If $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$, then the value of $x^2+y^2+z^2-2 x y z$ is
If $\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi$, then the value of $x^2+y^2+z^2-2 x y z$ is
3
2
1
5
Solution
Since, $0 \leq \cos ^{-1} x \leq \pi$
$\therefore \quad \cos ^{-1} x$ cannot be greater than $\pi$
$\therefore \quad \cos ^{-1} x=\cos ^{-1} y=\cos ^{-1} x=\pi$
Therefore, $x=y=\mathrm{z}=-1$
Putting these values in the given expression, we get
$\begin{aligned}
& x^2+y^2+z^2-2 x y z \\
& =(-1)^2+(-1)^2+(-1)^2-2(-1)(-1)(-1) \\
& =1+1+1-2(-1) \\
& =3+2 \\
& =5
\end{aligned}$