Question

If a vector $\vec p$ making angles $\alpha ,\beta ,{\text{ and }}\lambda $ respectively with the X, Y and Z axis respectively, then the value of ${\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\lambda $ =
$\left( a \right)$ 0
$\left( b \right)$ 1
$\left( c \right)$ 2
$\left( d \right)$ 3

Answer 1

Hint: In this particular type of question use the property that the dot product of any vector with the unit vector is equal to the product of modulus of vectors and multiply by the cosine of angle in which the unit vector has so use these concepts to reach the solution of the question.

Complete step by step answer:
Consider a line LL’ which makes an angle $\alpha ,\beta ,\lambda $ with the x, y and z axis respectively.
Now consider a vector, $\vec p$ be any non-zero vector along the line LL’
Let, $\vec p = {p_1}\hat i + {p_2}\hat j + {p_3}\hat k$.................. (1)
Now since $\hat i$ is a unit vector along the x-axis.
Therefore,
$\vec p.\hat i = \left| {\vec p} \right|.\left| {\hat i} \right|\cos \alpha $....................... (2)
Now from equation (1)
$ \Rightarrow \left| {\vec p} \right| = \sqrt {p_1^2 + p_2^2 + p_3^2} $
And
$\left| {\hat i} \right| = 1$
Now from equation (2) we have,
$ \Rightarrow \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} .\left( 1 \right)\cos \alpha $
$ \Rightarrow \left( {{p_1} + 0 + 0} \right) = \sqrt {p_1^2 + p_2^2 + p_3^2} .\left( 1 \right)\cos \alpha $, $\left[ {\because \hat i.\hat i = 1,\hat i.\hat j = 0,\hat i.\hat k = 0} \right]$
\[ \Rightarrow \cos \alpha = \dfrac{{{p_1}}}{{\sqrt {p_1^2 + p_2^2 + p_3^2} }}\]..................... (3)
Similarly,
\[ \Rightarrow \cos \beta = \dfrac{{{p_2}}}{{\sqrt {p_1^2 + p_2^2 + p_3^2} }}\]................. (4)
\[ \Rightarrow \cos \lambda = \dfrac{{{p_3}}}{{\sqrt {p_1^2 + p_2^2 + p_3^2} }}\].................. (5)
Now squaring and adding equations (3), (4), and (5) we have,
$ \Rightarrow {\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\lambda = {\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}$
$ \Rightarrow {\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\lambda = \left( {\dfrac{{p_1^2}}{{p_1^2 + p_2^2 + p_3^2}}} \right) + \left( {\dfrac{{p_2^2}}{{p_1^2 + p_2^2 + p_3^2}}} \right) + \left( {\dfrac{{p_3^2}}{{p_1^2 + p_2^2 + p_3^2}}} \right)$
$ \Rightarrow {\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\lambda = \left( {\dfrac{{p_1^2 + p_2^2 + p_3^2}}{{p_1^2 + p_2^2 + p_3^2}}} \right) = 1$
$ \Rightarrow {\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\lambda = 1$
Now as we know that, ${\sin ^2}x + {\cos ^2}x = 1$
Therefore, ${\cos ^2}x = 1 - {\sin ^2}x$ so use this property in the above equation we have,
$ \Rightarrow 1 - {\sin ^2}\alpha + 1 - {\sin ^2}\beta + 1 - {\sin ^2}\lambda = 1$
$ \Rightarrow {\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\lambda = 3 - 1 = 2$
So this is the required answer.

So, the correct answer is “Option C”.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the values of the standard trigonometric identities which is stated above, then simply substitute the values in equation and simplify as above we will get the required value of ${\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\lambda $.