Question

Unit vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular, and unit vector $\overrightarrow{c}$ is inclined at an angle $\theta$ to both $\overrightarrow{a}$ and $\overrightarrow{b}$ . If $\overrightarrow{c}=\alpha \overrightarrow{a}+\beta \overrightarrow{b}+\gamma \left( \overrightarrow{a}\times \overrightarrow{b} \right)$ , 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:- Use the formula that $\cos \phi =\dfrac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}$ , where $\phi$ is the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ . Also, the value of $|\overrightarrow{a}|$ is 1, provided $\overrightarrow{a}$ is a unit vector.

Complete step-by-step solution -
It is given in the question that $\overrightarrow{a}$ and $\overrightarrow{b}$ are perpendicular, and we know that the dot product of perpendicular vectors is zero.
$\therefore \overrightarrow{a}.\overrightarrow{b}=0.........(i)$
Now, as the angle between $\overrightarrow{a}$ and $\overrightarrow{c}$ is $\theta$ and the angle between $\overrightarrow{b}$ and $\overrightarrow{c}$ is also $\theta$ . So, the cosine of both the angles would be the same. We also know the $\cos \phi =\dfrac{\overrightarrow{a}.\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}$ , where $\phi$ is the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ .
$\therefore \overrightarrow{a}.\overrightarrow{c}=\overrightarrow{b}.\overrightarrow{c}$
$\Rightarrow \overrightarrow{a}.\left( \alpha \overrightarrow{a}+\beta \overrightarrow{b}+\gamma \left( \overrightarrow{a}\times \overrightarrow{b} \right) \right)=\overrightarrow{b}.\left( \alpha \overrightarrow{a}+\beta \overrightarrow{b}+\gamma \left( \overrightarrow{a}\times \overrightarrow{b} \right) \right)$
$\Rightarrow \alpha \overrightarrow{a}.\overrightarrow{a}+\beta \overrightarrow{a}.\overrightarrow{b}+\gamma \overrightarrow{a}.\left( \overrightarrow{a}\times \overrightarrow{b} \right)=\alpha \overrightarrow{a}.\overrightarrow{b}+\beta \overrightarrow{b}.\overrightarrow{b}+\gamma \overrightarrow{b}.\left( \overrightarrow{a}\times \overrightarrow{b} \right)$
Now we will substitute the required value from equation (i). On doing so, we get

$\Rightarrow \alpha \overrightarrow{a}.\overrightarrow{a}+0+\gamma \overrightarrow{a}.\left( \overrightarrow{a}\times \overrightarrow{b} \right)=0+\beta \overrightarrow{b}.\overrightarrow{b}+\gamma \overrightarrow{b}.\left( \overrightarrow{a}\times \overrightarrow{b} \right)$
We also know that $\overrightarrow{a}$ and $\overrightarrow{b}$ is always perpendicular to $\overrightarrow{a}\times \overrightarrow{b}$ and the dot product of perpendicular vectors is zero.
$\therefore \alpha \overrightarrow{a}.\overrightarrow{a}+0=\beta \overrightarrow{b}.\overrightarrow{b}$
Now we know that $\overrightarrow{a}.\overrightarrow{a}=|\overrightarrow{a}{{|}^{2}}$ . So, we get
$\alpha |\overrightarrow{a}{{|}^{2}}=\beta |\overrightarrow{b}{{|}^{2}}$
Also, it is given that $\overrightarrow{a}$ and $\overrightarrow{b}$ are unit vectors. Therefore, the magnitude of $\overrightarrow{a}$ and $\overrightarrow{b}$ is equal to 1.
$\alpha \times 1=\beta \times 1$
$\Rightarrow \alpha =\beta$
Therefore, the value of the $\alpha$ must be equal to the value of $\beta$ for the conditions given in the question to be satisfied.
Hence, the answer to the above question is option (a).

Note: It is important to remember the properties of the vector product and the scalar product for solving most of the problems related to vectors. Also, be careful about the calculations and the signs you are using while solving the calculation.