Question

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

Short answer

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

Step-by-step solution

Step 1:Given information

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

Step 2:Simplification

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

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

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

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

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

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

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

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

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

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

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

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

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