Let a unit vector OP make angle α, β, γ with the positive directions of the co-ordinate axes OX, OY,

Q: Wait, why do we have two possible unit vectors? How do I know which one to pick?

Great observation! A plane extends infinitely in both directions, so there are always two unit normal vectors pointing in opposite directions (like the two sides of a sheet of paper). The problem gives you a specific constraint: $\beta \in (0, \frac{\pi}{2})$. Since $\cos \beta$ is the y-component of the unit vector, this constraint forces the y-component to be positive. This acts as a filter to reject the 'wrong' direction.

Q: Can I use the dot product instead of the cross product?

You could, but it would be much harder! You would have to set up a system of equations where the dot product of your unknown vector $\vec{v} = (x, y, z)$ with $\vec{AB}$ and $\vec{AC}$ is zero. The cross product is the 'shortcut' designed specifically for this task—it gives you the normal vector in one clean operation.

Q: Why does $\cos \alpha < 0$ mean $\alpha$ is in the second quadrant?

Remember the unit circle! For any angle $\theta \in [0, \pi]$, the cosine function is positive in the first quadrant $(0, \frac{\pi}{2})$ and negative in the second quadrant $(\frac{\pi}{2}, \pi)$. Since our calculated $\cos \alpha$ is negative, the angle must be obtuse, placing it firmly in the second quadrant.

Q: Does the order of points in the cross product matter?

It matters for the direction! If you calculate $\vec{AC} \times \vec{AB}$ instead of $\vec{AB} \times \vec{AC}$, you will get the exact negative of the vector I found. However, because we have the $\beta$ constraint, the final unit vector will be the same regardless of your starting order. The math will 'correct' itself!

Q: Is it possible for $\beta$ to be exactly $\pi/2$?

The problem specifies $\beta \in (0, \pi/2)$, which is an open interval. If $\beta$ were $\pi/2$, the y-component would be zero, meaning the vector would be parallel to the XZ-plane. In our specific case, the y-component is non-zero, so we don't have to worry about that boundary condition.

Question

Let a unit vector $\vec{OP}$ make angle $\alpha, \beta, \gamma$ with the positive directions of the co-ordinate axes $OX, OY, OZ$ respectively, where $\beta \in (0, \pi/2)$. $\vec{OP}$ 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 ?

Options:

(A) $\alpha \in (\pi/2, \pi)$ and $\gamma \in (\pi/2, \pi)$

(B) $\alpha \in (0, \pi/2)$ and $\gamma \in (0, \pi/2)$

(C) $\alpha \in (\pi/2, \pi)$ and $\gamma \in (0, \pi/2)$

(D) $\alpha \in (0, \pi/2)$ and $\gamma \in (\pi/2, \pi)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional room. You have three points floating in the air: $A(1, 2, 3)$, $B(2, 3, 4)$, and $C(1, 5, 7)$.

These points are not just random coordinates; they are the anchors of a flat, invisible sheet—a plane—stretching out into the void. Our mission is to find a unit vector $\vec{OP}$ that stands perfectly perpendicular to this sheet.

There is a catch: this vector must obey a specific orientation constraint, defined by the angle $\beta$ it makes with the Y-axis.

### Constructing the Plane's DNA

To define the orientation of a plane, we need to know how it 'tilts.' We can capture this tilt by finding two vectors that lie flat on the surface of the plane.

Let's use point $A$ as our base. We define two vectors, $\vec{AB}$ and $\vec{AC}$, by subtracting the coordinates of $A$ from $B$ and $C$ respectively:

$$\vec{AB} = (2-1)\hat{i} + (3-2)\hat{j} + (4-3)\hat{k} = \hat{i} + \hat{j} + \hat{k}$$

$$\vec{AC} = (1-1)\hat{i} + (5-2)\hat{j} + (7-3)\hat{k} = 0\hat{i} + 3\hat{j} + 4\hat{k}$$

These two vectors are the 'DNA' of our plane. Any vector perpendicular to both $\vec{AB}$ and $\vec{AC}$ is, by definition, perpendicular to the entire plane.

### The Power of the Cross Product

Now, we invoke the cross product. This is the most elegant tool in our vector toolkit. By calculating $\vec{n} = \vec{AB} 	imes \vec{AC}$, we generate a vector that is orthogonal to the surface.

We set up our determinant:

$$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 0 & 3 & 4 \end{vmatrix}$$

Expanding this, we get $\vec{n} = \hat{i}(4 - 3) - \hat{j}(4 - 0) + \hat{k}(3 - 0)$, which simplifies beautifully to $\vec{n} = \hat{i} - 4\hat{j} + 3\hat{k}$. This vector is our compass needle, pointing directly away from the plane.

### Normalizing and the Constraint

We need a **unit vector**, so we divide $\vec{n}$ by its magnitude, $|\vec{n}| = \sqrt{1^2 + (-4)^2 + 3^2} = \sqrt{26}$. This gives us two candidates: $\pm \frac{1}{\sqrt{26}}(\hat{i} - 4\hat{j} + 3\hat{k})$.

Here is where the physics of the problem meets the constraint. We are told $\beta \in (0, \frac{\pi}{2})$. The direction cosine $\cos \beta$ is simply the y-component of our unit vector.

For $\beta$ to be in the first quadrant, $\cos \beta$ must be positive. If we chose the positive version of our vector, the y-component would be $-\frac{4}{\sqrt{26}}$, which is negative.

We must choose the negative sign to flip the vector, making the y-component positive:

$$\vec{OP} = -\frac{1}{\sqrt{26}}\hat{i} + \frac{4}{\sqrt{26}}\hat{j} - \frac{3}{\sqrt{26}}\hat{k}$$

### The Final Revelation

Now, look at the other components. The x-component is $\cos \alpha = -\frac{1}{\sqrt{26}}$, and the z-component is $\cos \gamma = -\frac{3}{\sqrt{26}}$. Both are negative!

In the world of trigonometry, a negative cosine value for an angle between $0$ and $\pi$ forces that angle into the second quadrant, specifically $(\frac{\pi}{2}, \pi)$.

We have successfully navigated the 3D space, respected the constraints, and arrived at the truth: both $\alpha$ and $\gamma$ must lie in the second quadrant. It is a beautiful result, showing how a simple geometric constraint can dictate the entire orientation of a vector in space.

Select Answer:

Visualized Solution

Visualizing the Plane and Points

Defining Vectors on the Plane

Calculating Vector $A B$

Calculating Vector $A C$

Setting up the Cross Product

Computing the Normal Vector $n$

Normalizing the Vector

The Two Possible Unit Vectors

Applying the Constraint on $β$

Selecting the Correct Vector

Determining Ranges for $α$ and $γ$

Final Conclusion

The Sigma Insight: Equation of a Plane

Analyzing the Setup

Constructing the Plane's DNA

The Power of the Cross Product

Normalizing and the Constraint

The Final Revelation

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