Scalar and Vector Products, with Mixed Practice — Worked Solutions

Ques­tion 1: Angle between two vec­tors#

The prob­lem#

Find the angle between $\hat i - \hat j$ and $\hat j - \hat k$.

Under­stand­ing the prob­lem#

Let $\vec a = \hat i - \hat j = (1, -1, 0)$ and $\vec b = \hat j - \hat k = (0, 1, -1)$. You need the angle $\theta$ between them, with $0 \le \theta \le \pi$.

The idea#

From $\vec a \cdot \vec b = \lvert\vec a\rvert\lvert\vec b\rvert\cos\theta$,

$$\displaystyle \cos\theta = \frac{\vec a \cdot \vec b}{\lvert\vec a\rvert\lvert\vec b\rvert}.$$

Step-by-step solu­tion#

Step 1. Com­pute the dot prod­uct com­po­nent by com­po­nent.

$$\vec a \cdot \vec b = (1)(0) + (-1)(1) + (0)(-1) = -1$$

Step 2. Com­pute the mag­ni­tudes.

$$\lvert\vec a\rvert = \sqrt{1 + 1 + 0} = \sqrt{2}, \qquad \lvert\vec b\rvert = \sqrt{0 + 1 + 1} = \sqrt{2}$$

Step 3. Find $\cos\theta$.

$$\displaystyle \cos\theta = \frac{-1}{\sqrt{2}\cdot\sqrt{2}} = -\frac{1}{2}$$

Step 4. The angle in $[0, \pi]$ with cosine $\displaystyle -\tfrac{1}{2}$ is $\displaystyle \tfrac{2\pi}{3}$.

$$\theta = 120^\circ$$

Check­ing the answer#

A neg­a­tive dot prod­uct means the angle is obtuse, and $120^\circ$ is obtuse ✓.

Answer#

$\theta = 120^\circ$ (that is, $\displaystyle \tfrac{2\pi}{3}$).

Com­mon mis­take to avoid#

Tak­ing $\displaystyle \cos\theta = \tfrac{1}{2}$ and answer­ing $60^\circ$ by drop­ping the minus sign.

Ques­tion 2: Pro­jec­tion of one vec­tor on another#

The prob­lem#

Find the pro­jec­tion of $3\hat i - \hat j + 4\hat k$ on $2\hat i + 3\hat j - 6\hat k$.

Under­stand­ing the prob­lem#

Let $\vec a = (3, -1, 4)$ and $\vec b = (2, 3, -6)$. The pro­jec­tion of $\vec a$ on $\vec b$ is the signed length of the "shadow" of $\vec a$ along the direc­tion of $\vec b$.

The idea#

Use the lesson's for­mula: pro­jec­tion of $\vec a$ on $\vec b$ $\displaystyle = \frac{\vec a \cdot \vec b}{\lvert\vec b\rvert}$. Note that we divide by the length of $\vec b$, the vec­tor we project onto.

Step-by-step solu­tion#

Step 1. Dot prod­uct.

$$\vec a \cdot \vec b = (3)(2) + (-1)(3) + (4)(-6) = 6 - 3 - 24 = -21$$

Step 2. Length of $\vec b$.

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

Step 3. Divide.

$$\displaystyle \text{projection} = \frac{-21}{7} = -3$$

Check­ing the answer#

The neg­a­tive sign means $\vec a$ makes an obtuse angle with $\vec b$, which agrees with the neg­a­tive dot prod­uct. Its size, $3$, is less than $\lvert\vec a\rvert = \sqrt{26} \approx 5.1$, as a pro­jec­tion's size must be.

Answer#

The pro­jec­tion is $-3$.

Ques­tion 3: Show­ing three vec­tors form a right-angled tri­an­gle#

The prob­lem#

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.

Under­stand­ing the prob­lem#

What must be shown: (i) the three vec­tors can be the sides of a tri­an­gle, and (ii) two of the sides are per­pen­dic­u­lar.

The idea#

Three vec­tors form a tri­an­gle when one of them is the sum of the other two (they close up head-to-tail). A right angle means two sides have dot prod­uct $0$.

Step-by-step solu­tion#

Step 1. Name the vec­tors: $\vec a = 2\hat i - \hat j + \hat k$, $\vec b = \hat i - 3\hat j - 5\hat k$, $\vec c = 3\hat i - 4\hat j - 4\hat k$.

Step 2. Add $\vec a$ and $\vec b$.

$$\vec a + \vec b = (2 + 1)\hat i + (-1 - 3)\hat j + (1 - 5)\hat k = 3\hat i - 4\hat j - 4\hat k = \vec c$$

Since $\vec c = \vec a + \vec b$, plac­ing $\vec a$ and then $\vec b$ head to tail, $\vec c$ joins the start to the end: the three vec­tors form a tri­an­gle.

Step 3. Test for a right angle between $\vec a$ and $\vec b$.

$$\vec a \cdot \vec b = (2)(1) + (-1)(-3) + (1)(-5) = 2 + 3 - 5 = 0$$

Step 4. A zero dot prod­uct between non-zero vec­tors means they are per­pen­dic­u­lar. So the angle between the sides $\vec a$ and $\vec b$ is $90^\circ$.

Check­ing the answer#

Pythago­ras: $\lvert\vec a\rvert^2 = 6$, $\lvert\vec b\rvert^2 = 35$, $\lvert\vec c\rvert^2 = 41$, and $6 + 35 = 41$ ✓.

Answer#

$\vec a + \vec b = \vec c$, so the vec­tors form a tri­an­gle, and $\vec a \cdot \vec b = 0$, so it is right-angled (the right angle is between $\vec a$ and $\vec b$).

Ques­tion 4: Angle and length from mag­ni­tudes and a dot prod­uct#

The prob­lem#

If $\lvert\vec a\rvert = 2$, $\lvert\vec b\rvert = 3$ and $\vec a \cdot \vec b = 3$, find the angle between them and $\lvert\vec a - \vec b\rvert$.

Under­stand­ing the prob­lem#

You do not have com­po­nents, only lengths and the dot prod­uct. You need the angle $\theta$ and the length of $\vec a - \vec b$.

The idea#

For the angle, use $\displaystyle \cos\theta = \frac{\vec a \cdot \vec b}{\lvert\vec a\rvert\lvert\vec b\rvert}$. For the length, expand $\lvert\vec a - \vec b\rvert^2 = (\vec a - \vec b)\cdot(\vec a - \vec b)$, just like Exam­ple 3 but with a minus sign.

Step-by-step solu­tion#

Step 1. Find $\cos\theta$.

$$\displaystyle \cos\theta = \frac{3}{2 \times 3} = \frac{1}{2} \;\Longrightarrow\; \theta = 60^\circ$$

Step 2. Expand the square of the length, using $\vec a\cdot\vec a = \lvert\vec a\rvert^2$ and $\vec a\cdot\vec b = \vec b\cdot\vec a$.

$$\lvert\vec a - \vec b\rvert^2 = \lvert\vec a\rvert^2 - 2\,\vec a\cdot\vec b + \lvert\vec b\rvert^2$$

Step 3. Sub­sti­tute.

$$\lvert\vec a - \vec b\rvert^2 = 4 - 2(3) + 9 = 7$$

Step 4. Take the square root.

$$\lvert\vec a - \vec b\rvert = \sqrt{7}$$

Check­ing the answer#

By the cosine rule for the tri­an­gle with sides $2$ and $3$ and included angle $60^\circ$: third side$^2 = 4 + 9 - 2(2)(3)\cos 60^\circ = 13 - 6 = 7$ ✓.

Answer#

The angle is $60^\circ$ and $\lvert\vec a - \vec b\rvert = \sqrt{7}$.

Ques­tion 5: Three vec­tors adding to zero#

The prob­lem#

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 $\vec b$.

Under­stand­ing the prob­lem#

$\vec a + \vec b + \vec c = \vec 0$ means $\vec c = -(\vec a + \vec b)$, so $\lvert\vec c\rvert = \lvert\vec a + \vec b\rvert$. You know all three lengths and want the angle between $\vec a$ and $\vec b$.

The idea#

Square $\lvert\vec a + \vec b\rvert$ and expand, as in Exam­ple 3; this brings in $\vec a\cdot\vec b = \lvert\vec a\rvert\lvert\vec b\rvert\cos\theta$, which you can solve for $\cos\theta$.

Step-by-step solu­tion#

Step 1. Rearrange and take lengths.

$$\vec a + \vec b = -\vec c \;\Longrightarrow\; \lvert\vec a + \vec b\rvert = \lvert\vec c\rvert = 7$$

Step 2. Square and expand.

$$\lvert\vec a + \vec b\rvert^2 = \lvert\vec a\rvert^2 + 2\lvert\vec a\rvert\lvert\vec b\rvert\cos\theta + \lvert\vec b\rvert^2$$

Step 3. Sub­sti­tute the lengths.

$$49 = 9 + 2(3)(5)\cos\theta + 25 = 34 + 30\cos\theta$$

Step 4. Solve for $\cos\theta$.

$$\displaystyle 30\cos\theta = 15 \;\Longrightarrow\; \cos\theta = \frac{1}{2} \;\Longrightarrow\; \theta = 60^\circ$$

Check­ing the answer#

The angle is acute and $\lvert\vec a + \vec b\rvert = 7$ is more than $\sqrt{3^2 + 5^2} = \sqrt{34} \approx 5.83$ (the value at $90^\circ$), which is what an acute angle gives ✓.

Answer#

The angle between $\vec a$ and $\vec b$ is $60^\circ$.

Com­mon mis­take to avoid#

Think­ing the answer is the tri­an­gle's inte­rior angle. The tri­an­gle formed has an inte­rior angle of $120^\circ$ between the sides of length $3$ and $5$, but the angle between the vec­tors $\vec a$ and $\vec b$ (placed tail to tail) is $60^\circ$.

Ques­tion 6: Cross prod­uct and a per­pen­dic­u­lar unit vec­tor#

The prob­lem#

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.

Under­stand­ing the prob­lem#

You need the vec­tor $\vec a \times \vec b$, which is per­pen­dic­u­lar to both $\vec a$ and $\vec b$, and then a vec­tor of length $1$ in that direc­tion.

The idea#

Com­pute the cross prod­uct with the deter­mi­nant, then divide by its length (as in Exam­ple 6).

Step-by-step solu­tion#

Step 1. Set up the deter­mi­nant.

$$\vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & 2 & 2 \\ 1 & 2 & -2 \end{vmatrix}$$

Step 2. Expand along the first row. Remem­ber the minus sign on the $\hat j$ term.

$$\begin{aligned} \vec a \times \vec b &= \hat i\left[(2)(-2) - (2)(2)\right] - \hat j\left[(3)(-2) - (2)(1)\right] + \hat k\left[(3)(2) - (2)(1)\right] \\ &= \hat i(-4 - 4) - \hat j(-6 - 2) + \hat k(6 - 2) \\ &= -8\hat i + 8\hat j + 4\hat k \end{aligned}$$

Step 3. Find its length.

$$\lvert\vec a \times \vec b\rvert = \sqrt{64 + 64 + 16} = \sqrt{144} = 12$$

Step 4. Divide to get a unit vec­tor.

$$\displaystyle \hat n = \frac{-8\hat i + 8\hat j + 4\hat k}{12} = \frac{1}{3}\left(-2\hat i + 2\hat j + \hat k\right)$$

Check­ing the answer#

Per­pen­dic­u­lar­ity: $(-8)(3) + (8)(2) + (4)(2) = -24 + 16 + 8 = 0$ and $(-8)(1) + (8)(2) + (4)(-2) = -8 + 16 - 8 = 0$ ✓. Length of $\hat n$: $\displaystyle \tfrac{1}{3}\sqrt{4 + 4 + 1} = 1$ ✓.

Answer#

$\vec a \times \vec b = -8\hat i + 8\hat j + 4\hat k$; a unit vec­tor per­pen­dic­u­lar to both is $\displaystyle \tfrac{1}{3}(-2\hat i + 2\hat j + \hat k)$ (its neg­a­tive also works).

Ques­tion 7: Area of a par­al­lel­o­gram#

The prob­lem#

Find the area of the par­al­lel­o­gram with adja­cent sides $\hat i + 2\hat j + 3\hat k$ and $3\hat i - 2\hat j + \hat k$.

Under­stand­ing the prob­lem#

The two given vec­tors are adja­cent sides of the par­al­lel­o­gram. You need its area.

The idea#

The area of the par­al­lel­o­gram on $\vec a$ and $\vec b$ is $\lvert\vec a \times \vec b\rvert$.

Step-by-step solu­tion#

Step 1. Com­pute the cross prod­uct.

$$\begin{aligned} \vec a \times \vec b &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 2 & 3 \\ 3 & -2 & 1 \end{vmatrix} \\ &= \hat i\left[(2)(1) - (3)(-2)\right] - \hat j\left[(1)(1) - (3)(3)\right] + \hat k\left[(1)(-2) - (2)(3)\right] \\ &= 8\hat i + 8\hat j - 8\hat k \end{aligned}$$

Step 2. Find its length.

$$\lvert\vec a \times \vec b\rvert = \sqrt{64 + 64 + 64} = \sqrt{192} = 8\sqrt{3}$$

Check­ing the answer#

$\vec a\cdot\vec b = 3 - 4 + 3 = 2$, so $\displaystyle \sin^2\theta = 1 - \tfrac{4}{14 \times 14} = \tfrac{192}{196}$, and $\displaystyle \lvert\vec a\rvert\lvert\vec b\rvert\sin\theta = 14 \cdot \tfrac{\sqrt{192}}{14} = \sqrt{192}$ ✓.

Answer#

Area $= 8\sqrt{3}$ square units.

Ques­tion 8: Area of a tri­an­gle from its ver­tices#

The prob­lem#

Find the area of the tri­an­gle with ver­tices $(1, 2, 3)$, $(2, -1, 1)$, $(1, 2, -4)$.

Under­stand­ing the prob­lem#

Call the ver­tices $A(1, 2, 3)$, $B(2, -1, 1)$, $C(1, 2, -4)$. You need the area of tri­an­gle $ABC$ in space.

The idea#

Form two side vec­tors from the same ver­tex, $\overrightarrow{AB}$ and $\overrightarrow{AC}$. The tri­an­gle's area is half the par­al­lel­o­gram's: $\displaystyle \tfrac{1}{2}\lvert\overrightarrow{AB} \times \overrightarrow{AC}\rvert$ (Exam­ple 5).

Step-by-step solu­tion#

Step 1. Side vec­tors (end minus start).

$$\overrightarrow{AB} = (2 - 1, -1 - 2, 1 - 3) = \hat i - 3\hat j - 2\hat k$$

$$\overrightarrow{AC} = (1 - 1, 2 - 2, -4 - 3) = -7\hat k$$

Step 2. Cross prod­uct.

$$\begin{aligned} \overrightarrow{AB} \times \overrightarrow{AC} &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -3 & -2 \\ 0 & 0 & -7 \end{vmatrix} \\ &= \hat i\left[(-3)(-7) - (-2)(0)\right] - \hat j\left[(1)(-7) - (-2)(0)\right] + \hat k\left[(1)(0) - (-3)(0)\right] \\ &= 21\hat i + 7\hat j + 0\hat k \end{aligned}$$

Step 3. Length.

$$\lvert 21\hat i + 7\hat j\rvert = \sqrt{441 + 49} = \sqrt{490} = 7\sqrt{10}$$

Step 4. Halve it.

$$\displaystyle \text{Area} = \frac{7\sqrt{10}}{2}$$

Check­ing the answer#

$\overrightarrow{AC}$ points straight along $-\hat k$ with length $7$. The dis­tance from $B$ to the line $AC$ (a ver­ti­cal line through $(1, 2)$) is the hor­i­zon­tal dis­tance $\sqrt{1^2 + 3^2} = \sqrt{10}$. So area $\displaystyle = \tfrac{1}{2} \times 7 \times \sqrt{10}$ ✓.

Answer#

Area $\displaystyle = \frac{7\sqrt{10}}{2}$ square units.

Ques­tion 9: A cross-prod­uct iden­tity#

The prob­lem#

Show that $(\vec a - \vec b) \times (\vec a + \vec b) = 2(\vec a \times \vec b)$.

Under­stand­ing the prob­lem#

What must be shown: expand­ing the left side gives $2(\vec a \times \vec b)$, for any vec­tors $\vec a$ and $\vec b$.

The idea#

The cross prod­uct dis­trib­utes over addi­tion, so expand like ordi­nary brack­ets, keep­ing the order of each prod­uct. Then use $\vec a \times \vec a = \vec 0$, $\vec b \times \vec b = \vec 0$ and $\vec b \times \vec a = -\vec a \times \vec b$.

Step-by-step solu­tion#

Step 1. Expand using the dis­trib­u­tive law.

$$(\vec a - \vec b) \times (\vec a + \vec b) = \vec a \times \vec a + \vec a \times \vec b - \vec b \times \vec a - \vec b \times \vec b$$

Step 2. A vec­tor crossed with itself is zero.

$$\vec a \times \vec a = \vec 0, \qquad \vec b \times \vec b = \vec 0$$

So the expres­sion becomes $\vec a \times \vec b - \vec b \times \vec a$.

Step 3. Reverse the order in the sec­ond term, which changes its sign: $\vec b \times \vec a = -\vec a \times \vec b$.

$$\vec a \times \vec b - (-\vec a \times \vec b) = \vec a \times \vec b + \vec a \times \vec b = 2(\vec a \times \vec b)$$

This is the right side, so the iden­tity holds.

Check­ing the answer#

Test with $\vec a = \hat i$, $\vec b = \hat j$: left side $(\hat i - \hat j) \times (\hat i + \hat j) = \hat i \times \hat j - \hat j \times \hat i = \hat k + \hat k = 2\hat k$; right side $2(\hat i \times \hat j) = 2\hat k$ ✓.

Answer#

$(\vec a - \vec b) \times (\vec a + \vec b) = \vec a \times \vec b - \vec b \times \vec a = 2(\vec a \times \vec b)$, as required.

Com­mon mis­take to avoid#

Treat­ing the cross prod­uct as com­mu­ta­tive and can­celling $\vec a \times \vec b$ against $\vec b \times \vec a$ to get $\vec 0$. They are neg­a­tives of each other, so they add, not can­cel.

Ques­tion 10: Zero cross prod­uct means par­al­lel#

The prob­lem#

Find $\lambda$ if $(2\hat i + 6\hat j + 27\hat k) \times (\hat i + \lambda\hat j + \mu\hat k) = \vec 0$.

Under­stand­ing the prob­lem#

The cross prod­uct of two non-zero vec­tors is $\vec 0$ exactly when they are par­al­lel. So $\hat i + \lambda\hat j + \mu\hat k$ must be a mul­ti­ple of $2\hat i + 6\hat j + 27\hat k$. You need $\lambda$ (and $\mu$ comes out too).

The idea#

Par­al­lel vec­tors have pro­por­tional com­po­nents:

$$\displaystyle \frac{1}{2} = \frac{\lambda}{6} = \frac{\mu}{27}.$$

Step-by-step solu­tion#

Step 1. State the con­di­tion: $\vec a \times \vec b = \vec 0$ with both vec­tors non-zero, so $\vec a \parallel \vec b$.

Step 2. Com­pare the $\hat i$ com­po­nents to find the ratio: $\vec b = k\vec a$ with $1 = 2k$, so $\displaystyle k = \tfrac{1}{2}$.

Step 3. Use the ratio on the other com­po­nents.

$$\displaystyle \lambda = 6k = 6 \times \frac{1}{2} = 3, \qquad \mu = 27k = \frac{27}{2}$$

Step 4. (Con­firm with the deter­mi­nant.) Set­ting each com­po­nent of the cross prod­uct to zero gives

$$6\mu - 27\lambda = 0, \qquad 27 - 2\mu = 0, \qquad 2\lambda - 6 = 0,$$

so $\lambda = 3$ and $\displaystyle \mu = \tfrac{27}{2}$, and the first equa­tion checks: $81 - 81 = 0$.

Check­ing the answer#

$\displaystyle \hat i + 3\hat j + \tfrac{27}{2}\hat k = \tfrac{1}{2}(2\hat i + 6\hat j + 27\hat k)$, a mul­ti­ple of the first vec­tor, so their cross prod­uct is $\vec 0$ ✓.

Answer#

$\lambda = 3$ (and $\displaystyle \mu = \tfrac{27}{2}$).