Q .50. - Calculus

Question

$\cos \beta =\frac{1}{4},\cos \gamma =\frac{1}{3}\text{. }$

Step-by-Step Solution

Answer

$\text{The value of the third direction cosine is }\pm \frac{\sqrt{119}}{12}$

Step 1:Given information

$\cos \beta =\frac{1}{4},\cos \gamma =\frac{1}{3}\text{. }$

Step 2:Simplification

$\text{ Consider the direction cosines }\cos \beta =\frac{1}{4},\cos \gamma =\frac{1}{3}\text{. }$

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

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

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

$\text{ Now, since }\cos \beta =\frac{1}{4},\cos \gamma =\frac{1}{3}\text{. }$

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

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

${\cos }^2\alpha +\frac{1}{16}+\frac{1}{9}=1$

${\cos }^2\alpha +\frac{9+16}{144}=1$

${\cos }^2\alpha +\frac{25}{144}=1$

${\cos }^2\alpha =1-\frac{25}{144}$

$\cos \alpha =\pm \frac{\sqrt{119}}{12}$

$\text{ Hence, the value of the third direction cosine is }\pm \frac{\sqrt{119}}{12}$