Key Concepts and Formulas

1. Direction Cosines: For a vector $\vec{v} = x\widehat{i} + y\widehat{j} + z\widehat{k}$, its direction cosines are $\cos\alpha = \frac{x}{|\vec{v}|}$, $\cos\beta = \frac{y}{|\vec{v}|}$, and $\cos\gamma = \frac{z}{|\vec{v}|}$, where $\alpha, \beta, \gamma$ are the angles the vector makes with the positive x, y, and z axes, respectively. For a unit vector, $|\vec{v}| = 1$, so the components are directly the direction cosines: $\widehat{v} = (\cos\alpha)\widehat{i} + (\cos\beta)\widehat{j} + (\cos\gamma)\widehat{k}$. The fundamental relation is $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$.

2. Normal Vector to a Plane: A vector perpendicular to a plane containing points $A, B, C$ can be found by computing the cross product of two vectors lying in the plane, such as $\overrightarrow{AB} \times \overrightarrow{AC}$.

3. Perpendicularity: A vector perpendicular to a plane is parallel to the normal vector of that plane.

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Step-by-Step Solution

Step 1: Define the unit vector $\widehat{OP}$ in terms of its direction cosines.

Let the unit vector be $\widehat{OP}$. We are given that it makes angles $\alpha, \beta, \gamma$ with the positive x, y, and z axes. By definition, the components of a unit vector are its direction cosines. Thus, we can write $\widehat{OP}$ as: $\widehat{OP} = (\cos\alpha)\widehat{i} + (\cos\beta)\widehat{j} + (\cos\gamma)\widehat{k}$ We are given the condition $\beta \in \left(0, \frac{\pi}{2}\right)$. This implies that $\cos\beta$ is positive.

Step 2: Find two vectors lying in the plane defined by the three given points.

Let the three points be $A(1,2,3)$, $B(2,3,4)$, and $C(1,5,7)$. We form two vectors in the plane, for example, $\overrightarrow{AB}$ and $\overrightarrow{AC}$. $\overrightarrow{AB} = B - A = (2-1)\widehat{i} + (3-2)\widehat{j} + (4-3)\widehat{k} = \widehat{i} + \widehat{j} + \widehat{k}$ $\overrightarrow{AC} = C - A = (1-1)\widehat{i} + (5-2)\widehat{j} + (7-3)\widehat{k} = 0\widehat{i} + 3\widehat{j} + 4\widehat{k}$

Step 3: Calculate a normal vector to the plane.

A vector perpendicular to the plane is given by the cross product of $\overrightarrow{AB}$ and $\overrightarrow{AC}$. Let this normal vector be $\vec{n}$. $\vec{n} = \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \widehat{i} & \widehat{j} & \widehat{k} \\ 1 & 1 & 1 \\ 0 & 3 & 4 \end{vmatrix}$ Expanding the determinant: $\vec{n} = (1 \cdot 4 - 1 \cdot 3)\widehat{i} - (1 \cdot 4 - 1 \cdot 0)\widehat{j} + (1 \cdot 3 - 1 \cdot 0)\widehat{k}$ $\vec{n} = (4-3)\widehat{i} - (4-0)\widehat{j} + (3-0)\widehat{k}$ $\vec{n} = \widehat{i} - 4\widehat{j} + 3\widehat{k}$

Step 4: Relate the unit vector $\widehat{OP}$ to the normal vector $\vec{n}$.

Since $\widehat{OP}$ is perpendicular to the plane, it must be parallel to the normal vector $\vec{n}$. As $\widehat{OP}$ is a unit vector, it must be the unit vector in the direction of $\vec{n}$ or $-\vec{n}$. First, calculate the magnitude of $\vec{n}$: $|\vec{n}| = \sqrt{1^2 + (-4)^2 + 3^2} = \sqrt{1 + 16 + 9} = \sqrt{26}$ Therefore, the possible forms for $\widehat{OP}$ are: $\widehat{OP} = \pm \frac{\vec{n}}{|\vec{n}|} = \pm \frac{\widehat{i} - 4\widehat{j} + 3\widehat{k}}{\sqrt{26}}$ This gives two possibilities:

1. $\widehat{OP}_1 = \frac{1}{\sqrt{26}}\widehat{i} - \frac{4}{\sqrt{26}}\widehat{j} + \frac{3}{\sqrt{26}}\widehat{k}$
2. $\widehat{OP}_2 = -\frac{1}{\sqrt{26}}\widehat{i} + \frac{4}{\sqrt{26}}\widehat{j} - \frac{3}{\sqrt{26}}\widehat{k}$

Step 5: Use the condition on $\beta$ to select the correct $\widehat{OP}$.

We know from Step 1 that $\cos\beta > 0$. Let's examine the coefficient of $\widehat{j}$ in both possibilities: For $\widehat{OP}_1$, $\cos\beta = -\frac{4}{\sqrt{26}}$, which is negative. This contradicts the given condition. For $\widehat{OP}_2$, $\cos\beta = \frac{4}{\sqrt{26}}$, which is positive. This satisfies the given condition. Therefore, the correct unit vector is: $\widehat{OP} = -\frac{1}{\sqrt{26}}\widehat{i} + \frac{4}{\sqrt{26}}\widehat{j} - \frac{3}{\sqrt{26}}\widehat{k}$

Step 6: Determine the ranges for $\alpha$ and $\gamma$.

From the components of the correct $\widehat{OP}$, we have: $\cos\alpha = -\frac{1}{\sqrt{26}}$ $\cos\gamma = -\frac{3}{\sqrt{26}}$ Since $\cos\alpha$ is negative, and direction angles are typically taken in the range $[0, \pi]$, $\alpha$ must be in the second quadrant. Thus, $\alpha \in \left(\frac{\pi}{2}, \pi\right)$. Similarly, since $\cos\gamma$ is negative, $\gamma$ must also be in the second quadrant. Thus, $\gamma \in \left(\frac{\pi}{2}, \pi\right)$.

Step 7: Match the derived ranges with the given options.

We found $\alpha \in \left(\frac{\pi}{2}, \pi\right)$ and $\gamma \in \left(\frac{\pi}{2}, \pi\right)$. This matches option (A).

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Common Mistakes & Tips

• Sign Errors in Cross Product: Carefully compute the cross product determinant to avoid sign errors in the normal vector components.
• Ambiguity of $\pm$: The $\pm$ sign arises because a vector can be parallel to a normal vector in two opposite directions. The given condition on $\beta$ is essential for resolving this ambiguity.
• Interpreting Cosine Signs: A negative cosine value for a direction angle (in the range $[0, \pi]$) implies the angle is between $\frac{\pi}{2}$ and $\pi$. A positive cosine value implies the angle is between $0$ and $\frac{\pi}{2}$.

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Summary

The problem required finding the direction cosines of a unit vector perpendicular to a plane. This was achieved by first calculating a normal vector to the plane using the cross product of vectors formed by the given points. The unit vector $\widehat{OP}$ was then identified as one of the two unit vectors parallel to the normal. The constraint on the angle $\beta$ was used to uniquely determine the correct direction of $\widehat{OP}$. Finally, the signs of the direction cosines $\cos\alpha$ and $\cos\gamma$ determined their respective ranges.

The final answer is \boxed{A}.