JEE Mathematics 2023 Question with Solution

Given: The vector $\overrightarrow{OP}$ is a unit vector perpendicular to the plane through the points $(1,2,3)$, $(2,3,4)$ and $(1,5,7)$.

Find: The correct interval for the angles $\alpha$ and $\gamma$, given that $\beta \in \left(0,\frac{\pi}{2}\right)$.

A normal vector to the plane is obtained from two direction vectors in the plane:

Their cross product gives a normal vector:

So a direction vector perpendicular to the plane is proportional to $(1,-4,3)$.

Since $\overrightarrow{OP}$ is perpendicular to the plane and $\beta \in \left(0,\frac{\pi}{2}\right)$, we need $\cos\beta > 0$. Therefore we take the opposite normal direction:

This gives direction cosines

Now $\cos\alpha < 0$, so $\alpha \in \left(\frac{\pi}{2}, \pi\right)$. Also $\cos\gamma < 0$, so $\gamma \in \left(\frac{\pi}{2}, \pi\right)$.

Therefore, the correct option is A.

The plane through the three given points can be written as

Expanding this determinant gives the plane equation

Hence a normal vector is $(1,-4,3)$, and the opposite normal is $(-1,4,-3)$.

Because $\beta \in \left(0,\frac{\pi}{2}\right)$, the $y$-component must be positive. So the required normal direction is $(-1,4,-3)$. Its direction cosines are proportional to $(-1,4,-3)$, hence

Thus $\alpha$ and $\gamma$ are both obtuse. Therefore the correct option is A.

The solution also contains a discrepancy where it displays "The Correct Option is B", but the worked conclusion and interval signs clearly give option 1, which maps to A.