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Question: A line makes \(\alpha ,\ \beta ,\gamma \) with the coordinate axes respectively, then \({{\sin }^{2}...

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

Explanation

Solution

Hint: Start by converting all the terms of the expression given in the question as a function of cosine using the identity sin⁡2x=1−cos⁡2x{{\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⁡2x=1−cos⁡2x{{\sin }^{2}}x=1-{{\cos }^{2}}x to convert all the terms of the expression to the required form.
sin⁡2α+sin⁡2β+sin⁡2γ{{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma
=1−cos⁡2α+1−cos⁡2β+1−cos⁡2γ=1-{{\cos }^{2}}\alpha +1-{{\cos }^{2}}\beta +1-{{\cos }^{2}}\gamma
=3−cos⁡2α−cos⁡2β−cos⁡2γ=3-{{\cos }^{2}}\alpha -{{\cos }^{2}}\beta -{{\cos }^{2}}\gamma
Now, we know that the direction cosines of a line are cos⁡α,cos⁡β,cos⁡γ\cos \alpha ,\cos \beta ,\cos \gamma where α, β,γ\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⁡2α+cos⁡2β+cos⁡2γ=1{{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma =1 . So, putting this value in our expression, we get
=3−(cos⁡2α+cos⁡2β+cos⁡2γ)=3-\left( {{\cos }^{2}}\alpha +{{\cos }^{2}}\beta +{{\cos }^{2}}\gamma \right)
=3−1=3-1
=2=2
Therefore, we can conclude that the value of the expression sin⁡2α+sin⁡2β+sin⁡2γ{{\sin }^{2}}\alpha +{{\sin }^{2}}\beta +{{\sin }^{2}}\gamma is equal to 2, provided α, β,γ\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⁡2x=cos⁡2x−1{{\sin }^{2}}x={{\cos }^{2}}x-1 instead of sin⁡2x=1−cos⁡2x{{\sin }^{2}}x=1-{{\cos }^{2}}x .