Miscellaneous · Q10: Parallelogram on $2\hat i-4\hat j+5\hat k$⁠, $\hat i-2\hat j-3\hat k$⁠: unit diagonal and area

Basics you need

Part (i): the unit vector along the diagonal

1. The diagonal

   $$\begin{aligned} \vec d = \vec a + \vec b &= (2 + 1)\hat i + (-4 - 2)\hat j \\ &\quad + (5 - 3)\hat k \\ &= 3\hat i - 6\hat j + 2\hat k. \end{aligned}$$
   Put $\vec a$ and $\vec b$ tail to tail at a corner O. Going along $\vec a$ and then along the side opposite $\vec b$ (which equals $\vec b$⁠) ends at the far corner, so the diagonal from O is $\vec a + \vec b$⁠. Add component by component.
   Parallelogram law (Section 10.4, Fig 10.9); sum by components (Section 10.5.1). Lesson: adding arrows, head to tail

2. Its length and the unit vector

   $$|\vec d| = \sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{49} = 7,$$ $$\hat d = \frac{1}{7}(3\hat i - 6\hat j + 2\hat k) = \frac37\hat i - \frac67\hat j + \frac27\hat k.$$
   $9 + 36 + 4 = 49$ is a perfect square, so the length is the whole number 7. Dividing $\vec d$ by its length keeps its direction and makes its length 1.
   Unit vector $\hat d = \frac{1}{|\vec d|}\vec d$ (Section 10.5). Lesson: components

3. The other diagonal

   A parallelogram has a second diagonal, joining the tips of $\vec a$ and $\vec b$⁠:

   $$\vec a - \vec b = \hat i - 2\hat j + 8\hat k, \quad |\vec a - \vec b| = \sqrt{69}.$$

   A unit vector along it is $\frac{1}{\sqrt{69}}(\hat i - 2\hat j + 8\hat k)$⁠.

   "Its diagonal" most naturally means the diagonal through the common point of the given sides, $\vec a + \vec b$⁠; that is the one the answer key gives. The second diagonal is shown so that you can answer either reading, and each unit vector could also be reversed.
   Difference of vectors and the triangle law (Section 10.4). Lesson: adding arrows, head to tail

Part (ii): the area

1. The cross product

   $$\begin{aligned} \vec a \times \vec b &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & -4 & 5 \\ 1 & -2 & -3 \end{vmatrix} \\ &= \hat i\big[(-4)(-3) - (5)(-2)\big] \\ &\quad - \hat j\big[(2)(-3) - (5)(1)\big] \\ &\quad + \hat k\big[(2)(-2) - (-4)(1)\big] \\ &= 22\hat i + 11\hat j + 0\hat k. \end{aligned}$$
   Expand along the first row: cover the row and column of $\hat i$ and take the 2 by 2 determinant left, and so on, with signs $+, -, +$⁠. The brackets are $12 + 10 = 22$⁠, $-6 - 5 = -11$ (and the minus sign in front makes $+11$⁠) and $-4 + 4 = 0$⁠.
   $\vec a \times \vec b$ in components (Section 10.6.3). Basics: a determinant measures a signed area

2. Its length is the area

   $$\begin{aligned} |\vec a \times \vec b| &= \sqrt{22^2 + 11^2} = \sqrt{484 + 121} \\ &= \sqrt{605} = \sqrt{121 \times 5} = 11\sqrt5. \end{aligned}$$

   So the area is $11\sqrt5$ square units.

   The parallelogram's area is base times height, $|\vec a| \times |\vec b|\sin\theta$⁠, and this is exactly $|\vec a \times \vec b|$⁠. Taking out the square factor: $605 = 121 \times 5 = 11^2 \times 5$⁠.
   Area of a parallelogram $= |\vec a \times \vec b|$ (Section 10.6.3, Observation 9); compare Example 25. Lesson: Examples 22–25

3. Check with $|\vec a||\vec b|\sin\theta$

   $|\vec a| = \sqrt{4 + 16 + 25} = \sqrt{45}$⁠, $|\vec b| = \sqrt{1 + 4 + 9} = \sqrt{14}$ and $\vec a\cdot\vec b = 2 + 8 - 15 = -5$⁠. So

   $$\begin{aligned} |\vec a|^2|\vec b|^2 - (\vec a\cdot\vec b)^2 &= 45 \times 14 - 25 \\ &= 630 - 25 = 605, \end{aligned}$$

   and $\sqrt{605}$ is the area again, since $|\vec a|^2|\vec b|^2\sin^2\theta = |\vec a|^2|\vec b|^2 - (\vec a\cdot\vec b)^2$⁠.

   This uses only dot products, so it checks the cross product by an independent route: $\sin^2\theta = 1 - \cos^2\theta$ and $\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}$⁠. The negative dot product shows the angle between the sides is obtuse, as drawn.
   $\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$ (Section 10.6.1); $|\vec a \times \vec b| = |\vec a||\vec b|\sin\theta$ (Section 10.6.3). Lesson: the dot product

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