Scalar and Vector Products, with Mixed Practice

Why this mat­ters#

You can mul­ti­ply two vec­tors in two dif­fer­ent ways, and each one answers a dif­fer­ent ques­tion. The scalar (dot) prod­uct gives you a plain num­ber, and it tells you about the angle between the vec­tors. The vec­tor (cross) prod­uct gives you a new vec­tor stand­ing per­pen­dic­u­lar to both, and its length turns out to be an area. We will define both, use them to find angles, pro­jec­tions, per­pen­dic­u­lar direc­tions and areas, and fin­ish with mixed prac­tice.

Scalar prod­uct#

$$\vec a \cdot \vec b = \lvert\vec a\rvert\lvert\vec b\rvert\cos\theta = a_1b_1 + a_2b_2 + a_3b_3.$$

• $\vec a \cdot \vec b = 0 \iff \vec a \perp \vec b$ (for non-zero vec­tors).
• $\vec a \cdot \vec a = \lvert\vec a\rvert^2$; $\hat i \cdot \hat i = 1$, $\hat i \cdot \hat j = 0$.
• It is com­mu­ta­tive and dis­trib­u­tive, so it behaves much like ordi­nary mul­ti­pli­ca­tion.
• The pro­jec­tion of $\vec a$ on $\vec b$ is $\displaystyle \frac{\vec a \cdot \vec b}{\lvert\vec b\rvert}$.

Vec­tor prod­uct#

$\vec a \times \vec b = \lvert\vec a\rvert\lvert\vec b\rvert\sin\theta\,\hat n$, where $\hat n$ is the unit vec­tor per­pen­dic­u­lar to both, cho­sen by the right-hand rule. Curl the fin­gers of your right hand from the first vec­tor to the sec­ond, and your thumb points along it. In com­po­nents,

$$\vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}.$$

• $\vec b \times \vec a = -\vec a \times \vec b$; $\vec a \times \vec a = \vec 0$; $\hat i \times \hat j = \hat k$, $\hat j \times \hat k = \hat i$, $\hat k \times \hat i = \hat j$.
• $\vec a \times \vec b = \vec 0 \iff$ the vec­tors are par­al­lel, or one of them is zero.
• $\lvert\vec a \times \vec b\rvert$ gives the area of the par­al­lel­o­gram on $\vec a$ and $\vec b$, and half of it gives the area of the tri­an­gle.

Mixed prac­tice for the chap­ter#

1. Find the angle between $\hat i - \hat j$ and $\hat j - \hat k$.
2. Find the pro­jec­tion of $3\hat i - \hat j + 4\hat k$ on $2\hat i + 3\hat j - 6\hat k$.
3. Show that $2\hat i - \hat j + \hat k$, $\hat i - 3\hat j - 5\hat k$ and $3\hat i - 4\hat j - 4\hat k$ form a right-angled tri­an­gle.
4. If $\lvert\vec a\rvert = 2$, $\lvert\vec b\rvert = 3$ and $\vec a \cdot \vec b = 3$, find the angle between them and $\lvert\vec a - \vec b\rvert$.
5. If $\vec a + \vec b + \vec c = \vec 0$ with $\lvert\vec a\rvert = 3$, $\lvert\vec b\rvert = 5$, $\lvert\vec c\rvert = 7$, find the angle between $\vec a$ and $\vec b$.
6. Find $\vec a \times \vec b$ for $\vec a = 3\hat i + 2\hat j + 2\hat k$, $\vec b = \hat i + 2\hat j - 2\hat k$, and a unit vec­tor per­pen­dic­u­lar to both.
7. Find the area of the par­al­lel­o­gram with adja­cent sides $\hat i + 2\hat j + 3\hat k$ and $3\hat i - 2\hat j + \hat k$.
8. Find the area of the tri­an­gle with ver­tices $(1, 2, 3)$, $(2, -1, 1)$, $(1, 2, -4)$.
9. Show that $(\vec a - \vec b) \times (\vec a + \vec b) = 2(\vec a \times \vec b)$.
10. Find $\lambda$ if $(2\hat i + 6\hat j + 27\hat k) \times (\hat i + \lambda\hat j + \mu\hat k) = \vec 0$.

Answers#