Vector Dot Product (Exercise 10.3)

This lesson works through the dot product questions of Exercise 10.3: finding angles between vectors, projections, unit and perpendicular vectors, the magnitude of a vector, and an angle inside a triangle.

Angle between two vectors from their magnitudes

Given $|\vec a| = \sqrt{3}$, $|\vec b| = 2$, and $\vec a \cdot \vec b = \sqrt{6}$, the angle uses

$\cos\theta = \frac{\vec a \cdot \vec b}{|\vec a|\,|\vec b|} = \frac{\sqrt{6}}{\sqrt{3}\cdot 2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}.$

Since $\cos\tfrac{\pi}{4} = \tfrac{1}{\sqrt 2}$, we get $\theta = \dfrac{\pi}{4}$.

Angle between two vectors given by components

Let $\vec a = \hat i - 2\hat j + 3\hat k$ and $\vec b = 3\hat i - 2\hat j + \hat k$. The dot product multiplies matching components and adds:

$\vec a \cdot \vec b = (1)(3) + (-2)(-2) + (3)(1) = 3 + 4 + 3 = 10.$

The magnitudes are

$|\vec a| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{14}, \qquad |\vec b| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{14}.$

Therefore

$\cos\theta = \frac{10}{\sqrt{14}\,\sqrt{14}} = \frac{10}{14} = \frac{5}{7}, \qquad \theta = \cos^{-1}\frac{5}{7}.$

Projection of one vector onto another

The projection of $\vec a$ on $\vec b$ is $\dfrac{\vec a \cdot \vec b}{|\vec b|}$.

Projection of $\hat i - \hat j$ on $\hat i + \hat j$

$\frac{(\hat i - \hat j)\cdot(\hat i + \hat j)}{|\hat i + \hat j|} = \frac{(1)(1) + (-1)(1)}{\sqrt{1^2 + 1^2}} = \frac{0}{\sqrt 2} = 0.$

The projection is $0$, so these vectors are perpendicular.

Projection of $\hat i + 3\hat j + 7\hat k$ on $7\hat i - \hat j + 8\hat k$

With $\vec a = \hat i + 3\hat j + 7\hat k$ and $\vec b = 7\hat i - \hat j + 8\hat k$,

$\vec a \cdot \vec b = (1)(7) + (3)(-1) + (7)(8) = 7 - 3 + 56 = 60,$

$|\vec b| = \sqrt{7^2 + (-1)^2 + 8^2} = \sqrt{114},$

so the projection is $\dfrac{60}{\sqrt{114}}$.

Unit vectors that are perpendicular

Let $\vec a = \tfrac{1}{7}\big(2\hat i + 3\hat j + 6\hat k\big)$ and $\vec b = \tfrac{1}{7}\big(3\hat i - 6\hat j + 2\hat k\big)$.

Magnitudes

$|\vec a| = \tfrac{1}{7}\sqrt{2^2 + 3^2 + 6^2} = \tfrac{1}{7}\sqrt{49} = 1,$

$|\vec b| = \tfrac{1}{7}\sqrt{3^2 + (-6)^2 + 2^2} = \tfrac{1}{7}\sqrt{49} = 1,$

so both are unit vectors.

Perpendicularity

$\vec a \cdot \vec b = \tfrac{1}{49}\big[(2)(3) + (3)(-6) + (6)(2)\big] = \tfrac{1}{49}\big[6 - 18 + 12\big] = 0.$

Since the dot product is $0$, $\vec a$ is perpendicular to $\vec b$.

Magnitude of a vector and expanding dot products

A difference of squares

Expanding $(\vec a + \vec b)\cdot(\vec a - \vec b)$ gives

$\vec a \cdot \vec a - \vec b \cdot \vec b = |\vec a|^2 - |\vec b|^2,$

using $\vec a \cdot \vec a = |\vec a|^2$ and $\vec b \cdot \vec b = |\vec b|^2$.

Expanding a product of combinations

For $(3\vec a - 5\vec b)\cdot(2\vec a + 7\vec b)$, expand term by term:

$6\,\vec a\cdot\vec a + 21\,\vec a\cdot\vec b - 10\,\vec b\cdot\vec a - 35\,\vec b\cdot\vec b.$

Since $\vec a\cdot\vec b = \vec b\cdot\vec a$, the middle terms combine:

$6\,|\vec a|^2 + 11\,(\vec a\cdot\vec b) - 35\,|\vec b|^2.$

Magnitude from a unit vector condition

For a unit vector $\hat a$, given $(\vec x - \hat a)\cdot(\vec x + \hat a) = 12$,

$|\vec x|^2 - |\hat a|^2 = 12.$

Since $|\hat a| = 1$, we get $|\vec x|^2 = 12 + 1 = 13$, so $|\vec x| = \sqrt{13}$.

Angle of a triangle from its vertices

For a triangle with vertices $A(1,2,3)$, $B(-1,0,0)$, $C(0,1,2)$, the angle at $B$ is the angle between $\vec{BA}$ and $\vec{BC}$. Using $\vec{PQ} = Q - P$ on the components,

$\vec{BA} = (2,\,2,\,3), \qquad \vec{BC} = (1,\,1,\,2).$

Then

$\vec{BA}\cdot\vec{BC} = (2)(1) + (2)(1) + (3)(2) = 2 + 2 + 6 = 10,$

$|\vec{BA}| = \sqrt{4 + 4 + 9} = \sqrt{17}, \qquad |\vec{BC}| = \sqrt{1 + 1 + 4} = \sqrt{6},$

so

$\cos\theta = \frac{10}{\sqrt{17}\,\sqrt{6}} = \frac{10}{\sqrt{102}}, \qquad \theta = \cos^{-1}\frac{10}{\sqrt{102}}.$

Key takeaways

• The angle between vectors comes from $\cos\theta = \dfrac{\vec a \cdot \vec b}{|\vec a|\,|\vec b|}$.
• The projection of $\vec a$ on $\vec b$ is $\dfrac{\vec a \cdot \vec b}{|\vec b|}$, and it is $0$ exactly when the vectors are perpendicular.
• A vector is a unit vector when its magnitude is $1$, and $\vec a \cdot \vec a = |\vec a|^2$ lets you expand dot products like ordinary algebra.