Question

A line makes $\alpha ,\ \beta ,\gamma $ with the coordinate axes respectively, then ${{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma $ is equal to.

Answer 1

Hint: Start by converting all the terms of the expression given in the question as a function of cosine using the identity ${{\sin }^{2}}x=1-{{\cos }^{2}}x$ . Then use the relation between the direction cosines of a line which states that the sum of the direction cosines of a line is equal to 1.

Complete step-by-step answer:
Let us start by simplification of the expression given in the question. We will use the identity ${{\sin }^{2}}x=1-{{\cos }^{2}}x$ to convert all the terms of the expression to the required form.
${{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma $
$=1-{{\cos }^{2}}\alpha +1-{{\cos }^{2}}\beta +1-{{\cos }^{2}}\gamma $
$=3-{{\cos }^{2}}\alpha -{{\cos }^{2}}\beta -{{\cos }^{2}}\gamma $
Now, we know that the direction cosines of a line are $\cos \alpha ,\cos \beta ,\cos \gamma $ where $\alpha ,\ \beta ,\gamma $ are the angles made with the coordinate axes respectively. We also know, that the sum of the squares of all three direction cosines of a line is equal to 1, i.e., ${{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1$ . So, putting this value in our expression, we get
$=3-\left( {{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma \right)$
$=3-1$
$=2$
Therefore, we can conclude that the value of the expression ${{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma $ is equal to 2, provided $\alpha ,\ \beta ,\gamma $ are the angle made with the coordinate axes, respectively.

Note: You should always remember that the direction cosines are the most important quantities related to a straight line in 3 dimensional space just like slope is for a 2 straight line in 2 dimensional space. Don’t get confused between the formulas of the line that we deal in coordinate geometry and the formulas of the line we are dealing in 3-D geometry, the basics remain the same, but the formulas are quite different. Also, be careful that you don’t confuse and take the identity as ${{\sin }^{2}}x={{\cos }^{2}}x-1$ instead of ${{\sin }^{2}}x=1-{{\cos }^{2}}x$ .