If a unit vector a makes angles π/3 with i, π/4 with j and θ ∈ (0, π) with k, then a value of θ is :

Q: Wait, why is the sum of the squares of the direction cosines always equal to 1?

Think of the unit vector $\vec{a}$ as having components $(x, y, z)$ along the $x$, $y$, and $z$ axes. Since it is a unit vector, its magnitude is $\sqrt{x^2 + y^2 + z^2} = 1$, which means $x^2 + y^2 + z^2 = 1$. Because the vector is a unit vector, its components are exactly the direction cosines: $x = \cos \alpha$, $y = \cos \beta$, and $z = \cos \gamma$. Thus, $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$. It is essentially the Pythagorean theorem in 3D!

Q: Why do we get two values for $\cos \theta$ when we take the square root?

When you solve $\cos^2 \theta = \frac{1}{4}$, you are looking for any number that, when squared, gives $\frac{1}{4}$. Both $\frac{1}{2}$ and $-\frac{1}{2}$ satisfy this. Geometrically, this means there are two possible cones of vectors that make the same angle with the $x$ and $y$ axes, one pointing 'up' and one pointing 'down' relative to the $xy$-plane. Both are valid solutions unless further constraints are given.

Q: Can we solve this using dot products instead of direction cosines?

Absolutely! If $\vec{a} = (x, y, z)$, then $\vec{a} \cdot \hat{i} = x = \cos(\frac{\pi}{3}) = \frac{1}{2}$. Similarly, $\vec{a} \cdot \hat{j} = y = \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$. Since $\vec{a}$ is a unit vector, $x^2 + y^2 + z^2 = 1$. You can solve for $z^2$ and then find $z = \cos \theta$. It leads to the exact same result and is a great way to verify your answer.

Question

If a unit vector $\vec{a}$ makes angles $\pi/3$ with $\hat{i}$, $\pi/4$ with $\hat{j}$ and $	heta \in (0, \pi)$ with $\hat{k}$, then a value of $	heta$ is :-

Options:

(A) $5\pi/12$

(B) $5\pi/6$

(C) $2\pi/3$

(D) $\pi/4$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in the center of a vast, empty room. You have three axes extending from you: the $x$-axis to your right, the $y$-axis in front of you, and the $z$-axis pointing straight up.

Now, imagine a single, perfect arrow—a unit vector $\vec{a}$—starting from your position and pointing somewhere into the room. This vector is special; its length is exactly $1$.

We are not just doing algebra; we are mapping a direction in three-dimensional space.

### The Language of Direction Cosines

When this vector $\vec{a}$ points into space, it makes specific angles with our axes. We call the angle with the $x$-axis $\alpha$, with the $y$-axis $\beta$, and with the $z$-axis $\gamma$.

The problem gives us $\alpha = \frac{\pi}{3}$ and $\beta = \frac{\pi}{4}$. The angle with the $z$-axis, $\gamma = 	heta$, is our mystery to solve.

To bridge the gap between geometry and algebra, we use the concept of **Direction Cosines**. These are simply the cosines of the angles the vector makes with the axes: $l = \cos \alpha$, $m = \cos \beta$, and $n = \cos \gamma$.

Think of these as the 'shadows' or projections of our unit vector onto each axis.

### The Master Identity

Here is where the magic happens. Because our vector is a unit vector, its components along the $x$, $y$, and $z$ axes are exactly its direction cosines.

By the Pythagorean theorem in three dimensions, the sum of the squares of these components must equal the square of the vector's magnitude. Since the magnitude is $1$, we get the fundamental identity:

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

This equation is the heartbeat of 3D vector geometry. It tells us that the orientation of any vector is constrained; you cannot choose all three angles independently.

### The Algebraic Dance

Now, let's substitute our known values into this identity:

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

Recall your trigonometry: $\cos \left( \frac{\pi}{3} ight) = \frac{1}{2}$ and $\cos \left( \frac{\pi}{4} ight) = \frac{1}{\sqrt{2}}$. Squaring these, we get:

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

This simplifies to:

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

Adding the fractions, $\frac{1}{4} + \frac{2}{4} = \frac{3}{4}$. So, our equation becomes:

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

Subtracting $\frac{3}{4}$ from both sides, we find:

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

### The Final Revelation

We are almost there. Taking the square root of both sides gives us $\cos 	heta = \pm \frac{1}{2}$.

This means $	heta$ could be $\frac{\pi}{3}$ or $\frac{2\pi}{3}$. Both values lie within the range $(0, \pi)$ specified by the problem.

Looking at our options, **$\frac{2\pi}{3}$** is the one that fits. We have successfully navigated the 3D space, used the power of direction cosines, and solved for the unknown angle.

Animated Solution for Mathematics - Vector Algebra: If a unit vector $a$ makes angles $π /3$ with $\hat{i}$ , $π /4$ with $\hat{j}$ and $θ \in (0, π)$ with $\hat{k}$ , then a value of $θ$ is :-

Select Answer:

Visualized Solution

Visualizing the Unit Vector $a$

Angles with $x$ and $y$ Axes

Angle with the $z$ -Axis

Concept of Direction Cosines

The Fundamental Identity

Substituting the Given Angles

Evaluating Trigonometric Values

Squaring and Adding

Isolating $cos^{2} θ$

Solving for $cos θ$

Finding the Value of $θ$

The Sigma Insight: Components of a Vector

Analyzing the Setup

The Language of Direction Cosines

The Master Identity

The Algebraic Dance

The Final Revelation

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Animated Solution for Mathematics - Vector Algebra: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a unit vector a makes angles π/3 with i^, π/4 with j^​ and θ∈(0,π) with k^, then a value of θ is :-