14.2 Dot product and vector projections

Dot Product

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Dot Product Calculation and Meaning

The dot product takes two vectors and produces a single scalar (a number, not a vector). You calculate it by multiplying corresponding components and adding the results.

• In 2D: $\vec{a} \cdot \vec{b} = a_xb_x + a_yb_y$
• In 3D: $\vec{a} \cdot \vec{b} = a_xb_x + a_yb_y + a_zb_z$

But the dot product isn't just arithmetic. There's a geometric formula that connects it to the angle between the two vectors:

$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$

where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$. This is what makes the dot product so useful: it encodes directional information.

• Positive dot product → vectors point in roughly the same direction (acute angle, $\theta < 90°$)
• Zero dot product → vectors are perpendicular ($\theta = 90°$)
• Negative dot product → vectors point in roughly opposite directions (obtuse angle, $\theta > 90°$)

Angle Determination Using the Dot Product

Since both formulas compute the same dot product, you can set them equal and solve for the angle:

1. Compute $\vec{a} \cdot \vec{b}$ using components.
2. Compute $|\vec{a}|$ and $|\vec{b}|$ (the magnitudes of each vector).
3. Plug into the rearranged formula:

$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$

1. Take the inverse cosine to get the angle:

$\theta = \arccos\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\right)$

Example: Find the angle between $\vec{a} = \langle 1, 2 \rangle$ and $\vec{b} = \langle 3, 1 \rangle$.

1. $\vec{a} \cdot \vec{b} = 1(3) + 2(1) = 5$
2. $|\vec{a}| = \sqrt{1^2 + 2^2} = \sqrt{5}$, and $|\vec{b}| = \sqrt{3^2 + 1^2} = \sqrt{10}$
3. $\cos\theta = \frac{5}{\sqrt{5}\cdot\sqrt{10}} = \frac{5}{\sqrt{50}} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$
4. $\theta = \arccos\left(\frac{1}{\sqrt{2}}\right) = 45°$

Perpendicularity check: Are $\vec{a} = \langle 3, 4 \rangle$ and $\vec{b} = \langle -4, 3 \rangle$ perpendicular?

• $\vec{a} \cdot \vec{b} = 3(-4) + 4(3) = -12 + 12 = 0$

The dot product is zero, so yes, these vectors are orthogonal (perpendicular). You don't even need to compute magnitudes when checking for perpendicularity; a dot product of zero is sufficient.

Vector Projections

Vector Projection and Scalar Projection

Projecting $\vec{a}$ onto $\vec{b}$ answers the question: how much of $\vec{a}$ points in the direction of $\vec{b}$? Think of it as the "shadow" $\vec{a}$ casts onto the line defined by $\vec{b}$ when light shines straight down perpendicular to that line.

There are two versions to know:

Scalar projection (a number): the signed length of that shadow.

$\text{comp}_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$

• Positive means $\vec{a}$ has a component in the same direction as $\vec{b}$
• Negative means $\vec{a}$ has a component in the opposite direction

Vector projection (a vector): the actual vector lying along $\vec{b}$ that represents that component.

$\text{proj}_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\vec{b}$

Notice the difference: the scalar projection divides by $|\vec{b}|$, while the vector projection divides by $|\vec{b}|^2$ and then multiplies by $\vec{b}$ itself. A good way to remember this: the vector projection takes the scalar projection and scales the unit vector $\frac{\vec{b}}{|\vec{b}|}$ by it, which is why you end up with $|\vec{b}|^2$ in the denominator.

Example: Project $\vec{a} = \langle 3, 4 \rangle$ onto $\vec{b} = \langle 5, 0 \rangle$.

1. $\vec{a} \cdot \vec{b} = 3(5) + 4(0) = 15$
2. $|\vec{b}|^2 = 5^2 + 0^2 = 25$
3. $\text{proj}_{\vec{b}}\vec{a} = \frac{15}{25}\langle 5, 0 \rangle = \langle 3, 0 \rangle$

This makes geometric sense: $\vec{b}$ points along the x-axis, so the projection of $\vec{a}$ onto it is just the x-component of $\vec{a}$.

Decomposing a Vector into Parallel and Perpendicular Components

Once you have the parallel component ($\text{proj}_{\vec{b}}\vec{a}$), you can find the perpendicular component by subtraction:

$\vec{a}_{\perp} = \vec{a} - \text{proj}_{\vec{b}}\vec{a}$

This decomposes $\vec{a}$ into two parts: one parallel to $\vec{b}$ and one perpendicular to it. You can verify your work by checking that $\vec{a}_{\perp} \cdot \vec{b} = 0$ (they should be orthogonal).

Continuing the example above: $\vec{a}_{\perp} = \langle 3, 4 \rangle - \langle 3, 0 \rangle = \langle 0, 4 \rangle$. Check: $\langle 0, 4 \rangle \cdot \langle 5, 0 \rangle = 0$. It works.

Applications in Geometric Problem-Solving

1. Work done by a force. If a force $\vec{F}$ acts on an object that moves along displacement $\vec{d}$, the work is $W = \vec{F} \cdot \vec{d}$. Only the component of force in the direction of motion contributes to work.

2. Resolving forces on inclined planes. Project the weight vector onto the direction parallel to the surface and the direction perpendicular to it. The parallel component is the force pulling the object down the slope; the perpendicular component equals the normal force.

3. Finding force components in a given direction. Use $\text{proj}_{\vec{d}}\vec{F}$ to extract the part of a force $\vec{F}$ that acts along a specific direction $\vec{d}$.

4. Distance from a point to a line. Project the vector from any point on the line to the given point onto the line's direction vector. The perpendicular component (what's left after subtracting the projection) has a magnitude equal to the distance from the point to the line.