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

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.