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Question: If \( \vec{a}=\hat{i}+\hat{j} \) and \( \vec{b}=\hat{j}-\hat{k} \) and \( \vec{c}=\vec{a}\times \vec...

If a⃗=i^+j^\vec{a}=\hat{i}+\hat{j} and b⃗=j^−k^\vec{b}=\hat{j}-\hat{k} and c⃗=a⃗×b⃗\vec{c}=\vec{a}\times \vec{b}. Find the value of cosθ\theta , where θ\theta is the angle made by the y-axis.
(A) 32\dfrac{\sqrt{3}}{2}
(B) 12\dfrac{1}{2}
(C) 0
(D) 13\dfrac{1}{\sqrt{3}}

Explanation

Solution

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 θ\theta and then calculate the value of cosθ\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
a⃗×b⃗=absin⁡θn^\vec{a}\times \vec{b}=ab\sin \theta \hat{n}

Complete step-by-step answer:
a⃗=i^+j^\vec{a}=\hat{i}+\hat{j}
b⃗=j^−k^\vec{b}=\hat{j}-\hat{k}
To find: the cosine of angleθ\theta when the cross product of the two vectors is taken
We have,
a⃗×b⃗=absin⁡θn^\vec{a}\times \vec{b}=ab\sin \theta \hat{n}
The cross product of a and b vectors is given by
a⃗×b⃗\vec{a}\times \vec{b} = (i^j^k^ 110 01−1 )\left( \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\\ 1 & 1 & 0 \\\ 0 & 1 & -1 \\\ \end{matrix} \right)
= −i^−j^+k^-\hat{i}-\hat{j}+\hat{k}
Magnitude of a vector a = ∣a⃗∣\left| {\vec{a}} \right|
= 12+12+0\sqrt{{{1}^{2}}+{{1}^{2}}+0}
= 2\sqrt{2}
Magnitude of vector b = ∣b⃗∣\left| {\vec{b}} \right|
= 02+12+(−1)2\sqrt{{{0}^{2}}+{{1}^{2}}+{{(-1)}^{2}}}
= 2\sqrt{2}
Magnitude of a⃗×b⃗\vec{a}\times \vec{b} = (−1)2+(−1)2+(1)2\sqrt{{{(-1)}^{2}}+{{(-1)}^{2}}+{{(1)}^{2}}}
= 3\sqrt{3}
Thus, considering the above formula we have,
a⃗×b⃗=absin⁡θn^\vec{a}\times \vec{b}=ab\sin \theta \hat{n}
Considering only the magnitude, we have
∣a⃗×b⃗∣=absin⁡θ\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
⇒32=sin⁡θ\Rightarrow \dfrac{\sqrt{3}}{2}=\sin \theta
⇒θ=60∘\Rightarrow \theta ={{60}^{\circ }}
Now we have to calculate the value of cosθ\theta
∴\therefore cosθ\theta =cos60 ∘^{\circ }
= 12\dfrac{1}{2}
Thus the value of cosθ\theta from the given question is calculated as 12\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.