The vector (1/3)(2i - 2j + k) is
(a) a unit vector
(b) makes an angle π/3 with the vector (2i - 4j +

Q: Wait, why do we reject the negative root when calculating magnitude?

Great question! The magnitude of a vector represents its length in space, which is a physical distance. By definition, distance is always a non-negative quantity. When we use the formula $|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$, the square root function inherently returns the principal (positive) root. So, we don't 'reject' a negative root; rather, the geometry of space dictates that length must be positive.

Q: How do I know if two vectors are parallel without just guessing the scalar?

You can use the ratio of their components! If $\vec{v} = v_x\hat{i} + v_y\hat{j} + v_z\hat{k}$ and $\vec{c} = c_x\hat{i} + c_y\hat{j} + c_z\hat{k}$ are parallel, then the ratios $\frac{v_x}{c_x} = \frac{v_y}{c_y} = \frac{v_z}{c_z}$ must all be equal to the same constant $k$. If these ratios are consistent, the vectors are parallel.

Q: Why does a zero dot product mean the vectors are perpendicular?

The dot product is defined as $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos \theta$, where $\theta$ is the angle between the vectors. If the dot product is $0$, then $|\vec{a}||\vec{b}| \cos \theta = 0$. Since the magnitudes of non-zero vectors are positive, the only way for the product to be zero is if $\cos \theta = 0$. This happens precisely when $\theta = 90^\circ$ or $\frac{\pi}{2}$ radians, which is the definition of perpendicularity.

Question

The vector $\frac{1}{3}(2\hat{i} - 2\hat{j} + \hat{k})$ is

Options:

(A) a unit vector

(B) makes an angle $\frac{\pi}{3}$ with the vector $(2\hat{i} - 4\hat{j} + 3\hat{k})$

(C) parallel to the vector $(-\hat{i} + \hat{j} - \frac{1}{2}\hat{k})$

(D) perpendicular to the vector $3\hat{i} + 2\hat{j} - 2\hat{k}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler of the mathematical realm! Today, we are not just solving a problem; we are exploring the fundamental language of physics: vectors.

Imagine you are standing in a three-dimensional room. You have a vector, $\vec{v} = \frac{1}{3}(2\hat{i} - 2\hat{j} + \hat{k})$, floating in front of you. It looks simple, but it holds secrets about length, direction, and orientation.

### The Quest for the Unit Vector

Our first mission is to determine if $\vec{v}$ is a unit vector. This means the vector must have a length of exactly one.

To find this, we use the 3D version of the Pythagorean theorem. We identify the components as $v_x = \frac{2}{3}$, $v_y = -\frac{2}{3}$, and $v_z = \frac{1}{3}$.

The magnitude is calculated as follows:

$$|\vec{v}| = \sqrt{\left(\frac{2}{3}ight)^2 + \left(-\frac{2}{3}ight)^2 + \left(\frac{1}{3}ight)^2}$$

Squaring these terms, we get:

$$|\vec{v}| = \sqrt{\frac{4}{9} + \frac{4}{9} + \frac{1}{9}} = \sqrt{\frac{9}{9}} = 1$$

The result is beautiful in its simplicity: the vector is indeed a unit vector. We have our first victory!

### The Dot Product and the Angle

Next, we investigate the relationship between our vector $\vec{v}$ and another vector $\vec{b} = 2\hat{i} - 4\hat{j} + 3\hat{k}$. The problem asks if the angle between them is $\frac{\pi}{3}$.

To find the angle, we use the dot product formula:

$$\cos 	heta = \frac{\vec{v} \cdot \vec{b}}{|\vec{v}| |\vec{b}|}$$

We calculate the dot product:

$$\vec{v} \cdot \vec{b} = \frac{1}{3} [ (2)(2) + (-2)(-4) + (1)(3) ] = \frac{1}{3} [ 4 + 8 + 3 ] = \frac{15}{3} = 5$$

Now, we find the magnitude of $\vec{b}$:

$$|\vec{b}| = \sqrt{2^2 + (-4)^2 + 3^2} = \sqrt{4 + 16 + 9} = \sqrt{29}$$

Thus, $\cos 	heta = \frac{5}{\sqrt{29}}$. Since $\cos \frac{\pi}{3} = 0.5$ and $\frac{5}{\sqrt{29}} \approx 0.928$, we can confidently say the angle is **not** $\frac{\pi}{3}$.

### The Parallelism Test

Now, let us look at vector $\vec{c} = -\hat{i} + \hat{j} - \frac{1}{2}\hat{k}$. Are $\vec{v}$ and $\vec{c}$ parallel?

Parallel vectors are essentially the same direction, just scaled by some factor $k$. If we look at $\vec{v} = \frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} + \frac{1}{3}\hat{k}$, we can factor out $-\frac{2}{3}$:

$$\vec{v} = -\frac{2}{3} \left( -\hat{i} + \hat{j} - \frac{1}{2}\hat{k} ight)$$

This shows that $\vec{v} = -\frac{2}{3}\vec{c}$. Because we can express one as a scalar multiple of the other, they are **perfectly parallel**.

### The Orthogonality Check

Finally, we test if $\vec{v}$ is perpendicular to $\vec{d} = 3\hat{i} + 2\hat{j} - 2\hat{k}$. The condition for perpendicularity is that their dot product must be zero.

Let us compute:

$$\vec{v} \cdot \vec{d} = \frac{1}{3} [ (2)(3) + (-2)(2) + (1)(-2) ] = \frac{1}{3} [ 6 - 4 - 2 ] = \frac{1}{3}(0) = 0$$

The dot product vanishes! This confirms that the vectors are indeed **perpendicular**. We have successfully navigated the properties of vectors, proving that magnitude, dot products, and scalar multiples are the keys to understanding spatial relationships.

Animated Solution for Mathematics - Vector Algebra: The vector $\frac{1}{3} (2 \hat{i} - 2 \hat{j} + \hat{k})$ is

Select Answer:

Visualized Solution

Defining the Vector $v$

The Magnitude Tool

Calculating Magnitude: Substitution

Atomic Compute: Squaring and Summing

Checking the Angle: The Dot Product Tool

Calculating the Dot Product

Evaluating $cos θ$

Checking Parallelism

Factoring to Prove Parallelism

Checking Orthogonality

Calculating the Final Dot Product

Summary and Key Takeaways

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Quest for the Unit Vector

The Dot Product and the Angle

The Parallelism Test

The Orthogonality Check

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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; } The vector 31​(2i^−2j^​+k^) is