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. \(\alpha \in \left(\frac{\pi}{2}, \pi\right)\) and \(\gamma \in \left(\frac{\pi}{2}, \pi\right)\)
• B. \(\alpha \in \left(0, \frac{\pi}{2}\right)\) 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. \(\alpha \in \left(0, \frac{\pi}{2}\right)\) and \(\gamma \in \left(0, \frac{\pi}{2}\right)\)

Hint:

Find a normal to the plane using the cross product of two vectors lying in it, then choose the sign so that the y-component (\(\cos\beta\)) is positive and read off the signs of the x and z components.

Answer 1

The Correct Option is A

Let A(1,2,3), B(2,3,4), C(1,5,7).
\(\vec{AB} = (1,1,1)\), \(\vec{AC} = (0,3,4)\).
Normal: \(\vec{AB} \times \vec{AC} = (1\cdot 4 - 1\cdot 3,\ 1\cdot 0 - 1\cdot 4,\ 1\cdot 3 - 1\cdot 0) = (1, -4, 3)\).
The unit vector along OP is \(\pm\frac{1}{\sqrt{26}}(1,-4,3)\).
Since \(\beta \in \left(0,\frac{\pi}{2}\right)\), \(\cos\beta \gt 0\), so the y-component must be positive. Hence \(\widehat{OP} = \frac{1}{\sqrt{26}}(-1, 4, -3)\).
So \(\cos\alpha = -\frac{1}{\sqrt{26}} \lt 0\) gives \(\alpha \in \left(\frac{\pi}{2},\pi\right)\), and \(\cos\gamma = -\frac{3}{\sqrt{26}} \lt 0\) gives \(\gamma \in \left(\frac{\pi}{2},\pi\right)\).
Answer: (1) \(\alpha \in \left(\frac{\pi}{2}, \pi\right)\) and \(\gamma \in \left(\frac{\pi}{2}, \pi\right)\)