Each of the angles β and γ that a given line makes with the positive y - and z -axes, respectively,

Q: Wait, why can't we just use $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$ directly without substituting?

You absolutely must use that identity! The substitution is simply the method to apply the given constraints ($\beta = \alpha/2$ and $\gamma = \alpha/2$) into that identity. Without substituting, you have three unknowns and only one equation. The substitution reduces the problem to a single variable, $\alpha$, which makes it solvable.

Q: Why do we get two different values for $\beta$? Is the line not unique?

Great observation! The quadratic equation $\cos^2 \alpha + \cos \alpha = 0$ naturally yields two roots: $\cos \alpha = 0$ and $\cos \alpha = -1$. Each root corresponds to a valid physical orientation of the line in 3D space that satisfies the given angle conditions. Both are mathematically valid, so we must include both in our sum.

Q: Can $\cos \alpha$ be negative? I thought direction angles were always between $0$ and $\pi$.

You are absolutely correct that direction angles $\alpha, \beta, \gamma$ lie in the interval $[0, \pi]$. The cosine of an angle in this interval can indeed be negative (specifically, when the angle is obtuse, i.e., between $\pi/2$ and $\pi$). So, $\cos \alpha = -1$ is perfectly valid and corresponds to $\alpha = \pi$.

Question

Each of the angles $\beta$ and $\gamma$ that a given line makes with the positive $y$ - and $z$ -axes, respectively, is half of the angle that this line makes with the positive $x$ -axes. Then the sum of all possible values of the angle $\beta$ is

Options:

(A) $3\pi/4$

(B) $\pi$

(C) $\pi/2$

(D) $3\pi/2$

Sigma Percentile · Accepted Answer

### The Geometry of Direction Angles

Imagine standing at the origin of a three-dimensional coordinate system. You have a line, let us call it $L$, shooting out from the origin into space.

This line makes specific angles with the positive $x, y,$ and $z$ axes. We call these the direction angles: $\alpha, \beta,$ and $\gamma$.

Visualizing these angles is the first step in mastering 3D geometry. They are the keys to understanding the orientation of any line in space.

### The Backbone of the Solution

Every line in 3D space is governed by a beautiful, fundamental identity involving its direction cosines:

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

This equation is the law of the land. It ensures that the line's orientation is consistent with the geometry of the axes.

Our problem gives us a specific constraint: the line makes angles $\beta$ and $\gamma$ with the $y$ and $z$ axes, respectively, and both are exactly half of the angle $\alpha$ it makes with the $x$-axis. Mathematically, we have:

$$\beta = \frac{\alpha}{2}, \quad \gamma = \frac{\alpha}{2}$$

### The Algebraic Bridge

Now, we substitute these conditions into our fundamental identity. The equation transforms into:

$$\cos^2 \alpha + \cos^2\left(\frac{\alpha}{2}ight) + \cos^2\left(\frac{\alpha}{2}ight) = 1$$

Combining the identical terms, we get:

$$\cos^2 \alpha + 2\cos^2\left(\frac{\alpha}{2}ight) = 1$$

This looks like a standard trigonometric equation, but we have a mix of $\alpha$ and $\alpha/2$. To solve this, we need a bridge. The half-angle identity is our perfect tool:

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

By applying this identity, we can unify the entire equation in terms of $\alpha$:

$$\cos^2 \alpha + (1 + \cos \alpha) = 1$$

### Solving the Quadratic

The equation simplifies beautifully. Subtracting $1$ from both sides, we are left with:

$$\cos^2 \alpha + \cos \alpha = 0$$

This is a classic quadratic form. Please, do not divide by $\cos \alpha$, as you might lose a potential solution! Instead, factor it:

$$\cos \alpha (\cos \alpha + 1) = 0$$

This gives us two distinct cases. Case 1: $\cos \alpha = 0$, which implies $\alpha = \pi/2$. Since $\beta = \alpha/2$, we find $\beta = \pi/4$.

Case 2: $\cos \alpha = -1$, which implies $\alpha = \pi$. Consequently, $\beta = \pi/2$.

### The Final Sum

We have discovered two possible values for $\beta$: $\pi/4$ and $\pi/2$. The problem asks for the sum of all possible values of $\beta$.

Adding them together:

$$	ext{Sum} = \frac{\pi}{4} + \frac{\pi}{2} = \frac{\pi}{4} + \frac{2\pi}{4} = \frac{3\pi}{4}$$

And there it is! Through the elegance of trigonometric identities and the rigor of 3D geometry, we have arrived at our destination. The final result is **$3\pi/4$**.

Select Answer:

Visualized Solution

Defining Direction Angles

The Given Relationship

The Fundamental Identity

Substitution

Grouping Terms

Half-Angle Identity

Applying the Identity

Simplifying the Equation

Factoring

Case 1: $cos α = 0$

Case 2: $cos α = - 1$

Final Sum of $β$ Values

The Sigma Insight: Direction Cosines and Direction Ratios

The Geometry of Direction Angles

The Backbone of the Solution

The Algebraic Bridge

Solving the Quadratic

The Final Sum

Similar Questions

If a line makes an angle of $π /4$ with the positive directions of each of $x$ -axis and $y$ -axis, then the angle that the line makes with the positive direction of the $z$ -axis is

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

A line makes the same angle $θ$ , with each of the $x$ and $z$ axis. If the angle $β$ , which it makes with $y$ -axis, is such that $sin^{2} β = 3 sin^{2} θ$ , then $cos^{2} θ$ equals

The angle between the lines whose direction cosines satisfy the equations $l + m + n = 0$ and $l^{2} = m^{2} + n^{2}$ is

The angle between the straight lines, whose direction cosines are given by the equations $2 l + 2 m - n = 0$ and $mn + n l + l m = 0$ , is :

Let $α$ be the angle between the lines whose direction cosines satisfy the equations $l + m - n = 0$ and $l^{2} + m^{2} - n^{2} = 0$ . Then the value of $sin^{4} α + cos^{4} α$ is :

Let the direction cosines of two lines satisfy the equations : $4 l + m - n = 0$ and $2 mn + 10 n l + 3 l m = 0$ . Then the cosine of the acute angle between these lines is :

If two straight lines whose direction cosines are given by the relations $l + m - n = 0, 3 l^{2} + m^{2} + c n l = 0$ are parallel, then the positive value of $c$ is :