Vectors, Components and the Section Formula — Worked Solutions

Ques­tion 1: Mag­ni­tude and unit vec­tor#

The prob­lem#

Find $\lvert\vec a\rvert$ and $\hat a$ for $\vec a = 6\hat i - 2\hat j + 3\hat k$.

Under­stand­ing the prob­lem#

$\lvert\vec a\rvert$ is the length of the vec­tor. $\hat a$ is the unit vec­tor in the same direc­tion: length $1$, same direc­tion as $\vec a$.

The idea#

Use $\lvert x\hat i + y\hat j + z\hat k \rvert = \sqrt{x^2 + y^2 + z^2}$ and $\displaystyle \hat a = \frac{\vec a}{\lvert\vec a\rvert}$.

Step-by-step solu­tion#

Step 1. Mag­ni­tude.

$$\lvert\vec a\rvert = \sqrt{6^2 + (-2)^2 + 3^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7$$

Step 2. Unit vec­tor: divide each com­po­nent by $7$.

$$\displaystyle \hat a = \frac{1}{7}(6\hat i - 2\hat j + 3\hat k) = \frac67\hat i - \frac27\hat j + \frac37\hat k$$

Check­ing the answer#

$\displaystyle \lvert\hat a\rvert^2 = \tfrac{36 + 4 + 9}{49} = 1$, so $\hat a$ really has length $1$. ✓

Answer#

$\lvert\vec a\rvert = 7$; $\displaystyle \hat a = \frac17(6\hat i - 2\hat j + 3\hat k)$.

Ques­tion 2: Direc­tion cosines of a join­ing vec­tor#

The prob­lem#

Find the direc­tion cosines of the vec­tor join­ing $(2, -1, 4)$ to $(5, 3, 4)$.

Under­stand­ing the prob­lem#

The vec­tor goes from $P(2, -1, 4)$ to $Q(5, 3, 4)$. Its direc­tion cosines are the cosines of the angles it makes with the $x$, $y$ and $z$ axes, which equal its com­po­nents divided by its length.

The idea#

$\overrightarrow{PQ} =$ (posi­tion vec­tor of $Q$) $-$ (posi­tion vec­tor of $P$). Then the direc­tion cosines are $\displaystyle \tfrac{x}{r}, \tfrac{y}{r}, \tfrac{z}{r}$.

Step-by-step solu­tion#

Step 1. Sub­tract coor­di­nates, end minus start.

$$\overrightarrow{PQ} = (5 - 2)\hat i + (3 - (-1))\hat j + (4 - 4)\hat k = 3\hat i + 4\hat j + 0\hat k$$

Step 2. Length.

$$\lvert\overrightarrow{PQ}\rvert = \sqrt{9 + 16 + 0} = 5$$

Step 3. Divide each com­po­nent by $5$.

$$\displaystyle l = \frac35, \quad m = \frac45, \quad n = 0$$

Check­ing the answer#

$\displaystyle l^2 + m^2 + n^2 = \tfrac{9}{25} + \tfrac{16}{25} + 0 = 1$. ✓ And $n = 0$ makes sense: the $z$-coor­di­nates are equal, so the vec­tor is per­pen­dic­u­lar to the $z$-axis.

Answer#

$\displaystyle \frac35, \frac45, 0$

Ques­tion 3: Sum and a com­bi­na­tion of two vec­tors#

The prob­lem#

Find $\vec a + \vec b$ and $3\vec a - 2\vec b$ for $\vec a = \hat i - 3\hat j + 2\hat k$, $\vec b = -2\hat i + \hat j + 4\hat k$.

Under­stand­ing the prob­lem#

Vec­tors in com­po­nent form are added and mul­ti­plied by num­bers com­po­nent by com­po­nent.

The idea#

Work sep­a­rately with the $\hat i$, $\hat j$ and $\hat k$ parts.

Step-by-step solu­tion#

Part (a): $\vec a + \vec b$

Step 1. Add match­ing com­po­nents.

$$\vec a + \vec b = (1 - 2)\hat i + (-3 + 1)\hat j + (2 + 4)\hat k = -\hat i - 2\hat j + 6\hat k$$

Part (b): $3\vec a - 2\vec b$

Step 1. Scale each vec­tor.

$$3\vec a = 3\hat i - 9\hat j + 6\hat k, \qquad 2\vec b = -4\hat i + 2\hat j + 8\hat k$$

Step 2. Sub­tract match­ing com­po­nents.

$$3\vec a - 2\vec b = (3 + 4)\hat i + (-9 - 2)\hat j + (6 - 8)\hat k = 7\hat i - 11\hat j - 2\hat k$$

Check­ing the answer#

Check the $\hat k$ part of (b) another way: $3(2) - 2(4) = 6 - 8 = -2$. ✓

Answer#

$\vec a + \vec b = -\hat i - 2\hat j + 6\hat k$; $3\vec a - 2\vec b = 7\hat i - 11\hat j - 2\hat k$.

Ques­tion 4: A vec­tor of given length in a given direc­tion#

The prob­lem#

Find a vec­tor of mag­ni­tude $7$ in the direc­tion of $2\hat i - 3\hat j + 6\hat k$.

Under­stand­ing the prob­lem#

You need a vec­tor point­ing the same way as $\vec v = 2\hat i - 3\hat j + 6\hat k$ but with length exactly $7$.

The idea#

Take the unit vec­tor $\hat v$ (length 1, same direc­tion) and mul­ti­ply by $7$.

Step-by-step solu­tion#

Step 1. Length of $\vec v$.

$$\lvert\vec v\rvert = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$$

Step 2. Required vec­tor.

$$\displaystyle 7\hat v = 7 \cdot \frac{2\hat i - 3\hat j + 6\hat k}{7} = 2\hat i - 3\hat j + 6\hat k$$

The given vec­tor already has length $7$, so it is itself the answer.

Check­ing the answer#

Its length is $7$ (Step 1) and it points along $\vec v$. ✓

Answer#

$2\hat i - 3\hat j + 6\hat k$ (the given vec­tor already has mag­ni­tude $7$).

Ques­tion 5: Equal vec­tors#

The prob­lem#

Find $x, y$ if $x\hat i + 3\hat j$ and $4\hat i + y\hat j$ are equal.

Under­stand­ing the prob­lem#

Two vec­tors are equal exactly when their cor­re­spond­ing com­po­nents are equal.

The idea#

Com­pare the $\hat i$ parts and the $\hat j$ parts.

Step-by-step solu­tion#

Step 1. $\hat i$ com­po­nents: $x = 4$.

Step 2. $\hat j$ com­po­nents: $3 = y$.

Check­ing the answer#

With $x = 4$, $y = 3$ both vec­tors are $4\hat i + 3\hat j$. ✓

Answer#

$x = 4$, $y = 3$

Ques­tion 6: Two collinear vec­tors#

The prob­lem#

Show that $2\hat i - 3\hat j + 4\hat k$ and $-4\hat i + 6\hat j - 8\hat k$ are collinear.

Under­stand­ing the prob­lem#

Two vec­tors are collinear (par­al­lel) when one is a scalar mul­ti­ple of the other: $\vec b = \lambda\vec a$ for some num­ber $\lambda$.

The idea#

Divide each com­po­nent of $\vec b$ by the match­ing com­po­nent of $\vec a$. If all three ratios are the same, that com­mon ratio is $\lambda$.

Step-by-step solu­tion#

Step 1. Ratios of com­po­nents.

$$\displaystyle \frac{-4}{2} = -2, \qquad \frac{6}{-3} = -2, \qquad \frac{-8}{4} = -2$$

Step 2. All three ratios equal $-2$, so

$$-4\hat i + 6\hat j - 8\hat k = -2\,(2\hat i - 3\hat j + 4\hat k)$$

and the vec­tors are collinear (point­ing in oppo­site direc­tions, since $\lambda < 0$).

Check­ing the answer#

$-2 \times 2 = -4$, $-2 \times (-3) = 6$, $-2 \times 4 = -8$. ✓

Answer#

The sec­ond vec­tor is $-2$ times the first, so they are collinear.

Ques­tion 7: Three collinear points#

The prob­lem#

Show that $A(2, 6, 3)$, $B(1, 2, 7)$, $C(3, 10, -1)$ are collinear.

Under­stand­ing the prob­lem#

Three points lie on one line when $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are collinear vec­tors. Because both start at $A$, being par­al­lel forces them onto the same line.

The idea#

Find $\overrightarrow{AB}$ and $\overrightarrow{AC}$ and show one is a mul­ti­ple of the other.

Step-by-step solu­tion#

Step 1. $\overrightarrow{AB}$ = posi­tion vec­tor of $B$ minus that of $A$.

$$\overrightarrow{AB} = (1 - 2)\hat i + (2 - 6)\hat j + (7 - 3)\hat k = -\hat i - 4\hat j + 4\hat k$$

Step 2. $\overrightarrow{AC}$.

$$\overrightarrow{AC} = (3 - 2)\hat i + (10 - 6)\hat j + (-1 - 3)\hat k = \hat i + 4\hat j - 4\hat k$$

Step 3. Com­pare: $\overrightarrow{AC} = -\overrightarrow{AB}$. The two vec­tors are par­al­lel and share the point $A$, so $A$, $B$, $C$ lie on one line.

Check­ing the answer#

$\overrightarrow{AC} = -\overrightarrow{AB}$ also says $A$ is the mid­point of $BC$: $\displaystyle \tfrac{B + C}{2} = \left(\tfrac{4}{2}, \tfrac{12}{2}, \tfrac{6}{2}\right) = (2, 6, 3) = A$. ✓

Answer#

$\overrightarrow{AC} = -\overrightarrow{AB}$, so $A$, $B$, $C$ are collinear ($A$ is in fact the mid­point of $BC$).

Ques­tion 8: Sec­tion for­mula, inter­nal and exter­nal#

The prob­lem#

Find the point divid­ing the join of $(1, -2, 3)$ and $(3, 4, -5)$ inter­nally in the ratio $2 : 3$, and exter­nally in $2 : 3$.

Under­stand­ing the prob­lem#

Let $A(1, -2, 3)$ with posi­tion vec­tor $\vec a$, and $B(3, 4, -5)$ with posi­tion vec­tor $\vec b$. The ratio $m : n = 2 : 3$ is mea­sured from $A$ towards $B$. Inter­nal divi­sion gives a point between $A$ and $B$; exter­nal divi­sion gives a point on the line $AB$ out­side the seg­ment.

The idea#

Inter­nal: $\displaystyle \frac{m\vec b + n\vec a}{m + n}$. Exter­nal: $\displaystyle \frac{m\vec b - n\vec a}{m - n}$. Here $m = 2$, $n = 3$.

Step-by-step solu­tion#

Part (a): inter­nal divi­sion

Step 1. Com­pute $2\vec b + 3\vec a$ com­po­nent by com­po­nent.

$$2\vec b + 3\vec a = (6 + 3)\hat i + (8 - 6)\hat j + (-10 + 9)\hat k = 9\hat i + 2\hat j - \hat k$$

Step 2. Divide by $m + n = 5$.

$$\displaystyle \frac{9\hat i + 2\hat j - \hat k}{5} \;\Rightarrow\; \left(\frac95, \frac25, -\frac15\right)$$

Part (b): exter­nal divi­sion

Step 1. Com­pute $2\vec b - 3\vec a$.

$$2\vec b - 3\vec a = (6 - 3)\hat i + (8 + 6)\hat j + (-10 - 9)\hat k = 3\hat i + 14\hat j - 19\hat k$$

Step 2. Divide by $m - n = -1$.

$$\displaystyle \frac{3\hat i + 14\hat j - 19\hat k}{-1} = -3\hat i - 14\hat j + 19\hat k \;\Rightarrow\; (-3, -14, 19)$$

Check­ing the answer#

Inter­nal point $P$: $\displaystyle \overrightarrow{AP} = \left(\tfrac45, \tfrac{12}{5}, -\tfrac{16}{5}\right)$ and $\overrightarrow{AB} = (2, 6, -8)$, so $\displaystyle \overrightarrow{AP} = \tfrac25\overrightarrow{AB}$ — exactly $2$ parts out of $5$. ✓
Exter­nal point $Q$: $\overrightarrow{AQ} = (-4, -12, 16) = -2\overrightarrow{AB}$ and $\overrightarrow{BQ} = (-6, -18, 24) = -3\overrightarrow{AB}$, so $AQ : BQ = 2 : 3$. ✓

Answer#

Inter­nally: $\displaystyle \left(\tfrac95, \tfrac25, -\tfrac15\right)$; exter­nally: $(-3, -14, 19)$.

Com­mon mis­take to avoid#

In the exter­nal for­mula the denom­i­na­tor is $m - n = -1$; for­get­ting it flips every sign of the answer.

Ques­tion 9: Going round a tri­an­gle gives the zero vec­tor#

The prob­lem#

In tri­an­gle $ABC$, show that $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \vec 0$.

Under­stand­ing the prob­lem#

This is a proof. Trav­el­ling $A \to B \to C \to A$ brings you back where you started, so the total dis­place­ment should be zero. You must show this with the laws of vec­tor addi­tion.

The idea#

Use the tri­an­gle law $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$, and the fact that $\overrightarrow{CA}$ is the neg­a­tive of $\overrightarrow{AC}$.

Step-by-step solu­tion#

Step 1. By the tri­an­gle law, the first two vec­tors com­bine.

$$\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$$

Step 2. So the sum becomes

$$\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \overrightarrow{AC} + \overrightarrow{CA}$$

Step 3. $\overrightarrow{CA}$ has the same length as $\overrightarrow{AC}$ but the oppo­site direc­tion, so $\overrightarrow{CA} = -\overrightarrow{AC}$.

$$\overrightarrow{AC} + \overrightarrow{CA} = \overrightarrow{AC} - \overrightarrow{AC} = \vec 0$$

Check­ing the answer#

With posi­tion vec­tors: $(\vec b - \vec a) + (\vec c - \vec b) + (\vec a - \vec c) = \vec 0$. ✓

Answer#

$\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \overrightarrow{AC} + \overrightarrow{CA} = \vec 0$.

Ques­tion 10: Pos­si­ble direc­tion angles?#

The prob­lem#

Can a vec­tor have direc­tion angles $45^\circ, 60^\circ, 120^\circ$?

Under­stand­ing the prob­lem#

Direc­tion angles $\alpha, \beta, \gamma$ are the angles a vec­tor makes with the $x$, $y$, $z$ axes. They can­not be cho­sen freely: their cosines must sat­isfy $l^2 + m^2 + n^2 = 1$.

The idea#

Com­pute $\cos^2 45^\circ + \cos^2 60^\circ + \cos^2 120^\circ$ and see whether it equals $1$.

Step-by-step solu­tion#

Step 1. Cosines.

$$\displaystyle \cos 45^\circ = \frac{1}{\sqrt2}, \quad \cos 60^\circ = \frac12, \quad \cos 120^\circ = -\frac12$$

Step 2. Sum of squares.

$$\displaystyle \frac12 + \frac14 + \frac14 = 1$$

Step 3. The con­di­tion holds, so such a vec­tor exists — for exam­ple, the unit vec­tor $\displaystyle \tfrac{1}{\sqrt2}\hat i + \tfrac12\hat j - \tfrac12\hat k$.

Check­ing the answer#

$\displaystyle \left\lvert \tfrac{1}{\sqrt2}\hat i + \tfrac12\hat j - \tfrac12\hat k \right\rvert = \sqrt{\tfrac12 + \tfrac14 + \tfrac14} = 1$, and its com­po­nents are exactly the three cosines. ✓

Answer#

Yes: $\cos^2 45^\circ + \cos^2 60^\circ + \cos^2 120^\circ = 1$, e.g. $\displaystyle \tfrac{1}{\sqrt2}\hat i + \tfrac12\hat j - \tfrac12\hat k$.

Com­mon mis­take to avoid#

The squares remove the sign, so the neg­a­tive cosine of $120^\circ$ is not a prob­lem; do not reject the angles just because one cosine is neg­a­tive.