A line AB in three-dimensional space makes angles 45° and 120° with the positive x-axis and the posi

Q: Wait, why is $\cos(120^\circ)$ negative? I thought cosine was always positive.

That is a common point of confusion! Remember that the cosine function is positive in the first and fourth quadrants, but negative in the second and third quadrants. Since $120^\circ$ is in the second quadrant ($90^\circ < 120^\circ < 180^\circ$), the cosine value must be negative. Specifically, $\cos(120^\circ) = -\frac{1}{2}$. Keep your unit circle handy!

Q: Why do we use the identity $l^2 + m^2 + n^2 = 1$?

This identity is the definition of direction cosines for a unit vector. If you represent the line as a unit vector $\hat{u} = (l, m, n)$, then its magnitude must be 1. Since the magnitude of a vector is $\sqrt{l^2 + m^2 + n^2}$, squaring both sides gives $l^2 + m^2 + n^2 = 1$. It is the geometric equivalent of the Pythagorean theorem in 3D space.

Q: Why do we reject the negative root when solving for $\cos \theta$?

The problem explicitly states that $\theta$ is an acute angle. By definition, an acute angle lies between $0^\circ$ and $90^\circ$. In this range, the cosine function is always positive. If we had chosen the negative root, we would have found an obtuse angle, which contradicts the problem statement.

Question

A line $AB$ in three-dimensional space makes angles $45^\circ$ and $120^\circ$ with the positive $x$-axis and the positive $y$-axis respectively. If $AB$ makes an acute angle $	heta$ with the positive $z$-axis, then $	heta$ equals

Options:

(A) $45^\circ$

(B) $60^\circ$

(C) $75^\circ$

(D) $30^\circ$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing at the origin of a three-dimensional coordinate system. You have a line, $AB$, stretching out into space.

To define its orientation, we need to know how it leans relative to the $x$, $y$, and $z$ axes. These angles, $\alpha$, $\beta$, and $\gamma$, are the keys to unlocking the line's identity.

In this problem, we are given $\alpha = 45^\circ$ and $\beta = 120^\circ$. Our mission is to find the angle $	heta$ (or $\gamma$) that the line makes with the $z$-axis.

### The Magic Identity

In the world of 3D geometry, we use direction cosines to simplify our lives. These are defined as $l = \cos \alpha$, $m = \cos \beta$, and $n = \cos \gamma$.

The beauty of these values lies in a fundamental identity: the sum of their squares is always unity. This is expressed as:

$$ \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1 $$

This is not just a formula; it is a profound geometric constraint that ensures our line exists in a valid 3D space.

### The Calculation

Now, let us apply this to our specific problem. We substitute our known angles into the identity:

$$ \cos^2(45^\circ) + \cos^2(120^\circ) + \cos^2(	heta) = 1 $$

First, we evaluate the trigonometric functions. We know that $\cos(45^\circ) = \frac{1}{\sqrt{2}}$, so $\cos^2(45^\circ) = \frac{1}{2}$.

Next, for $\cos(120^\circ)$, we recognize it as being in the second quadrant, where cosine is negative, giving us $-\frac{1}{2}$. Squaring this yields $\cos^2(120^\circ) = \frac{1}{4}$.

Substituting these back into our equation, we get:

$$ \frac{1}{2} + \frac{1}{4} + \cos^2(	heta) = 1 $$

Adding the fractions, $\frac{1}{2} + \frac{1}{4} = \frac{3}{4}$. Thus, the equation simplifies to:

$$ \frac{3}{4} + \cos^2(	heta) = 1 $$

### The Final Decision

Isolating $\cos^2(	heta)$, we find:

$$ \cos^2(	heta) = 1 - \frac{3}{4} = \frac{1}{4} $$

Taking the square root of both sides gives $\cos(	heta) = \pm \frac{1}{2}$.

Here is where we must be careful. The problem states that $	heta$ is an acute angle. Since $\cos(	heta)$ must be positive for an acute angle ($0^\circ < 	heta < 90^\circ$), we reject the negative value.

Therefore, $\cos(	heta) = \frac{1}{2}$. Looking at our standard trigonometric values, we know that $\cos(60^\circ) = \frac{1}{2}$.

Thus, the final answer is **$	heta = 60^\circ$**.

Animated Solution for Mathematics - Three Dimensional Geometry: A line $A B$ in three-dimensional space makes angles $4 5^{\circ}$ and $12 0^{\circ}$ with the positive $x$ -axis and the positive $y$ -axis respectively. If $A B$ makes an acute angle $θ$ with the positive $z$ -axis, then $θ$ equals

Select Answer:

Visualized Solution

Visualizing the Line $A B$

Angle with the $x$ -axis

Angle with the $y$ -axis

Angle with the $z$ -axis

Direction Cosines

The Fundamental Identity

Substituting the Angles

Evaluating the Cosines

Squaring the Values

Adding the Fractions

Isolating $cos^{2} θ$

Taking the Square Root

Applying the Constraint

Final Answer

The Sigma Insight: Direction Cosines and Direction Ratios

Analyzing the Setup

The Magic Identity

The Calculation

The Final Decision

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