Vectors, Components and the Section Formula

Why this mat­ters#

If some­one tells you an auto-rick­shaw went a cer­tain dis­tance, you still do not know where it ended up. You need the direc­tion too. Quan­ti­ties like dis­place­ment, veloc­ity and force carry both a size and a direc­tion, and we call them vec­tors. In this chap­ter you will meet the dif­fer­ent kinds of vec­tors, add them with the tri­an­gle law, mul­ti­ply them by scalars, write them in com­po­nents as $a\hat i + b\hat j + c\hat k$, and use the sec­tion for­mula in vec­tor form.

Vec­tors#

A vec­tor is a directed line seg­ment $\overrightarrow{AB}$ with mag­ni­tude $\lvert\overrightarrow{AB}\rvert$ and a direc­tion. You will come across these types: zero (mag­ni­tude $0$), unit (mag­ni­tude $1$), equal (same mag­ni­tude and direc­tion), collinear (par­al­lel), neg­a­tive of $\vec a$ (same size, oppo­site direc­tion).

Direc­tion cosines. Sup­pose $\vec r$ makes angles $\alpha, \beta, \gamma$ with the axes. Then $l = \cos\alpha$, $m = \cos\beta$, $n = \cos\gamma$ are its direc­tion cosines, and $l^2 + m^2 + n^2 = 1$.

Addi­tion and scalar mul­ti­ples#

Tri­an­gle law: $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$. Par­al­lel­o­gram law: if two vec­tors start from the same point, their sum is the diag­o­nal of the par­al­lel­o­gram they make. Vec­tor addi­tion is both com­mu­ta­tive and asso­cia­tive. For any scalar, $\lvert k\vec a \rvert = \lvert k \rvert\lvert\vec a\rvert$; and the direc­tion flips when $k < 0$. To get the unit vec­tor along $\vec a$, just divide by its length: $\displaystyle \hat a = \tfrac{\vec a}{\lvert\vec a\rvert}$.

Com­po­nents#

Let $\hat i, \hat j, \hat k$ be unit vec­tors along the axes. Then the posi­tion vec­tor of $P(x, y, z)$ is $\vec r = x\hat i + y\hat j + z\hat k$, $\lvert\vec r\rvert = \sqrt{x^2 + y^2 + z^2}$, and its direc­tion cosines are $\displaystyle \tfrac{x}{r}, \tfrac{y}{r}, \tfrac{z}{r}$.

$\overrightarrow{PQ}$ = posi­tion vec­tor of $Q$ − posi­tion vec­tor of $P$. In other words, the end point minus the start­ing point. Adding and scal­ing vec­tors is done one com­po­nent at a time.

Collinear­ity. $\vec a$ and $\vec b$ are collinear when $\vec b = \lambda\vec a$ for some num­ber. Three points $A, B, C$ lie on one line when $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are collinear.

Sec­tion for­mula#

This is the same sec­tion for­mula you know from coor­di­nate geom­e­try, now in vec­tor lan­guage. The point divid­ing $\overrightarrow{AB}$ (posi­tion vec­tors $\vec a$, $\vec b$) inter­nally in $m : n$ has posi­tion vec­tor $\displaystyle \frac{m\vec b + n\vec a}{m + n}$; and exter­nally, it is $\displaystyle \frac{m\vec b - n\vec a}{m - n}$. The mid­point is sim­ply $\displaystyle \tfrac{\vec a + \vec b}{2}$.

Prac­tice#

1. Find $\lvert\vec a\rvert$ and $\hat a$ for $\vec a = 6\hat i - 2\hat j + 3\hat k$.
2. Find the direc­tion cosines of the vec­tor join­ing $(2, -1, 4)$ to $(5, 3, 4)$.
3. Find $\vec a + \vec b$ and $3\vec a - 2\vec b$ for $\vec a = \hat i - 3\hat j + 2\hat k$, $\vec b = -2\hat i + \hat j + 4\hat k$.
4. Find a vec­tor of mag­ni­tude $7$ in the direc­tion of $2\hat i - 3\hat j + 6\hat k$.
5. Find $x, y$ if $x\hat i + 3\hat j$ and $4\hat i + y\hat j$ are equal.
6. Show that $2\hat i - 3\hat j + 4\hat k$ and $-4\hat i + 6\hat j - 8\hat k$ are collinear.
7. Show that $A(2, 6, 3)$, $B(1, 2, 7)$, $C(3, 10, -1)$ are collinear.
8. Find the point divid­ing the join of $(1, -2, 3)$ and $(3, 4, -5)$ inter­nally in the ratio $2 : 3$, and exter­nally in $2 : 3$.
9. In tri­an­gle $ABC$, show that $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \vec 0$.
10. Can a vec­tor have direc­tion angles $45^\circ, 60^\circ, 120^\circ$?

Answers#