Vectors and the dot product
Tonight's idea
A vector \(\mathbf{a} = (a_1, a_2)\) is an arrow with length \(\lvert\mathbf{a}\rvert = \sqrt{a_1^2 + a_2^2}\). Tonight is about multiplying two vectors, and the answer is not a vector but a number:
\(\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2\)
Here is what it measures. Put the arrows tail to tail with angle \(\theta\) between them. The third side of the triangle is \(\mathbf{a} - \mathbf{b}\), and the cosine rule gives \(\lvert\mathbf{a} - \mathbf{b}\rvert^2 = \lvert\mathbf{a}\rvert^2 + \lvert\mathbf{b}\rvert^2 - 2\lvert\mathbf{a}\rvert\lvert\mathbf{b}\rvert\cos\theta\). In components the same left side is \((a_1 - b_1)^2 + (a_2 - b_2)^2 = \lvert\mathbf{a}\rvert^2 + \lvert\mathbf{b}\rvert^2 - 2\,\mathbf{a}\cdot\mathbf{b}\). Compare:
\(\mathbf{a} \cdot \mathbf{b} = \lvert\mathbf{a}\rvert\,\lvert\mathbf{b}\rvert\cos\theta\)
So the dot product is "length times length times how aligned they are". At right angles it is zero: a perpendicular test that needs no angles, just a multiply and an add. Past \(90°\) it goes negative.
Shadow of \(\mathbf{a}\) on the line of \(\mathbf{b}\): length \(\dfrac{\mathbf{a}\cdot\mathbf{b}}{\lvert\mathbf{b}\rvert}\).
A force \(\mathbf{F}\) moving something through \(\mathbf{d}\) does work \(\mathbf{F}\cdot\mathbf{d}\), because only the part of the force along the motion counts.
Watch one: the angle between two arrows
\(\mathbf{a} = (1, 2)\), \(\mathbf{b} = (3, 1)\). Dot product \(3 + 2 = 5\). Lengths \(\sqrt5\) and \(\sqrt{10}\). So \(\cos\theta = \dfrac{5}{\sqrt{50}} = \dfrac{1}{\sqrt2}\) and \(\theta = 45°\) exactly, which is not obvious from the picture. The projection (the shadow) of \(\mathbf{a}\) onto \(\mathbf{b}\) is \(\dfrac{5}{10}(3, 1) = (1.5, 0.5)\); check \((1, 2) - (1.5, 0.5) = (-0.5, 1.5)\) is perpendicular to \(\mathbf{b}\): \(-1.5 + 1.5 = 0\).
Your turn
- Compute each dot product and say whether the angle is acute, right or obtuse: \((2, 3)\cdot(4, -1)\); \((1, 5)\cdot(-10, 2)\); \((-3, 1)\cdot(2, 4)\).
- Find the angle between \((1, 0)\) and \((1, \sqrt3)\) exactly. Then the angle between \((2, 1)\) and \((1, -3)\) to the nearest tenth of a degree.
- Find \(k\) so that \((k, 2)\) is perpendicular to \((3, -6)\). Then find a unit vector (length 1) perpendicular to \((3, 4)\). How many answers are there?
- A force \(\mathbf{F} = (6, 2)\) newtons moves a crate through \(\mathbf{d} = (4, 3)\) metres. Find the work done and the component of the force along the direction of motion, in newtons.
Stretch: the rhombus in one line
A parallelogram with adjacent sides \(\mathbf{a}\) and \(\mathbf{b}\) has diagonals \(\mathbf{a} + \mathbf{b}\) and \(\mathbf{a} - \mathbf{b}\). Expand \((\mathbf{a} + \mathbf{b})\cdot(\mathbf{a} - \mathbf{b})\). Deduce that a rhombus has perpendicular diagonals, and that a parallelogram with perpendicular diagonals is a rhombus.
Answers
- \(8 - 3 = 5 \gt 0\): acute. \(-10 + 10 = 0\): right angle. \(-6 + 4 = -2 \lt 0\): obtuse.
- Dot product 1, lengths 1 and 2, \(\cos\theta = \tfrac12\), \(\theta = 60°\). Second: dot product \(-1\), lengths \(\sqrt5\) and \(\sqrt{10}\), \(\cos\theta = -\tfrac{1}{\sqrt{50}}\), \(\theta \approx 98.1°\).
- \(3k - 12 = 0\), so \(k = 4\). \((-4, 3)\) and \((4, -3)\) are perpendicular to \((3, 4)\) with length 5, so the unit vectors are \(\left(-\tfrac45, \tfrac35\right)\) and \(\left(\tfrac45, -\tfrac35\right)\). Exactly two.
- Work \(= 24 + 6 = 30\) J. \(\lvert\mathbf{d}\rvert = 5\), so the component along the motion is \(\tfrac{30}{5} = 6\) N.
Stretch. \((\mathbf{a} + \mathbf{b})\cdot(\mathbf{a} - \mathbf{b}) = \lvert\mathbf{a}\rvert^2 - \lvert\mathbf{b}\rvert^2\), since the cross terms cancel. Equal sides give 0, so the diagonals are perpendicular. Conversely, perpendicular diagonals force \(\lvert\mathbf{a}\rvert = \lvert\mathbf{b}\rvert\), so all four sides are equal.