Question

$\left|\begin{array}{ccc}1& \cos \alpha & \cos \beta \\ \cos \alpha & 1& \cos \gamma \\ \cos \beta & \cos \gamma & 1\end{array}\right|=\left|\begin{array}{ccc}0& \cos \alpha & \cos \beta \\ \cos \alpha & 0& \cos \beta \\ \cos \beta & \cos \gamma & 0\end{array}\right|$ if ${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =$

• A. $1$
• B. $2$
• C. $\frac{3}{2}$
• D. $\frac{1}{2}$

Answer 1

The correct option is A $1$

MISTAKE:in the determinant having $0$'s $3^{rd}$ element in the second row must have $\cos \gamma$ instead of $\cos \beta$

$\left|\begin{array}{ccc}1& \cos \alpha & \cos \beta \\ \cos \alpha & 1& \cos \gamma \\ \cos \beta & \cos \gamma & 1\end{array}\right|=1(1-{\cos }^2\gamma )-\cos \alpha (\cos \alpha -\cos \beta \cos \gamma )+\cos \beta (\cos \alpha \cos \gamma -\cos \beta )$

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

$\left|\begin{array}{ccc}0& \cos \alpha & \cos \beta \\ \cos \alpha & 0& \cos \gamma \\ \cos \beta & \cos \gamma & 0\end{array}\right|=0(0-{\cos }^2\gamma )-\cos \alpha \left(\cos \alpha \right(0)-\cos \beta \cos \gamma )+\cos \beta (\cos \alpha \cos \gamma -\cos \beta (0\left)\right)$

$=2\cos \alpha \cos \beta \cos \gamma$

Given $\left|\begin{array}{ccc}1& \cos \alpha & \cos \beta \\ \cos \alpha & 1& \cos \gamma \\ \cos \beta & \cos \gamma & 1\end{array}\right|=\left|\begin{array}{ccc}0& \cos \alpha & \cos \beta \\ \cos \alpha & 0& \cos \gamma \\ \cos \beta & \cos \gamma & 0\end{array}\right|$

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

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