Three Dimensional Geometry Question 18
Question 18 - 30 January - Shift 1
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 ?
(1) $\alpha \in(\frac{\pi}{2}, \pi)$ and $\gamma \in(\frac{\pi}{2}, \pi)$
(2) $\alpha \in(0, \frac{\pi}{2})$ and $\gamma \in(0, \frac{\pi}{2})$
(3) $\alpha \in(\frac{\pi}{2}, \pi)$ and $\gamma \in(0, \frac{\pi}{2})$
(4) $\alpha \in(0, \frac{\pi}{2})$ and $\gamma \in(\frac{\pi}{2}, \pi)$
Show Answer
Answer: (1)
Solution:
Formula: Equation of a plane (6.7), Direction ratio and direction cosines
Equation of plane :-
$ \begin{aligned} & \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 \end{aligned} $
D.R’s of normal of plane $<1,-4,3>$
D.C’s of $\quad\langle \pm \frac{1}{\sqrt{26}}, \mp \frac{4}{\sqrt{26}}, \pm \frac{3}{\sqrt{26}}\rangle$
$\cos \beta=\frac{4}{\sqrt{26}}$
$\cos \alpha=\frac{-1}{\sqrt{26}} \quad \frac{\pi}{2}<\alpha<\pi$
$\cos \gamma=\frac{-3}{\sqrt{26}} \quad \frac{\pi}{2}<\gamma<\pi$
Ans. : (1)
