Q.Compute the magnitude of the following vectors: $\vec{a} = \hat{i} + \hat{j} + \hat{k}$; $\vec{b} = 2\hat{i} - 7\hat{j} - 3\hat{k}$; $\vec{c} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} - \frac{1}{\sqrt{3}}\hat{k}$

Concept: Vector Magnitude — For a vector $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$, its magnitude is $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$.

Step 1: For $\vec{a} = \hat{i} + \hat{j} + \hat{k}$,

$x = 1, y = 1, z = 1$

$|\vec{a}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}$

Step 2: For $\vec{b} = 2\hat{i} - 7\hat{j} - 3\hat{k}$,

$x = 2, y = -7, z = -3$

$|\vec{b}| = \sqrt{2^2 + (-7)^2 + (-3)^2} = \sqrt{4 + 49 + 9} = \sqrt{62}$

Step 3: For $\vec{c} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} - \frac{1}{\sqrt{3}}\hat{k}$,

$x = \frac{1}{\sqrt{3}}, y = \frac{1}{\sqrt{3}}, z = -\frac{1}{\sqrt{3}}$

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

The magnitude of a vector is its length — the distance from its tail to its head when placed at the origin. For a vector expressed in component form, say $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$, the magnitude $|\vec{v}|$ is given by:

This is a direct extension of the Pythagorean theorem into three dimensions. Each component squared contributes to the total squared length; the square root brings it back to the original scale. There’s no trick here — just careful arithmetic.

Let’s apply this to each vector.

1. For $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ Components: $x = 1$, $y = 1$, $z = 1$.

2. For $\vec{b} = 2\hat{i} - 7\hat{j} - 3\hat{k}$ Components: $x = 2$, $y = -7$, $z = -3$. Squaring each: $2^2 = 4$, $(-7)^2 = 49$, $(-3)^2 = 9$.

3. For $\vec{c} = \frac{1}{\sqrt{3}}\hat{i} + \frac{1}{\sqrt{3}}\hat{j} - \frac{1}{\sqrt{3}}\hat{k}$ Components: $x = \frac{1}{\sqrt{3}}$, $y = \frac{1}{\sqrt{3}}$, $z = -\frac{1}{\sqrt{3}}$. Square each: $\left(\frac{1}{\sqrt{3}}\right)^2 = \frac{1}{3}$, and the same for the other two.

Method: Computing the magnitude of a vector from its components

Use this whenever a vector is given in $\hat{i},\hat{j},\hat{k}$ form and its length is required.

Steps

Step 1: Read off the components

For $\vec{v}=x\hat{i}+y\hat{j}+z\hat{k}$, identify $x,y,z$ with their signs.

Step 2: Apply the 3-D Pythagoras formula

Step 3: Square each component (sign disappears)

Because every term is squared, negative components contribute positively — e.g. $(-7)^2=49$. Keep any fractional components exact.

Step 4: Simplify and check for a unit vector

Add the squares and take the root. If the sum of squares equals $1$, the vector is a unit vector (magnitude $1$) — a useful sanity check.

Common Mistakes

Mistake 1: Losing the sign-squaring on negative components

Why it's wrong: for $\vec b=2\hat i-7\hat j-3\hat k$, some students write $-7^2$ or forget a term; correctly $(-7)^2=49$ and $(-3)^2=9$, giving $\sqrt{4+49+9}=\sqrt{62}$. Correct approach: square first, so every component is positive.

Mistake 2: Mishandling the $\tfrac{1}{\sqrt3}$ components

Why it's wrong: $\left(\tfrac{1}{\sqrt3}\right)^2=\tfrac13$, and three of them sum to $1$, so $|\vec c|=1$ (a unit vector). Skipping the squaring or misreading the fraction hides this. Correct approach: square each fraction exactly before adding.