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Question: If \(\hat u\) and \(\hat v\) are the unit vectors and \(\theta \) is the acute angle between them, t...

If u^\hat u and v^\hat v are the unit vectors and θ\theta is the acute angle between them, then 2u^×3v^2\hat u \times 3\hat v is a unit vector for
A.Exactly two values of θ\theta
B.More than two values of θ\theta
C.No value of θ\theta
D.Exactly one value of θ\theta

Explanation

Solution

At first, we will learn about the cross product, and its formula i.e. a⃗×b⃗=(∣a⃗∣∣b⃗∣sin⁡β)n^\vec a \times \vec b = \left( {\left| {\vec a} \right|\left| {\vec b} \right|\sin \beta } \right)\hat n.Using this we will simplify the given expression, from where we will get an equation in θ\theta or in the function of θ\theta , therefore solving that equation we will get our answer.
We know that the magnitude of the cross product of 2 vectors a and b is given by ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ\left| {\vec a \times \vec b} \right| = \left| {\vec a} \right|\left| {\vec b} \right|\sin \theta where θ\theta is the angle between the two vectors. Then we can find the magnitude of the cross product of the given vectors. Then we can equate it to unity and solve for the angle θ\theta . Then the number of solutions of θ\theta will give the required option.

Complete step-by-step answer:
We are given that u^\hat u and v^\hat v are two unit vectors and θ\theta is the acute angle between them

We know that the cross product of two vectors is also a vector perpendicular to both the vectors. We know that the magnitude of the cross product of 2 vectors a and b is given by ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ\left| {\vec a \times \vec b} \right| = \left| {\vec a} \right|\left| {\vec b} \right|\sin \theta where θ\theta is the angle between the two vectors.
We are given the vectors u^\hat u and v^\hat v
We can find the magnitude of 2u^×3v^2\hat u \times 3\hat v.
⇒∣2u^×3v^∣=∣2u^∣∣3v^∣sin⁡θ\Rightarrow \left| {2\hat u \times 3\hat v} \right| = \left| {2\hat u} \right|\left| {3\hat v} \right|\sin \theta
We know that ∣ca⃗∣=c∣a⃗∣\left| {c\vec a} \right| = c\left| {\vec a} \right|. So, we get,
⇒∣2u^×3v^∣=2∣u^∣×3∣v^∣×sin⁡θ\Rightarrow \left| {2\hat u \times 3\hat v} \right| = 2\left| {\hat u} \right| \times 3\left| {\hat v} \right| \times \sin \theta
Since it is given that u^\hat u and v^\hat v is the unit vectors i.e. ∣u^∣=∣v^∣=1\left| {\hat u} \right| = \left| {\hat v} \right| = 1. So, we get,
⇒∣2u^×3v^∣=2×3×sin⁡θ\Rightarrow \left| {2\hat u \times 3\hat v} \right| = 2 \times 3 \times \sin \theta
On simplification we get,
⇒∣2u^×3v^∣=6sin⁡θ\Rightarrow \left| {2\hat u \times 3\hat v} \right| = 6\sin \theta … (1)
We need to find the angle when the cross product will give a unit vector. For 2u^×3v^2\hat u \times 3\hat v to be a unit vector, its modulus should be equal to 1.
i.e. ⇒∣2u^×3v^∣=1 \Rightarrow \left| {2\hat u \times 3\hat v} \right| = 1 …. (2)
On equating (1) and (2), we get,
⇒1=6sin⁡θ\Rightarrow 1 = 6\sin \theta
On dividing the equation by 6 we get,
⇒sin⁡θ=16\Rightarrow \sin \theta = \dfrac{1}{6}
Taking both side expression as a function of sin⁡−1{\sin ^{ - 1}}
⇒θ=sin⁡−116\Rightarrow \theta = {\sin ^{ - 1}}\dfrac{1}{6}
Therefore θ\theta has only one value which makes it an acute angle.
Therefore, the correct answer is option D.

Note: While solving the equation for θ\theta most of the students take the value of θ\theta as θ=sin⁡−116\theta = {\sin ^{ - 1}}\dfrac{1}{6} and θ=π−sin⁡−116\theta = \pi - {\sin ^{ - 1}}\dfrac{1}{6}, but it is mentioned the question that the θ\theta is an acute angle so it cannot take the value θ=π−sin⁡−116\theta = \pi - {\sin ^{ - 1}}\dfrac{1}{6}, and will have only one solution, so remember to apply all the given data to get a more accurate answer.
The vector product of two vectors will give another vector. The resultant vector will be perpendicular to both the vectors. It will be zero if the vectors are parallel. The scalar product of two vectors will give a scalar quantity. It will be zero if the vectors are perpendicular to each other.