrow y } \right|\cos \left( {\dfrac{\pi }{3}} \right)\]
We know that:
\[\cos \left( {\dfrac{\pi }{3}} \right) = \dfrac{1}{2}\]
Hence putting in the respective values we get:-
\[\overrightarrow x .\overrightarrow y = \left( {\sqrt 2 } \right)\left( {\sqrt 2 } \right)\left( {\dfrac{1}{2}} \right)\]
Solving it further we get:-
\[
  \overrightarrow x .\overrightarrow y = 2\left( {\dfrac{1}{2}} \right) \\
   \Rightarrow \overrightarrow x .\overrightarrow y = 1...................\left( 3 \right) \\
 \]
Putting the values from equation 2 and 3 in equation 1 we get:-
\[\overrightarrow a = {k_1}\left[ {\left( 1 \right)\overrightarrow y - \left( 1 \right)\overrightarrow z } \right]\]
Solving it further we get:-
\[\overrightarrow a = {k_1}\left[ {\overrightarrow y - \overrightarrow z } \right]\]…………………………………(4)
Now we will find the value of \[\overrightarrow a \cdot \overrightarrow {y\;} \].
\[\overrightarrow a \cdot \overrightarrow {y\;} = {k_1}(\overrightarrow {y\;} \cdot \overrightarrow {y\;} - \overrightarrow {y\;} \cdot \overrightarrow z )\]
Now applying the formula:-
\[\overrightarrow c .\overrightarrow d = \left| {\overrightarrow c } \right|\left| {\overrightarrow d } \right|\cos \theta \]
We know that a vector is parallel to itself i.e. angle is \[{0^ \circ }\]
Hence we get:
\[\overrightarrow a \cdot \overrightarrow {y\;} = {k_1}\left[ {\left| {\overrightarrow {y\;} } \right|\left| {\overrightarrow {y\;} } \right|\cos {0^ \circ } - \left| {\overrightarrow {y\;} } \right|\left| {\overrightarrow z } \right|\cos \left( {\dfrac{\pi }{3}} \right)} \right]\]
We know that:-
\[\cos {0^ \circ } = 1\] and \[\cos \left( {\dfrac{\pi }{3}} \right) = \dfrac{1}{2}\]
Hence putting in the respective values we get:-
\[\overrightarrow a \cdot \overrightarrow {y\;} = {k_1}\left[ {\left( {\sqrt 2 } \right)\left( {\sqrt 2 } \right)\left( 1 \right) - \left( {\sqrt 2 } \right)\left( {\sqrt 2 } \right)\left( {\dfrac{1}{2}} \right)} \right]\]
Simplifying it further we get:-
\[
  \overrightarrow a \cdot \overrightarrow {y\;} = {k_1}(2 - 1) \\
   \Rightarrow {k_1} = \overrightarrow a \cdot \overrightarrow {y\;} \\
 \]
Putting this value in equation 4 we get:-
\[\overrightarrow a = \left( {\overrightarrow a \cdot \overrightarrow {y\;} } \right)\left[ {\overrightarrow y - \overrightarrow z } \right]\]…………………………….(5)
Similarly, we can obtain
\[\overrightarrow b = (\overrightarrow b \cdot \overrightarrow z )(\overrightarrow z - \overrightarrow x )\]………………………………….(6)
Now we will find the value of \[\overrightarrow a .\overrightarrow b \]
From equation 5 and 6 we get:-
\[
  \overrightarrow a .\overrightarrow b = \left( {\overrightarrow a \cdot \overrightarrow {y\;} } \right)\left[ {\overrightarrow y - \overrightarrow z } \right].(\overrightarrow b \cdot \overrightarrow z )(\overrightarrow z - \overrightarrow x ) \\
   \Rightarrow \overrightarrow a .\overrightarrow b = \left( {\overrightarrow a \cdot \overrightarrow {y\;} } \right)\left( {\overrightarrow b \cdot \overrightarrow z } \right)\left[ {\left( {\overrightarrow y - \overrightarrow z } \right).\left( {\overrightarrow z - \overrightarrow x } \right)} \right] \\
 \]
Solving it further we get:-
\[\overrightarrow a .\overrightarrow b = \left( {\overrightarrow a \cdot \overrightarrow {y\;} } \right)\left( {\overrightarrow b \cdot \overrightarrow z } \right)\left[ {\left( {\overrightarrow y .\overrightarrow z - \overrightarrow x .\overrightarrow y - \overrightarrow z .\overrightarrow z + \overrightarrow z .\overrightarrow x } \right)} \right]\]
Applying the following formula
\[\overrightarrow c .\overrightarrow d = \left| {\overrightarrow c } \right|\left| {\overrightarrow d } \right|\cos \theta \]
We get:-
\[\overrightarrow a .\overrightarrow b = \left( {\overrightarrow a \cdot \overrightarrow {y\;} } \right)\left( {\overrightarrow b \cdot \overrightarrow z } \right)\left[ {\left| {\overrightarrow {y\;} } \right|\left| {\overrightarrow z } \right|\cos \left( {\dfrac{\pi }{3}} \right) - \left| {\overrightarrow x } \right|\left| {\overrightarrow y } \right|\cos \left( {\dfrac{\pi }{3}} \right) - \left| {\overrightarrow z } \right|\left| {\overrightarrow z } \right|\cos {0^ \circ } + \left| {\overrightarrow x } \right|\left| {\overrightarrow z } \right|\cos \left( {\dfrac{\pi }{3}} \right)} \right]\]
Putting in the respective values we get:-
\[\overrightarrow a .\overrightarrow b = \left( {\overrightarrow a \cdot \overrightarrow {y\;} } \right)\left( {\overrightarrow b \cdot \overrightarrow z } \right)\left[ {\left( {\sqrt 2 } \right)\left( {\sqrt 2 } \right)\left( {\dfrac{1}{2}} \right) - \left( {\sqrt 2 } \right)\left( {\sqrt 2 } \right)\left( {\dfrac{1}{2}} \right) - \left( {\sqrt 2 } \right)\left( {\sqrt 2 } \right)\left( 1 \right) + \left( {\sqrt 2 } \right)\left( {\sqrt 2 } \right)\left( {\dfrac{1}{2}} \right)} \right]\]
Simplifying it we get:-
\[
  \overrightarrow a .\overrightarrow b = \left( {\overrightarrow a \cdot \overrightarrow {y\;} } \right)\left( {\overrightarrow b \cdot \overrightarrow z } \right)\left[ {1 - 1 - 2 + 1} \right] \\
   \Rightarrow \overrightarrow a .\overrightarrow b = - \left( {\overrightarrow a \cdot \overrightarrow {y\;} } \right)\left( {\overrightarrow b \cdot \overrightarrow z } \right) \\
 \]

So, the correct answer is “Option c”.

Note: Students should take a note that when a vector is perpendicular to two other vectors then it is parallel to the vector product of those two vectors i.e. it is in the direction of the vector product of those two vectors.
Also, students might make mistakes in evaluating the vector product so the formula used should be correct and should be wisely used.