Question

If $ \vec{a}=\hat{i}+\hat{j} $ and $ \vec{b}=\hat{j}-\hat{k} $ and $ \vec{c}=\vec{a}\times \vec{b} $. Find the value of cos\[\theta \], where \[\theta \] is the angle made by the y-axis.
(A) $ \dfrac{\sqrt{3}}{2} $
(B) $ \dfrac{1}{2} $
(C) 0
(D) $ \dfrac{1}{\sqrt{3}} $

Answer 1

Hint: The vector c is the cross product of vectors a and b. The three vectors are related by a certain relation. From that particular relation, we can calculate the angle \[\theta \] and then calculate the value of cos\[\theta \] for that particular angle. This is how we approach this problem with the help of the given vectors in question.

Formula used:
The cross product is given by
 $ \vec{a}\times \vec{b}=ab\sin \theta \hat{n} $

Complete step-by-step answer:
 $ \vec{a}=\hat{i}+\hat{j} $
 $ \vec{b}=\hat{j}-\hat{k} $
To find: the cosine of angle\[\theta \] when the cross product of the two vectors is taken
We have,
 $ \vec{a}\times \vec{b}=ab\sin \theta \hat{n} $
The cross product of a and b vectors is given by
 $ \vec{a}\times \vec{b} $ = \[\left( \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   1 & 1 & 0 \\
   0 & 1 & -1 \\
\end{matrix} \right)\]
            = $ -\hat{i}-\hat{j}+\hat{k} $
Magnitude of a vector a = $ \left| {\vec{a}} \right| $
                                         = $ \sqrt{{{1}^{2}}+{{1}^{2}}+0} $
                                         = $ \sqrt{2} $
Magnitude of vector b = $ \left| {\vec{b}} \right| $
                                          = $ \sqrt{{{0}^{2}}+{{1}^{2}}+{{(-1)}^{2}}} $
                                          = $ \sqrt{2} $
Magnitude of $ \vec{a}\times \vec{b} $ = $ \sqrt{{{(-1)}^{2}}+{{(-1)}^{2}}+{{(1)}^{2}}} $
                                      = $ \sqrt{3} $
Thus, considering the above formula we have,
 $ \vec{a}\times \vec{b}=ab\sin \theta \hat{n} $
Considering only the magnitude, we have
 $ \left| \vec{a}\times \vec{b} \right|=ab\sin \theta $
Plugging the values of magnitudes of cross product of a and b vectors, magnitude of a vector and b vector, we have
 $ \Rightarrow \dfrac{\sqrt{3}}{2}=\sin \theta $
 $ \Rightarrow \theta ={{60}^{\circ }} $
Now we have to calculate the value of cos\[\theta \]
 $ \therefore $ cos\[\theta \]=cos60 $ ^{\circ } $
               = $ \dfrac{1}{2} $
Thus the value of cos\[\theta \] from the given question is calculated as $ \dfrac{1}{2} $.

So, the correct answer is “Option B”.

Additional Information: The cross product of two vectors is defined as a vector that is perpendicular or orthogonal to both the vectors, with a direction given by the right-hand rule and a magnitude equal to the area of the parallelogram that the vectors span. The direction of cross product cannot be arbitrary. Thus cross product is a vector product in contrast to dot product which is a scalar product.

Note: The angle made by the resultant vector of the cross product of two vectors with the y axis is to be calculated in the given problem which can simply be calculated following the very definition of cross product of two vectors. It is to be noted that the cosine of the angle is asked and not the sine of the angle which is available from the initial relationship of cross product.