Let O be the origin, and $\vec{OX}, \vec{OY}, \vec{OZ}$ be three unit vectors in the directions of the sides $\vec{QR}, \vec{RP}, \vec{PQ}$ respectively, of a ΔPQR. If the triangle PQR varies, then the minimum value of $\cos (P+Q) + \cos (Q+ R) + \cos (R+P)$, is
Answer & explanation
Correct answer: option 2
In ΔPQR, $P+Q+R =π$.
We have, $\vec{OX}.\vec{OY} = \cos (π-R) = - \cos R$
$\vec{OY}.\vec{OZ} = \cos (π-P) = - \cos P$
$\vec{OZ}.\vec{OX} = \cos (π-Q) = - \cos Q$
$∴\cos (P+Q) + \cos (Q+ R) + \cos (R+P)$
$= \cos (π-R) + \cos (π-Р)+ cos (π-Q)$
$=-(\cos P+\cos Q+\cos R)$
$=\vec{OX}.\vec{OY}+\vec{OY}.\vec{OZ}+\vec{OZ}.\vec{OX}$
$=\frac{1}{2}\left[\left|\vec{OX}+\vec{OY}+\vec{OZ}\right|^2-\left\{|\vec{OX}|^2+|\vec{OY}|^2+|\vec{OZ}|^2\right\}\right]$
$=\frac{1}{2}\left[\left|\vec{OX}+\vec{OY}+\vec{OZ}\right|^2-3\right]=\frac{1}{2}\left\{|\vec{OX}|^2+|\vec{OY}|^2+|\vec{OZ}|^2\right\}-\frac{3}{2}$
$≥-\frac{3}{2}$ $\left[∵\left\{|\vec{OX}|^2+|\vec{OY}|^2+|\vec{OZ}|^2\right\}≥0\right]$