Question

If $\vec{a},\vec{b},\vec{c}$ are unit vectors such that $\vec{a}\cdot \vec{b}=\vec{a}\cdot \vec{c}=0$ and the angle between $\vec{b}$ and $\vec{c}$ is $\dfrac{\pi }{6}$ .Prove that $\vec{a}=\pm 2\left( \vec{b}\times \vec{c} \right)$\[\]

Answer 1

Hint: We first use the cross product formula to find the angle between $\vec{a}$ and $\vec{b}$, $\vec{a}$ and $\vec{c}$ using the dot product formula $\vec{a}\cdot \vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\cos \theta $ . Then we take the cross product of
 $\vec{b}$ and $\vec{c}$$\left( \vec{b}\times \vec{c}=\left| {\vec{b}} \right|\left| {\vec{c}} \right|\sin \theta \hat{n} \right)$. We use the information we obtained from angles to conclude that $\vec{a}$ is either in direction of $\hat{n}$ or opposite to the direction of $\hat{n}$.\[\]

Complete step-by-step solution:
We know that the dot product of two vectors $\vec{a}$ and $\vec{b}$ is denoted as $\vec{a}\cdot \vec{b}$ and is given by $\vec{a}\cdot \vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\cos \theta =ab\cos \theta $ where $\theta $ is the smaller angle between the vectors $\vec{a}$ and $\vec{b}$. The magnitude of the vector $\vec{a}$ here is symbolized as $\left| {\vec{a}} \right|$ or $a$.
 The cross product between two vectors is denoted as $\vec{a}\times \vec{b}$ and is given by $\vec{a}\times \vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\sin \theta \hat{n}=ab\sin \theta \hat{n}$ where $\hat{n}$ is a vector perpendicular to both $\vec{a}$ and $\vec{b}$ and in a direction according to right hand rule. \[\]
It is given in the question that $\vec{a}$, $\vec{b}$ and $\vec{c}$ are unit vectors which means the magnitude of the vectors is 1. We write i symbols as $a=b=c=1$. Let us denote the angle between $\vec{a}$ and $\vec{b}$ as $\alpha $ and also denote the angle between $\vec{b}$ and $\vec{c}$ as $\beta $. We are given that $\vec{a}\cdot \vec{b}=\vec{a}\cdot \vec{c}=0$. So we have
\[ \begin{align}
  & \vec{a}\cdot \vec{b}=\vec{a}\cdot \vec{c}=0 \\
 & \Rightarrow ab\cos \alpha =ac\cos \beta =0 \\
 & \Rightarrow 1\left( 1 \right)\cos \beta =1\left( 1 \right)\cos \beta =0 \\
 & \Rightarrow \cos \alpha =\cos \beta =0 \\
 & \Rightarrow \alpha =\beta =\dfrac{\pi }{2} \\
\end{align} \]
So the unit vector $\vec{a}$ is perpendicular to $\vec{b}$ and also perpendicular to $\vec{c}$. So we can say vector $\vec{a}$ is perpendicular to the plane containing $\vec{b}$ and $\vec{c}$. Now we take the cross product of $\vec{b}$ and $\vec{c}$ . The angle between them is given to us in the question as $\theta =\dfrac{\pi }{6}$. So we have,
\[\vec{b}\times \vec{c}=bc\sin \theta \hat{n}=1\left( 1 \right)\sin \left( \dfrac{\pi }{6} \right)\hat{n}=\dfrac{1}{2}\hat{n}\]
Where $\hat{n}$ is a unit vector perpendicular to both $\vec{b}$ , $\vec{c}$ and also is in a direction according to right hand rule. The unit vector that is in opposite direction of $\hat{n}$ we can write as $-\hat{n}$. We are given in the question that $\vec{a}$ is unit vector and we have obtained that $\vec{a}$ is perpendicular to both $\vec{b}$ and $\vec{c}$ . So the direction of $\vec{a}$ can be in either along the direction of $\hat{n}$ or the direction $-\hat{n}$. So we have,
\[\begin{align}
  & \vec{b}\times \vec{c}=\dfrac{1}{2}\left( \pm \vec{a} \right) \\
 & \Rightarrow \pm \vec{a}=2\left( \vec{b}\times \vec{c} \right) \\
 & \Rightarrow \vec{a}=\pm 2\left( \vec{b}\times \vec{c} \right) \\
\end{align}\]
Hence it is proved. \[\]

Note: We take note that none of the vectors are zero vectors. We also note that $\vec{a}$ is perpendicular to the plane containing $\vec{b}$ and $\vec{c}$ means $\vec{a}$ is perpendicular to $l\vec{b}+m\vec{c}$ where $l,m$ are real scalars. The right hand rule means if we point our forefinger in the direction of $\vec{a}$ and the middle finger in the direction of $\vec{b}$, then the thumb gives the direction of $\vec{a}\times \vec{b}$.