Question

Let the unit vectors a and b be perpendicular and the unit vector c be inclined at an angle $\theta$ to both a and b. If $\vec c = x\vec a + y\vec b + z(\vec a \times \vec b)$, then find the values of x, y and z.
A. $x = \cos \theta ,y = \sin \theta$ and $z = \cos 2\theta$
B. $x = \sin \theta ,y = \cos \theta$ and $z = - \cos 2\theta$
C. $x = y = \cos \theta$ and ${z^2} = \cos 2\theta$
D. $x = y = \cos \theta$ and ${z^2} = - \cos 2\theta$

Answer 1

We will first find the dot products of $\vec a,\vec b$ and $\vec c$ with each other. Then take the $\vec c = x\vec a + y\vec b + z(\vec a \times \vec b)$ 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 z.

Complete step-by-step answer:
We know that $\vec a.\vec b = |\vec a|.|\vec b|.\cos \theta$, where $\theta$ is the angle between $\vec a$ and $\vec b$.
Therefore, $\vec a.\vec a = |\vec a|.|\vec a|.\cos {0^ \circ } = |\vec a{|^2} = 1$ (Because we are given that a is a unit vector) …….(1)
Similarly, $\vec b.\vec b = |\vec b|.|\vec b|.\cos {0^ \circ } = |\vec b{|^2} = 1$ ……………(2)
And, $\vec a.\vec b = |\vec a|.|\vec b|.\cos {90^ \circ } = 0$ (Because they are perpendicular to each other) …….(3)
$\vec a.\vec c = |\vec a|.|\vec c|.\cos \theta = \cos \theta$ (Because both the vectors a and c are unit vectors) ……….(4)
$\vec b.\vec c = |\vec b|.|\vec c|.\cos \theta = \cos \theta$ (Because both the vectors b and c are unit vectors) ……….(5)
We also know that if we have two vectors then the resultant of their cross product is always perpendicular to both the vectors.
Therefore, $\vec a$ is perpendicular to $\vec a \times \vec b$ and $\vec b$ is perpendicular to $\vec a \times \vec b$ as well.
So, $\vec a.(\vec a \times \vec b) = 0$ and $\vec b.(\vec a \times \vec b) = 0$. ………(6)
Now, we are given that $\vec c = x\vec a + y\vec b + z(\vec a \times \vec b)$.
So, $\vec b.\vec c = \vec b\\{ x\vec a + y\vec b + z(\vec a \times \vec b)\\}$
This can be written as $\vec b.\vec c = x(\vec b.\vec a) + y(\vec b.\vec b) + z\\{ \vec b.(\vec a \times \vec b)\\}$
Now, using (2), (3) and (6), we will get:-
$\Rightarrow \vec b.\vec c = y$ ……….(7)
Now, $\vec a.\vec c = \vec a\\{ x\vec a + y\vec b + z(\vec a \times \vec b)\\}$
This can be written as $\vec a.\vec c = x(\vec a.\vec a) + y(\vec a.\vec b) + z\\{ \vec a.(\vec a \times \vec b)\\}$
Now, using (1), (3) and (6), we will get:-
$\Rightarrow \vec a.\vec c = x$ ……….(8).
Using (4), (5), (7) and (8), we will get:- $x = y = \cos \theta$
Putting $x = y$ in $\vec c = x\vec a + y\vec b + z(\vec a \times \vec b)$, we will get:-
$\Rightarrow \vec c = x(\vec a + \vec b) + z(\vec a \times \vec b)$ …….(9)
Let $\vec p = \vec a + \vec b$ and $\vec q = \vec a \times \vec b$. ……….(10)
Now, $p = \vec a + \vec b$ always lies in the same place as vectors a and b and $q = \vec a \times \vec b$ always lies in a plane perpendicular to them. Therefore, $p = \vec a + \vec b$ and $q = \vec a \times \vec b$ are perpendicular to each other.
Consider $|\vec p| = |\vec a + \vec b| = \sqrt {|\vec a{|^2} + |\vec b{|^2} + 2\vec a.\vec b\cos \theta }$
Using (1), (2) and (3), we will get:-
$|\vec p| = \sqrt {1 + 1} = \sqrt 2$
$|x\vec p| = x\left( {\sqrt {1 + 1} } \right) = \sqrt 2 x$ …………….(11)
Consider $\vec q = \vec a \times \vec b$, so, $|\vec q| = |\vec a|.|\vec b|\sin \theta$.
$\Rightarrow |\vec q| = \sin {90^ \circ } = 1$ (Using (1), (2) and the fact that the vectors a and b are perpendicular)
So, $|z\vec q| = z$ …………..(12)
Since, $\vec c = x\vec a + y\vec b + z(\vec a \times \vec b) = x\vec p + z\vec q$
So, $|\vec c| = |x\vec p + z\vec q| = \sqrt {|x\vec p{|^2} + |y\vec q{|^2} + 2|x\vec p|.|y\vec q|.\cos \varphi }$
Since $p = \vec a + \vec b$ and $q = \vec a \times \vec b$ are perpendicular to each other, therefore, $\cos \varphi = 0$.
Hence, $|\vec c| = \sqrt {|x\vec p{|^2} + |y\vec q{|^2}}$
Now, using (11) and (12), we will get:-
$|\vec c| = \sqrt {2{x^2} + {z^2}}$
Squaring both sides, we will get:-
${c^2} = 2{x^2} + {z^2}$
Now, since the vector c is a unit vector. We will get:-
$\Rightarrow 1 = 2{x^2} + {z^2}$
$\Rightarrow {z^2} = 1 - 2{x^2}$
Now, we found out that $x = \cos \theta$.
$\Rightarrow {z^2} = 1 - 2{\cos ^2}\theta = - \cos 2\theta$

So, the correct answer is “Option D”.

Additional Information: Being a vector operation, the Cross Product is extremely important in all sorts of sciences (particularly physics), engineering, and mathematics. One important example of the Cross Product is in torque or moment

Note: 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.