A line makes the same angle θ, with each of the x and z axis. If the angle β, which it makes with y-

Q: Why is the sum of the squares of direction cosines equal to 1?

Think of a unit vector $$\hat{u} = (l, m, n)$$. The magnitude of a unit vector is 1, so $$\sqrt{l^2 + m^2 + n^2} = 1$$, which implies $$l^2 + m^2 + n^2 = 1$$. Since $$l, m, n$$ are the cosines of the angles with the axes, this is the geometric foundation of 3D space.

Q: Can I use vectors instead of direction cosines?

Absolutely! If the line is represented by vector $$\vec{r} = (x, y, z)$$, then the direction cosines are just the components of the unit vector $$\hat{r} = \frac{\vec{r}}{|\vec{r}|}$$. It is the same math, just a different perspective. Both methods lead to the same elegant result.

Q: Why convert sines to cosines? Can't we just solve for sines?

You could, but the fundamental identity $$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$$ is already in terms of cosines. Converting everything to cosines makes the algebra much cleaner and less prone to errors.

Question

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

Options:

(A) $2/5$

(B) $1/5$

(C) $3/5$

(D) $2/3$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine standing at the origin of a three-dimensional coordinate system. You have a line shooting out into space, and your goal is to understand its orientation.

The line makes an angle $	heta$ with the $x$-axis and the same angle $	heta$ with the $z$-axis. With the $y$-axis, it makes a different angle, $\beta$.

### The Fundamental Law of 3D Geometry

To tackle this, we invoke the direction cosine identity. If a line makes angles $\alpha, \beta, \gamma$ with the $x, y, z$ axes respectively, its direction cosines are $l = \cos \alpha$, $m = \cos \beta$, and $n = \cos \gamma$.

The fundamental law states that the sum of the squares of these direction cosines is unity:

$$l^2 + m^2 + n^2 = 1$$

This holds because the direction cosines are the components of a unit vector along the line. Since the magnitude of a unit vector is always $1$, the sum of the squares of its components must be $1$ by the Pythagorean theorem extended into three dimensions.

### Translating the Problem

We are given that the angle with the $x$-axis is $	heta$ (so $\alpha = 	heta$) and the angle with the $z$-axis is also $	heta$ (so $\gamma = 	heta$). Substituting these into our identity, we get:

$$\cos^2 	heta + \cos^2 \beta + \cos^2 	heta = 1$$

Combining the terms, we arrive at our first crucial equation:

$$2 \cos^2 	heta + \cos^2 \beta = 1$$

### The Constraint: Unlocking the Algebra

The problem provides a second piece of information: $\sin^2 \beta = 3 \sin^2 	heta$. To solve for $\cos^2 	heta$, we bridge the gap between sines and cosines using the identity $\sin^2 A = 1 - \cos^2 A$.

Applying this to both sides of our constraint:

$$(1 - \cos^2 \beta) = 3(1 - \cos^2 	heta)$$

Expanding the right side, we have:

$$1 - \cos^2 \beta = 3 - 3 \cos^2 	heta$$

Rearranging this to isolate $\cos^2 \beta$, we find:

$$\cos^2 \beta = 3 \cos^2 	heta - 2$$

### The Final Synthesis

We now have a system of two equations:

$$2 \cos^2 	heta + \cos^2 \beta = 1$$

$$\cos^2 \beta = 3 \cos^2 	heta - 2$$

Substituting the second equation into the first, the complexity collapses:

$$2 \cos^2 	heta + (3 \cos^2 	heta - 2) = 1$$

Combining the $\cos^2 	heta$ terms yields:

$$5 \cos^2 	heta - 2 = 1$$

Adding $2$ to both sides gives $5 \cos^2 	heta = 3$. Finally, dividing by $5$, we arrive at the result:

$$\cos^2 	heta = \frac{3}{5}$$

This is the beauty of mathematics—taking a complex 3D orientation and reducing it to a single, elegant fraction. The final value is **$\cos^2 	heta = \frac{3}{5}$**.

Animated Solution for Mathematics - Three Dimensional Geometry: 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

Select Answer:

Visualized Solution

Visualizing the Line in 3D

Angles with Coordinate Axes

The Direction Cosine Identity

Substituting the Given Angles

Simplifying the Identity

Using the Given Condition

Converting Sines to Cosines

Applying the Conversion

Isolating $cos^{2} β$

Final Substitution

Grouping Terms

Solving for $cos^{2} θ$

The Final Result

The Sigma Insight: Direction Cosines and Direction Ratios

Analyzing the Setup

The Fundamental Law of 3D Geometry

Translating the Problem

The Constraint

The Final Synthesis

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