Q .51. - Calculus

Question

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

$\text{ 51. }\cos \alpha =\frac{1}{2},\cos \gamma =\frac{\sqrt{3}}{4}\text{. }$

Step-by-Step Solution

Answer

$\text{ the value of the third directicn cosine is }\pm \frac{3}{4}$

Step 1:Given information

$\cos \alpha =\frac{1}{2},\cos \gamma =\frac{\sqrt{3}}{4}\text{. }$

Step 2:Simplification

$\text{ Consider the drection cosines }\cos \alpha =\frac{1}{2},\cos \gamma =\frac{\sqrt{3}}{4}\text{. }$

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

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

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

width="234" height="45" style="max-width: none; vertical-align: -15px;" $\mathrm{Nos},\mathrm{sinc}e \cos \alpha =\frac{1}{2},\cos \gamma =\frac{\sqrt{3}}{4}$

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

${\left(\frac{1}{2}\right)}^2+{\cos }^2\beta +{\left(\frac{\sqrt{3}}{4}\right)}^2=1$

$\frac{1}{4}+{\cos }^2\beta +\frac{3}{16}=1$

$\frac{4+3}{16}+{\cos }^2\beta =1$

$\frac{7}{16}+{\cos }^2\beta =1$

${\cos }^2\beta =1-\frac{7}{16}$

$=\frac{16-7}{16}$$=\pm \frac{9}{16}$

$\cos \gamma =\pm \sqrt{\frac{9}{16}}$

$=\pm \frac{3}{4}$

$\text{ the value of the third directicn cosine is }\left.\pm \frac{3}{4}\right]$