Vectors Supplement – Dot Product and Projections (Study Notes)

The Dot Product

• The dot product of two nonzero vectors a and b is a scalar quantity representing the product of the vectors in a directional sense.
• Notation: $\mathbf{a} \cdot \mathbf{b}$ or $a \cdot b$.
• Two main formulas:
  • Magnitude–angle form: $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| \, |\mathbf{b}| \, \cos \theta$ where the angle between a and b is $0 \le \theta \le \pi$.
  • Component form: If $\mathbf{a} = (a<em>x, a</em>y)$ and $\mathbf{b} = (b<em>x, b</em>y)$, then$\mathbf{a} \cdot \mathbf{b} = a<em>x b</em>x + a<em>y b</em>y$.
• Special cases of the dot product occur when the angle is:
  • $\theta = 0$
  • $\theta = \frac{\pi}{2}$
  • $\theta = \pi$
• The angle between two vectors can be found from the dot product: $\theta = \arccos\left( \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| \, |\mathbf{b}|} \right)$
• The orthogonal complement of a nonzero vector a = (ax, ay) is a vector orthogonal to a with the same magnitude. A standard choice is: -$\mathbf{a}<em>{\perp} = (-a</em>y, a_x)$
  • The vector $(ay, -ax)$ is also orthogonal to a, but it does not have a standard name.

Problem-solving examples

• Problem 1: Find $\mathbf{a} \cdot \mathbf{b}$ for $\mathbf{a} = (4, 7)$ and $\mathbf{b} = (3, -2)$
  • Calculation: $\mathbf{a} \cdot \mathbf{b} = 4\cdot 3 + 7\cdot(-2) = 12 - 14 = -2$
  • Answer: $-2$
• Problem 2: Find $\mathbf{a} \cdot \mathbf{b}$ when $|\mathbf{a}| = 7, \; |\mathbf{b}| = 2$ and the angle between them is $\theta = \frac{\pi}{6}$
  • Calculation: $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}|\,|\mathbf{b}| \cos \theta = 7\cdot 2 \cos\left(\frac{\pi}{6}\right) = 14 \cdot \frac{\sqrt{3}}{2} = 7\sqrt{3}$

Orthogonal complement in detail

• The orthogonal complement of a vector a = (ax, ay) is any vector perpendicular to a.
• A standard orthogonal companion is:
  -$\mathbf{a}<em>{\perp} = (-a</em>y, a_x)$
• The vector $(ay, -ax)$ is also orthogonal to a but lacks a special name.
• Problem 3: Find the orthogonal complement to ((-2, 8)).
  • A perpendicular direction is given by $\mathbf{a}<em>{\perp} = (-8, -2)$ since for a = (-2, 8): $\mathbf{a}</em>{\perp} = (-a<em>y, a</em>x) = (-8, -2)$.
  • Length of a is $|\mathbf{a}| = \sqrt{(-2)^2 + 8^2} = \sqrt{4 + 64} = \sqrt{68} = 2\sqrt{17}$.
  • A vector of length 2 orthogonal to a is obtained by scaling $\mathbf{a}<em>{\perp}$ to length 2: $\mathbf{v} = \frac{2}{|\mathbf{a}|}\, \mathbf{a}</em>{\perp} = \frac{2}{2\sqrt{17}} (-8, -2) = \left(-\frac{8}{\sqrt{17}}, -\frac{2}{\sqrt{17}}\right)$
    (Additionally, the opposite direction is also valid.)
• Problem 4 (orthogonality condition): Find the value(s) of $x$ such that the vectors
  • $\mathbf{a} = (2x^2, -1)$ and $\mathbf{b} = (2, -6x)$ are orthogonal.
  • Dot product: $\mathbf{a} \cdot \mathbf{b} = (2x^2)(2) + (-1)(-6x) = 4x^2 + 6x$
  • Set to zero: $4x^2 + 6x = 0 \Rightarrow 2x(2x + 3) = 0$
  • Solutions: $x = 0 \quad \text{or} \quad x = -\frac{3}{2}$

Angles between vectors

• The dot product formula also yields the angle: $\theta = \arccos\left( \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| \, |\mathbf{b}|} \right)$
• Problem 5: Find the angle between $\mathbf{a} = (-1, 4)$ and $\mathbf{b} = (2, 5)$
  • $\mathbf{a} \cdot \mathbf{b} = (-1)(2) + 4(5) = -2 + 20 = 18$
  • $|\mathbf{a}| = \sqrt{(-1)^2 + 4^2} = \sqrt{17}, \quad |\mathbf{b}| = \sqrt{2^2 + 5^2} = \sqrt{29}$
  • $\cos \theta = \dfrac{18}{\sqrt{17}\sqrt{29}} = \dfrac{18}{\sqrt{493}}$
  • $\theta = \arccos\left( \dfrac{18}{\sqrt{493}} \right)$
• Problem 6: Triangle with vertices A(1,5), B(2,1), C(4,6); find $\angle BCA$ (angle at C)
  • Vectors: $\overrightarrow{CB} = B - C = (-2, -5)$, $\overrightarrow{CA} = A - C = (-3, -1)$
  • Dot product: $\overrightarrow{CB} \cdot \overrightarrow{CA} = (-2)(-3) + (-5)(-1) = 6 + 5 = 11$
  • Magnitudes: $|\overrightarrow{CB}| = \sqrt{(-2)^2 + (-5)^2} = \sqrt{29}, \quad |\overrightarrow{CA}| = \sqrt{(-3)^2 + (-1)^2} = \sqrt{10}$
  • $\cos \theta = \dfrac{11}{\sqrt{29}\sqrt{10}} = \dfrac{11}{\sqrt{290}}$
  • $\theta = \arccos\left( \dfrac{11}{\sqrt{290}} \right)$

Work and applications of the dot product

• Work done by a force when moving an object: the work is the dot product of the force vector F and the displacement vector d, i.e.,
  • Displacement: $\mathbf{d} = \overrightarrow{PQ} = Q - P$
  • Work: $W = \mathbf{F} \cdot \mathbf{d}$
• Units:
  • Metric: force in Newtons (N), distance in meters (m), work in joules (J) or N·m
  • Imperial: force in pounds (lb), distance in feet (ft), work in ft·lb
• Problem 7: A force $\mathbf{F} = (-4, 2)$ N moves an object from $P(3,6)$ to $Q(2,9)$. If distance is measured in meters, how much work is done?
  • Displacement: $\mathbf{d} = Q - P = (2-3, 9-6) = (-1, 3)$
  • Work: $W = \mathbf{F} \cdot \mathbf{d} = (-4)(-1) + (2)(3) = 4 + 6 = 10 \text{ J}$
• Problem 8: A horizontal force of 30 N is used to push a box up to the top of a ramp of length 20 m. The ramp makes an angle of 70° with the horizon.
  • Displacement along the ramp: $\mathbf{d} = (20\cos 70^\circ,\; 20\sin 70^\circ)$
  • Force: $\mathbf{F} = (30, 0)$
  • Work: $W = \mathbf{F} \cdot \mathbf{d} = 30 \cdot (20\cos 70^\circ) = 600 \cos 70^\circ \approx 2.05\times 10^2 \text{ J}$

Scalar and Vector Projections

• Scalar projection of b onto a (length of the shadow of b on a):
  • $\operatorname{comp}_{\mathbf{a}} \mathbf{b} = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|}$
• Vector projection of b onto a (the actual projected vector along a):
  • $\operatorname{proj}_{\mathbf{a}} \mathbf{b} = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|^2} \, \mathbf{a} = \left( \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|^2} \right) \mathbf{a}$
• Alternative equivalent form: $\operatorname{proj}_{\mathbf{a}} \mathbf{b} = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|} \cdot \dfrac{\mathbf{a}}{|\mathbf{a}|}$

Problems: projections

• Problem 9: Find the scalar and vector projections of $\mathbf{a} = (3, -1)$ onto $\mathbf{b} = (-2, 5)$
  • Dot product: $\mathbf{a} \cdot \mathbf{b} = 3(-2) + (-1)(5) = -6 - 5 = -11$
  • Magnitude of a: $|\mathbf{a}| = \sqrt{3^2 + (-1)^2} = \sqrt{10}$
  • Scalar projection: $\operatorname{comp}_{\mathbf{a}} \mathbf{b} = \dfrac{-11}{\sqrt{10}}$
  • Vector projection: $\operatorname{proj}_{\mathbf{a}} \mathbf{b} = \dfrac{-11}{|\mathbf{a}|^2} \mathbf{a} = \dfrac{-11}{10} (3, -1) = \left(-\dfrac{33}{10}, \dfrac{11}{10}\right)$
• Problem 10: Find the shortest distance from the point (3, 6) to the line y = \tfrac{1}{2}x + 1 using scalar projections.
  • A convenient approach uses the line in standard form and distance from a point to a line:
  • Line: y = \tfrac{1}{2}x + 1 ⇒ x - 2y + 2 = 0 with A=1, B=-2, C=2
  • Distance: $\text{dist} = \dfrac{|A x<em>0 + B y</em>0 + C|}{\sqrt{A^2 + B^2}} = \dfrac{|1\cdot 3 + (-2)\cdot 6 + 2|}{\sqrt{1^2 + (-2)^2}} = \dfrac{|-7|}{\sqrt{5}} = \dfrac{7}{\sqrt{5}} \approx 3.13$
  • If using a projection approach: take a point Q on the line, e.g., Q=(0,1). The displacement from Q to P is r = P - Q = (3,5). A normal vector to the line is n = (-1, 2) (perpendicular to the direction (2,1)). The distance is |r · n| / |n| = |(-3) + 10| / \sqrt{5} = 7/\sqrt{5}.

What do scalar and vector projections represent?

• Scalar projection: the length of the component of b along the direction of a (how much of b lies in the direction of a).
• Vector projection: the actual vector along a that represents the component of b in the direction of a.
• These concepts connect to many applications in physics (work, force components) and geometry (shadow length, decomposition of vectors).

Quick connections and checks

• Projections rely on the dot product and the magnitudes of the vectors.
• The results scale predictably with changes in vector magnitudes and angles.
• In physics, the work calculation W = F · d emphasizes that only the component of the force along the displacement contributes to work; the perpendicular component does no work.

Summary of key formulas

• Dot product (two forms):
  • $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}|\,|\mathbf{b}| \cos \theta$
  • $\mathbf{a} \cdot \mathbf{b} = a<em>x b</em>x + a<em>y b</em>y$
• Angle from dot product: $\theta = \arccos\left( \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}| \, |\mathbf{b}|} \right)$
• Orthogonal complement: $\mathbf{a}<em>{\perp} = (-a</em>y, a_x)$
• Work: $W = \mathbf{F} \cdot \mathbf{d}$
• Scalar projection: $\operatorname{comp}_{\mathbf{a}} \mathbf{b} = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|}$
• Vector projection: $\operatorname{proj}_{\mathbf{a}} \mathbf{b} = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|^2} \mathbf{a}$