Q .49. - Calculus

Question

In Exercises 49-51, two direction cosines are given. Use Exercise 48 to find the third direction cosine.

$\cos \alpha =\frac{1}{2},\cos \beta =\frac{1}{2}\text{. }$

Step-by-Step Solution

Answer

$\text{The value of the third direction cosine is }\pm \frac{\sqrt{2}}{2}\text{. }$

Step 1:Given information

$\text{ Consider the direction cosines }\cos \alpha =\frac{1}{2},\cos \beta =\frac{1}{2}\text{. }$

Step 2:Simplification

$\text{ The objective is to find the third direction cosine using the result Consider the direction cosines }$

${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$

$\text{ Also, }{\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$

$\text{ Now, since }\cos \alpha =\frac{1}{2},\cos \beta =\frac{1}{2}\text{. }$

$\text{ Substitute the values in (1) }$

${\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2+{\cos }^2\gamma =1$

$\frac{1}{4}+\frac{1}{4}+{\cos }^2\gamma =1$

$\frac{2}{4}+{\cos }^2\gamma =1$

$\frac{1}{2}+{\cos }^2\gamma =1$

${\cos }^2\gamma =1-\frac{1}{2}$

$\cos \gamma =\pm \frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}$

$=\pm \frac{\sqrt{2}}{2}$

$\text{ Hence, the value of the third direction cosine is }\pm \frac{\sqrt{2}}{2}\text{. }$