Question

Let a unit vector $\widehat{ OP }$ make angles $\alpha, \beta, \gamma$ with the positive directions of the co-ordinate axes $OX$, $OY , OZ$ respectively, where $\beta \in\left(0, \frac{\pi}{2}\right)$. If $\widehat{ OP }$ is perpendicular to the plane through points $(1,2$, $3),(2,3,4)$ and $(1,5,7)$, then which one of the following is true ?

• A. $\alpha \in\left(0, \frac{\pi}{2}\right)$ and $\gamma \in\left(0, \frac{\pi}{2}\right)$
• B. $\alpha \in\left(\frac{\pi}{2}, \pi\right)$ and $\gamma \in\left(\frac{\pi}{2}, \pi\right)$
• C. $\alpha \in\left(\frac{\pi}{2}, \pi\right)$ and $\gamma \in\left(0, \frac{\pi}{2}\right)$
• D. $\alpha \in\left(0, \frac{\pi}{2}\right)$ annd $\gamma \in\left(\frac{\pi}{2}, \pi\right)$

Answer 1

Answer: (B)

Equation of plane :-
$ \begin{vmatrix} x-1 & y-2 & z-3 \\ 1 & 1 & 1 \\ 0 & 3 & 4 \end{vmatrix}=0 $
$ \Rightarrow[x-1]-4[y-2]+3[z-3]=0$
$ \Rightarrow x-4 y+3 z=2$
D.R's of normal of plane $<1,-4,3>$
D.C's of
$\left\langle\pm \frac{1}{\sqrt{26}}, \mp \frac{4}{\sqrt{26}}, \pm \frac{3}{\sqrt{26}}\right\rangle$
$ \cos \beta=\frac{4}{\sqrt{26}} $
$\cos \alpha=\frac{-1}{\sqrt{26}} \frac{\pi}{2} < \alpha < \pi $
$\cos \gamma=\frac{-3}{\sqrt{26}} \frac{\pi}{2} < \gamma < \pi$