Question

Unit vectors $\vec a$ and $\vec b$ are perpendicular and a unit vector $\vec c$ is inclined at an angle $\theta $ to both $\vec a$ and $\vec b$. If $\vec c = \alpha \vec a + \beta \vec b + \gamma \left( {\vec a \times \vec b} \right)$, then
A) $\alpha = \beta $
B) ${\gamma ^2} = 1 - 2{\alpha ^2}$
C) ${\gamma ^2} = - \cos 2\theta $
D) ${\beta ^2} = \dfrac{{1 + \cos 2\theta }}{2}$

Answer 1

Hint: We will first find the dot products of $\vec a,\vec b$ and $\vec c$with each other. Then take the $\vec c = \alpha \vec a + \beta \vec b + \gamma \left( {\vec a \times \vec b} \right)$ and take its dot product with $\vec a,\vec b$ to reach to the values of $\alpha $ and $\beta $. Then use their values in $\vec c$ and do some modifications to get the value of $\gamma $.

Complete step-by-step answer:
We know that \[\vec a \cdot \vec b = \left| {\vec a} \right| \cdot \left| {\vec b} \right| \cdot \cos \theta \], where $\theta $ is the angle between $\vec a$ and $\vec b$.
Since, $\vec a$, $\vec b$ and $\vec c$ are unit vectors.
The dot product of $\vec a$ with $\vec a$ is,
$ \Rightarrow \vec a \cdot \vec a = \left| {\vec a} \right| \cdot \left| {\vec a} \right| \cdot \cos 0$
Since, $\vec a$ is unit vectors.
$ \Rightarrow \vec a \cdot \vec a = {\left| {\vec a} \right|^2} = 1$...........….. (1)
The dot product of $\vec b$ with $\vec b$ is,
\[ \Rightarrow \vec b \cdot \vec b = \left| {\vec b} \right| \cdot \left| {\vec b} \right| \cdot \cos 0\]
Since, $\vec b$ is unit vectors.
\[ \Rightarrow \vec b \cdot \vec b = {\left| {\vec b} \right|^2} = 1\]..............….. (2)
Now, the dot product of $\vec a$ with $\vec c$ is,
\[ \Rightarrow \vec a \cdot \vec c = \left| {\vec a} \right| \cdot \left| {\vec c} \right| \cdot \cos \theta \]
Since, $\vec a$ and $\vec c$ are unit vectors.
\[ \Rightarrow \vec a \cdot \vec c = \cos \theta \].................….. (3)
Now, the dot product of $\vec b$ with $\vec c$ is,
\[ \Rightarrow \vec b \cdot \vec c = \left| {\vec b} \right| \cdot \left| {\vec c} \right| \cdot \cos \theta \]
Since, $\vec b$ and $\vec c$ are unit vectors.
\[ \Rightarrow \vec b \cdot \vec c = \cos \theta \]..............….. (4)
Since, $\vec a$ and $\vec b$ are parallel to each other.
\[ \Rightarrow \vec a \cdot \vec b = \left| {\vec a} \right| \cdot \left| {\vec b} \right| \cdot \cos 90\]
Substitute the values,
\[ \Rightarrow \vec a \cdot \vec b = 0\]............….. (5)
We also know that if we have two vectors then the resultant of their cross product is always perpendicular to both the vectors.
So, $\vec a$ is perpendicular to \[\left( {\vec a \times \vec b} \right)\],
\[ \Rightarrow \vec a \cdot \left( {\vec a \times \vec b} \right) = 0\]...........….. (6)
So, $\vec b$ is perpendicular to \[\left( {\vec a \times \vec b} \right)\] as well,
\[ \Rightarrow \vec b \cdot \left( {\vec a \times \vec b} \right) = 0\].........….. (7)
Now, we have $\vec c = \alpha \vec a + \beta \vec b + \gamma \left( {\vec a \times \vec b} \right)$. So,
\[ \Rightarrow \vec b \cdot \vec c = \vec b \cdot \left\{ {\alpha \vec a + \beta \vec b + \gamma \left( {\vec a \times \vec b} \right)} \right\}\]
This can be written as,
\[ \Rightarrow \vec b \cdot \vec c = \alpha \left( {\vec b \cdot \vec a} \right) + \beta \left( {\vec b \cdot \vec b} \right) + \gamma \left\{ {\vec b \cdot \left( {\vec a \times \vec b} \right)} \right\}\]
Substitute the values from equation (2), (4), (5), and (7)
$ \Rightarrow \cos \theta = \beta $..............….. (8)
Now, find the dot product of \[\vec a\] and \[\vec c\].
\[ \Rightarrow \vec a \cdot \vec c = \vec a \cdot \left\{ {\alpha \vec a + \beta \vec b + \gamma \left( {\vec a \times \vec b} \right)} \right\}\]
This can be written as,
\[ \Rightarrow \vec a \cdot \vec c = \alpha \left( {\vec a \cdot \vec a} \right) + \beta \left( {\vec a \cdot \vec b} \right) + \gamma \left\{ {\vec a \cdot \left( {\vec a \times \vec b} \right)} \right\}\]
Substitute the values from equation (1), (4), (5) and (7)
$ \Rightarrow \cos \theta = \alpha $............….. (9)
Compare equation (8) and (9),
$\therefore \alpha = \beta $
Substitute the value in $\vec c$,
\[ \Rightarrow \vec c = \alpha \vec a + \alpha \vec b + \gamma \left( {\vec a \times \vec b} \right)\]
Now, the dot product of $\vec c$ with $\vec c$,
\[ \Rightarrow \vec c \cdot \vec c = {\left( {\alpha \vec a} \right)^2} + {\left( {\alpha \vec b} \right)^2} + {\left\{ {\gamma \left( {\vec a \times \vec b} \right)} \right\}^2}\]
Square the terms,
\[ \Rightarrow \vec c \cdot \vec c = {\alpha ^2}\left( {\vec a \cdot \vec a} \right) + {\alpha ^2}\left( {\vec b \cdot \vec b} \right) + {\gamma ^2}{\left( {\vec a \times \vec b} \right)^2}\]
Substitute the values as all are unit vectors,
$ \Rightarrow 1 = {\alpha ^2} + {\alpha ^2} + {\gamma ^2}$
Add the terms on the right side,
$ \Rightarrow 1 = 2{\alpha ^2} + {\gamma ^2}$
Move $2{\alpha ^2}$ on the other side,
$\therefore {\gamma ^2} = 1 - 2{\alpha ^2}$.........….. (10)
Now, substitute the value of $\alpha $ from equation (9),
$ \Rightarrow {\gamma ^2} = 1 - 2{\cos ^2}\theta $
Take -1 common from right side,
$ \Rightarrow {\gamma ^2} = - \left( {2{{\cos }^2}\theta - 1} \right)$
Substitute the value $\cos 2\theta = 2{\cos ^2}\theta - 1$,
$\therefore {\gamma ^2} = - \cos 2\theta $..........….. (11)
Compare equation (10) and (11),
$ \Rightarrow 1 - 2{\alpha ^2} = - \cos 2\theta $
Substitute the value $\beta $ in place of $\alpha $,
$ \Rightarrow 1 - 2{\beta ^2} = - \cos 2\theta $
Simplify the terms,
$ \Rightarrow 2{\beta ^2} = 1 + \cos 2\theta $
Divide both sides by 2,
$\therefore {\beta ^2} = \dfrac{{1 + \cos 2\theta }}{2}$

Hence, option (a), (b), (c) and (d) all are correct.

Note: A vector is an object that has both magnitude and direction. The students must know the difference between the dot product and cross product. The dot product results in a scalar quantity whereas the cross-product results in a vector only.