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# Let $\vec{a}$ be a vector perpendicular to unit vectors $\vec{b}$ and $\vec{c}$ and if the angle between $\vec{b}$ and $\vec{c}$ is $\alpha$, then $\vec{b}\times \vec{c}$ īs(a) $\pm \left( \cos \alpha \right)\vec{a}$(b) $\pm \left( cosec\alpha \right)\vec{a}$(c) $\pm \left( \sin \alpha \right)\vec{a}$(d) $\pm \tan \alpha$

Last updated date: 24th Jul 2024
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Hint: Since the vector $\vec{a}$ is given to be perpendicular to both of the unit vectors $\vec{b}$ and $\vec{c}$, so it will also be perpendicular to the plane containing the vectors $\vec{b}$ and $\vec{c}$. This means that the vector $\vec{a}$ must be parallel to the vector $\vec{b}\times \vec{c}$ . So the vector $\vec{b}\times \vec{c}$ can be written as a scalar multiple of the vector $\vec{a}$. Further, the magnitude of the cross product $\vec{b}\times \vec{c}$ is given by $\left| \vec{b}\times \vec{c} \right|=\left| {\vec{b}} \right|\left| {\vec{c}} \right|\sin \alpha$. Equating this magnitude with the scalar multiplied by the magnitude of $\vec{a}$, we will get the value of the scalar.

Complete step-by-step solution:
According to the question, the vector $\vec{a}$ is perpendicular to the unit vectors $\vec{b}$ and $\vec{c}$. This means that it will also be perpendicular to the plane containing the vectors $\vec{b}$ and $\vec{c}$. Now, we know that the cross product of two vectors is perpendicular to the plane containing the two vectors. So we can say that the vector $\vec{a}$ is parallel to the vector $\vec{b}\times \vec{c}$. So we can write
$\Rightarrow \vec{b}\times \vec{c}=k\vec{a}.......(i)$
Taking the magnitudes of the vectors on both the sides, we get
$\Rightarrow \left| \vec{b}\times \vec{c} \right|=\left| k\vec{a} \right|$
Now, since the angle between the vectors $\vec{b}$ and $\vec{c}$ is given to be equal to $\alpha$, so we can write
$\Rightarrow \left| \vec{b}\times \vec{c} \right|=\left| {\vec{b}} \right|\left| {\vec{c}} \right|\sin \alpha$
Equating the above two equations, we get
\begin{align} & \Rightarrow \left| k\vec{a} \right|=\left| {\vec{b}} \right|\left| {\vec{c}} \right|\sin \alpha \\ & \Rightarrow k\left| {\vec{a}} \right|=\left| {\vec{b}} \right|\left| {\vec{c}} \right|\sin \alpha \\ \end{align}
Since the vectors $\vec{b}$ and $\vec{c}$ are given to be unit vectors, so their magnitudes are equal to one. So we substitute $\left| {\vec{b}} \right|=1$ and $\left| {\vec{c}} \right|=1$ in the above equation to get
\begin{align} & \Rightarrow \left| k \right|\left| {\vec{a}} \right|=\left( 1 \right)\left( 1 \right)\sin \alpha \\ & \Rightarrow \left| k \right|\left| {\vec{a}} \right|=\sin \alpha \\ & \Rightarrow \left| k \right|=\sin \alpha \\ & \Rightarrow k=\pm \sin \alpha \\ \end{align}
Substituting this value in (i) we get
$\Rightarrow \vec{b}\times \vec{c}=\pm \left( \sin \alpha \right)\vec{a}$
Hence, the correct answer is option (c).

Note: In this question, we were not given any information regarding the magnitude of the vector $\vec{a}$. So we simply assumed its magnitude to be equal to one. In other words, we assumed the vector $\vec{a}$ as a unit vector.