Question

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

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

Short answer

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

Step-by-step solution

Step 1:Given information

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

Step 2:Simplification

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

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

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

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

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

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

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

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

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

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

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

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

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