Question

If a vector $\overrightarrow P $ making angles $\alpha,\beta $ and $\gamma $ respectively with the X, Y, and Z axes respectively.
Then ${\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\gamma = $
A. 0
B. 1
C. 2
D. 3

Answer 1

Hint: Let the given vector $\overrightarrow P $ be \[{P_1}\hat i + {P_2}\hat j + {P_3}\hat k\]. Find the relations for the $\cos \alpha ,\cos \beta $ and $\cos \gamma $ by finding the dot product of the vector $\overrightarrow P $ with the X,Y and Z axes respectively. Use the values of $\cos \alpha ,\cos \beta $ and $\cos \gamma $ to find the value of ${\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\gamma $

Complete step by step solution:

Let us assume the three dimensional $\overrightarrow P $ be \[{P_1}\hat i + {P_2}\hat j + {P_3}\hat k\]. Then the magnitude of the vector $\overrightarrow P $ will be
$\left| P \right| = \sqrt {{P_1}^2 + {P_2}^2 + {P_3}^2} $
It is known that if the angle between two vectors, $\overrightarrow A $ and $\overrightarrow B $ is $\theta $, then the given relation is valid.
$\cos \theta = \dfrac{{\overrightarrow A .\overrightarrow B }}{{\left| A \right|\left| B \right|}}$.
As we are given in the question, the angle between the vector $\overrightarrow P $ and $x$ axis is $\alpha $.
The unit vector in the direction of the $x$ axis is given by $\hat i$. The magnitude of the unit vector is 1.
Substituting the value $\alpha $ for $\theta $, value $\overrightarrow P $ for $\overrightarrow A $ and value $\hat i$ for $\overrightarrow B $ in the relation $\cos \theta = \dfrac{{\overrightarrow A .\overrightarrow B }}{{\left| A \right|\left| B \right|}}$, we get
$\cos \alpha = \dfrac{{\overrightarrow P .\hat i}}{{\left| P \right|\left| {\hat i} \right|}}$
Substituting the value \[{P_1}\hat i + {P_2}\hat j + {P_3}\hat k\] for $\overrightarrow P $, the value $\sqrt {{P_1}^2 + {P_2}^2 + {P_3}^2} $ for $\left| P \right|$ and the value 1 for $\left| {\hat i} \right|$ in the above relation, we get
$\cos \alpha = \dfrac{{\left( {{P_1}\hat i + {P_2}\hat j + {P_3}\hat k} \right).\hat i}}{{\sqrt {{P_1}^2 + {P_2}^2 + {P_3}^2} }}$
The equation can be simplified as
$\cos \alpha = \dfrac{{{P_1}}}{{\sqrt {{P_1}^2 + {P_2}^2 + {P_3}^2} }}$, as $\hat i.\hat i = 1$
Similarly, the relations can be derived for the $\cos \beta $ and $\cos \gamma $.
$\cos \beta = \dfrac{{{P_2}}}{{\sqrt {{P_1}^2 + {P_2}^2 + {P_3}^2} }}$
$\cos \gamma = \dfrac{{{P_3}}}{{\sqrt {{P_1}^2 + {P_2}^2 + {P_3}^2} }}$
We can simplify the given relation ${\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\gamma $ by using the trigonometric identity ${\sin ^2}\theta = 1 - {\cos ^2}\theta $, we get
$
  1 - {\cos ^2}\alpha + 1 - {\cos ^2}\beta + 1 - {\cos ^2}\gamma \\
   \Rightarrow 3 - \left( {{{\cos }^2}\alpha + {{\cos }^2}\beta + {{\cos }^2}\gamma } \right) \\
$
Substituting the values of $\cos \alpha ,\cos \beta $ and $\cos \gamma $ in the desired relation $3 - \left( {{{\cos }^2}\alpha + {{\cos }^2}\beta + {{\cos }^2}\gamma } \right)$
$3 - \left( {{{\left( {\dfrac{{{P_1}}}{{\sqrt {{P_1}^2 + {P_2}^2 + {P_3}^2} }}} \right)}^2} + {{\left( {\dfrac{{{P_2}}}{{\sqrt {{P_1}^2 + {P_2}^2 + {P_3}^2} }}} \right)}^2} + {{\left( {\dfrac{{{P_3}}}{{\sqrt {{P_1}^2 + {P_2}^2 + {P_3}^2} }}} \right)}^2}} \right)$
Simplifying the above expression, we get
$
  3 - \left( {\dfrac{{{P_1}^2}}{{{P_1}^2 + {P_2}^2 + {P_3}^2}} + \dfrac{{{P_2}^2}}{{{P_1}^2 + {P_2}^2 + {P_3}^2}} + \dfrac{{{P_3}^2}}{{{P_1}^2 + {P_2}^2 + {P_3}^2}}} \right) \\
   \Rightarrow 3 - \left( {\dfrac{{{P_1}^2 + {P_2}^2 + {P_3}^2}}{{{P_1}^2 + {P_2}^2 + {P_3}^2}}} \right) \\
   \Rightarrow 3 - \left( 1 \right) \\
   \Rightarrow 2 \\
$
Thus the option C is the correct option.

Note: If the angle between two vectors, $\overrightarrow A $ and $\overrightarrow B $ is $\theta $, then we can say $\cos \theta = \dfrac{{\overrightarrow A .\overrightarrow B }}{{\left| A \right|\left| B \right|}}$. The dot product of the two vectors \[{A_1}\hat i + {A_2}\hat j + {A_3}\hat k\] and \[{B_1}\hat i + {B_2}\hat j + {B_3}\hat k\] can be simplified as ${A_1}{B_1} + {A_2}{B_2} + {A_3}{B_3}$. Thus the value $\left( {{P_1}\hat i + {P_2}\hat j + {P_3}\hat k} \right).\hat i$ is simplified as ${P_1} + 0 + 0$ = ${P_1}$.