Key Concepts and Formulas

• Direction Cosines of a Line: For a line in three-dimensional space, the direction cosines are the cosines of the angles it makes with the positive directions of the $x$, $y$, and $z$-axes. If these angles are $\alpha$, $\beta$, and $\gamma$ respectively, then the direction cosines are $\ell = \cos \alpha$, $m = \cos \beta$, and $n = \cos \gamma$.
• Fundamental Identity of Direction Cosines: A crucial property relating the direction cosines is that the sum of their squares is always equal to 1. This is expressed as: $\ell^2 + m^2 + n^2 = 1$ This identity is fundamental for determining an unknown direction cosine or angle when the other two are known.

Step-by-Step Solution

1. Identify Given Information and Define Direction Cosines: We are given the angles that the line $AB$ makes with the positive $x$-axis and $y$-axis, and we need to find the acute angle it makes with the positive $z$-axis.

   • Angle with positive $x$-axis: $\alpha = 45^\circ$
   • Angle with positive $y$-axis: $\beta = 120^\circ$
   • Angle with positive $z$-axis: $\gamma = \theta$, where $\theta$ is an acute angle ($0^\circ < \theta < 90^\circ$).

   Our goal is to find the value of $\theta$. We will use the definition of direction cosines:

   • $\ell = \cos \alpha = \cos 45^\circ$
   • $m = \cos \beta = \cos 120^\circ$
   • $n = \cos \gamma = \cos \theta$

2. Calculate the Known Direction Cosines: Next, we compute the numerical values for $\ell$ and $m$:

   • For $\ell$: We know the standard trigonometric value $\cos 45^\circ = \frac{1}{\sqrt{2}}$. So, $\ell = \frac{1}{\sqrt{2}}$.
   • For $m$: We evaluate $\cos 120^\circ$. Since $120^\circ$ is in the second quadrant, its cosine will be negative. We use the identity $\cos(180^\circ - x) = -\cos x$: $\cos 120^\circ = \cos(180^\circ - 60^\circ) = -\cos 60^\circ = -\frac{1}{2}$. So, $m = -\frac{1}{2}$. Explanation: Accurate calculation of these values, especially paying attention to the sign for angles in different quadrants, is essential for the correctness of the solution.

3. Apply the Fundamental Identity of Direction Cosines: Now, we substitute the calculated values of $\ell$ and $m$, along with $n = \cos \theta$, into the fundamental identity $\ell^2 + m^2 + n^2 = 1$: $\left(\frac{1}{\sqrt{2}}\right)^2 + \left(-\frac{1}{2}\right)^2 + (\cos \theta)^2 = 1$ Squaring the terms: $\frac{1}{2} + \frac{1}{4} + \cos^2 \theta = 1$

4. Solve for the Unknown Direction Cosine ($n = \cos \theta$): Combine the numerical fractions on the left side: $\frac{2}{4} + \frac{1}{4} + \cos^2 \theta = 1$ $\frac{3}{4} + \cos^2 \theta = 1$ Isolate $\cos^2 \theta$ by subtracting $\frac{3}{4}$ from both sides: $\cos^2 \theta = 1 - \frac{3}{4}$ $\cos^2 \theta = \frac{1}{4}$ Take the square root of both sides to find $\cos \theta$: $\cos \theta = \pm \sqrt{\frac{1}{4}}$ $\cos \theta = \pm \frac{1}{2}$

5. Determine the Acute Angle $\theta$: We have two possible values for $\cos \theta$: $\frac{1}{2}$ and $-\frac{1}{2}$. The problem statement specifies that $AB$ makes an acute angle $\theta$ with the positive $z$-axis. An acute angle is an angle between $0^\circ$ and $90^\circ$. For angles in this range, the cosine function is always positive. Therefore, we must choose the positive value for $\cos \theta$: $\cos \theta = \frac{1}{2}$ Now, we find the angle $\theta$ whose cosine is $\frac{1}{2}$: $\theta = \cos^{-1}\left(\frac{1}{2}\right)$ $\theta = 60^\circ$ This value of $\theta = 60^\circ$ is indeed an acute angle, satisfying the condition given in the problem.

Common Mistakes & Tips

• Sign of Cosine: Always be careful with the signs of trigonometric functions based on the quadrant of the angle. $\cos 120^\circ$ being negative is a common point of error.
• Acute Angle Condition: The "acute angle" condition is crucial. If it were not specified, $\theta$ could also be $120^\circ$ (since $\cos 120^\circ = -1/2$), which is an obtuse angle. This condition helps in selecting the correct value of $\theta$.
• Direction Cosines vs. Direction Ratios: Remember that direction cosines are unique for a given direction (up to sign for a line) and always satisfy $\ell^2 + m^2 + n^2 = 1$. Direction ratios are any set of numbers proportional to the direction cosines and do not necessarily satisfy this identity.

Summary

This problem is a direct application of the fundamental property of direction cosines in three-dimensional geometry. By calculating the known direction cosines for the angles with the $x$ and $y$-axes, and then using the identity $\ell^2 + m^2 + n^2 = 1$, we can solve for the cosine of the angle with the $z$-axis. The condition that the angle is acute helps us select the correct value from the possible solutions. The final answer is $60^\circ$.

The final answer is \boxed{\text{60^\circ}}, which corresponds to option (B).