Vector Magnitude and Unit Vectors (Exercise 10.2)

This lesson works through the first questions of Exercise 10.2: finding the magnitude of a vector, recognising equal vectors, reading off scalar and vector components, adding vectors, and building unit vectors of any magnitude.

Magnitude of a vector

The magnitude of $\vec a = a_1\hat i + a_2\hat j + a_3\hat k$ is $|\vec a| = \sqrt{a_1^2 + a_2^2 + a_3^2}$.

$\vec a = \hat i + \hat j + \hat k$

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

$\vec b = 2\hat i - 7\hat j + 3\hat k$

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

Showing a vector is a unit vector

For $\vec c = -\tfrac{1}{\sqrt3}\hat i - \tfrac{1}{\sqrt3}\hat j - \tfrac{1}{\sqrt3}\hat k$,

$|\vec c| = \sqrt{\tfrac{1}{3} + \tfrac{1}{3} + \tfrac{1}{3}} = \sqrt{\tfrac{3}{3}} = 1,$

so $\vec c$ is a unit vector.

Equal vectors and their components

If $2\hat i + 3\hat j = x\hat i + y\hat j$, then equal vectors must have equal corresponding components, so

$x = 2, \qquad y = 3.$

Scalar and vector components between two points

For a vector from initial point $P(2,1)$ to terminal point $Q(-5,7)$, use $\vec{PQ} = (x_2 - x_1)\hat i + (y_2 - y_1)\hat j$:

$\vec{PQ} = (-5 - 2)\hat i + (7 - 1)\hat j = -7\hat i + 6\hat j + 0\hat k.$

Vector components: $-7\hat i$, $6\hat j$, $0\hat k$. Scalar components: $-7$, $6$, $0$. The coefficients of $\hat i, \hat j, \hat k$ are the scalar components, and the full terms are the vector components.

Adding vectors component by component

Given $\vec a = \hat i - 2\hat j + \hat k$, $\vec b = -2\hat i + 4\hat j + 5\hat k$, and $\vec c = \hat i - 6\hat j - 7\hat k$, add matching components:

$\vec a + \vec b + \vec c = (1 - 2 + 1)\hat i + (-2 + 4 - 6)\hat j + (1 + 5 - 7)\hat k.$

$\vec a + \vec b + \vec c = 0\hat i - 4\hat j - \hat k = -4\hat j - \hat k.$

Unit vector in a given direction

The unit vector along $\vec a$ is $\hat a = \dfrac{\vec a}{|\vec a|}$.

Direction of $\hat i + \hat j + 2\hat k$

$|\vec a| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{6}, \qquad \hat a = \tfrac{1}{\sqrt6}\hat i + \tfrac{1}{\sqrt6}\hat j + \tfrac{2}{\sqrt6}\hat k.$

Direction of $\vec{PQ}$ for $P(1,2,3)$, $Q(4,5,6)$

Here $\vec{PQ} = (4-1)\hat i + (5-2)\hat j + (6-3)\hat k = 3\hat i + 3\hat j + 3\hat k$, with

$|\vec{PQ}| = \sqrt{3^2 + 3^2 + 3^2} = \sqrt{27} = 3\sqrt3,$

so

$\hat{PQ} = \frac{3\hat i + 3\hat j + 3\hat k}{3\sqrt3} = \tfrac{1}{\sqrt3}\hat i + \tfrac{1}{\sqrt3}\hat j + \tfrac{1}{\sqrt3}\hat k.$

Direction of $\vec a + \vec b$

For $\vec a = 2\hat i - \hat j + 2\hat k$ and $\vec b = -\hat i + \hat j - \hat k$,

$\vec a + \vec b = (2 - 1)\hat i + (-1 + 1)\hat j + (2 - 1)\hat k = \hat i + 0\hat j + \hat k,$

$|\vec a + \vec b| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt2,$

so the unit vector is $\tfrac{1}{\sqrt2}\hat i + \tfrac{1}{\sqrt2}\hat k$.

A vector of given magnitude in a direction

To find a vector of magnitude $8$ in the direction of $5\hat i - \hat j + 2\hat k$, first build the unit vector:

$|\vec a| = \sqrt{5^2 + (-1)^2 + 2^2} = \sqrt{25 + 1 + 4} = \sqrt{30}, \qquad \hat a = \frac{5\hat i - \hat j + 2\hat k}{\sqrt{30}}.$

Then multiply by $8$:

$8\,\hat a = \frac{8(5\hat i - \hat j + 2\hat k)}{\sqrt{30}} = \tfrac{40}{\sqrt{30}}\hat i - \tfrac{8}{\sqrt{30}}\hat j + \tfrac{16}{\sqrt{30}}\hat k.$

Key takeaways

• The magnitude of a vector is the square root of the sum of the squares of its components, and a vector is a unit vector when this equals $1$.
• Equal vectors have equal corresponding components; the coefficients of $\hat i, \hat j, \hat k$ are the scalar components.
• A unit vector is $\hat a = \dfrac{\vec a}{|\vec a|}$, and multiplying it by any number gives a vector of that magnitude in the same direction.