Vector space of real numbers

Unit 7 Vector space of real numbers

Key unit competence: Apply operations on vectors in three dimensions in solving mathematical and real-life problems.

Key ideas

• ${\mathbb{R}}^3=\{(x,y,z)\}$ is a vector space of dimension 3 with standard basis $\vec{i}=(1,0,0)$, $\vec{j}=(0,1,0)$, $\vec{k}=(0,0,1)$. Three vectors form a basis iff their determinant is non-zero.
• Magnitude $|\vec{u}|=\sqrt{x^2+y^2+z^2}$. Dot product $\vec{u}\cdot \vec{v}=x_1{x}_2+y_1{y}_2+z_1{z}_2=|\vec{u}||\vec{v}|\cos \theta$; orthogonal iff $\vec{u}\cdot \vec{v}=0$.
• Cross (vector) product $\vec{u}\times \vec{v}=\left|\begin{array}{ccc}\vec{i}& \vec{j}& \vec{k}\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|$: a vector perpendicular to both, with $|\vec{u}\times \vec{v}|=|\vec{u}||\vec{v}|\sin \theta$ (area of the parallelogram). $\vec{v}\times \vec{u}=-\vec{u}\times \vec{v}$.
• Mixed (triple) product $[\vec{u},\vec{v},\vec{w}]=\vec{u}\cdot (\vec{v}\times \vec{w})=\det (\vec{u},\vec{v},\vec{w})$: volume of the parallelepiped; zero iff the vectors are coplanar.
• Applications: area of triangle $=\frac{1}{2}|\overrightarrow{AB}\times \overrightarrow{AC}|$; volume of tetrahedron $=\frac{1}{6}|[\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD}]|$; parallel vectors have $\vec{u}\times \vec{v}=\vec{0}$.

Formulas to know

Common misconceptions

• "$\vec{u}\times \vec{v}=\vec{v}\times \vec{u}$." No: $\vec{v}\times \vec{u}=-(\vec{u}\times \vec{v})$.
• "The cross product is a number." No: it is a vector; the dot product is the number.

Worked examples

Practice questions

Easier and harder questions are mixed, just like in a real exam.

1. Compute $\vec{u}\times \vec{v}$ for $\vec{u}=(2,-1,3)$, $\vec{v}=(1,4,-2)$ and verify it is perpendicular to $\vec{u}$.

   Show answer

   $((-1)(-2)-3\cdot 4,3\cdot 1-2(-2),2\cdot 4-(-1)\cdot 1)=(-10,7,9)$. Check: $-20-7+27=0$ ✓.

   Common mistake: Sign error in the middle component.

2. Prove that $|\vec{u}\times \vec{v}{|}^2+(\vec{u}\cdot \vec{v}{)}^2=|\vec{u}{|}^2|\vec{v}{|}^2$.

   Show answer

   $|\vec{u}{|}^2|\vec{v}{|}^2{\sin }^2\theta +|\vec{u}{|}^2|\vec{v}{|}^2{\cos }^2\theta =|\vec{u}{|}^2|\vec{v}{|}^2$.

   Common mistake: Forgetting to square the dot product.

3. Find the angle between $\vec{a}=(2,2,1)$ and $\vec{b}=(1,-2,2)$.

   Show answer

   $\vec{a}\cdot \vec{b}=2-4+2=0$: $90^{\circ }$.

   Common mistake: Forgetting the third component.

4. Find $x$ so that $(x,2,-1)$ and $(3,x,4)$ are orthogonal.

   Show answer

   $3x+2x-4=0$, $x=\frac{4}{5}$.

   Common mistake: Multiplying components crosswise.

5. Show that $(1,1,0)$, $(0,1,1)$, $(1,0,1)$ form a basis of ${\mathbb{R}}^3$.

   Show answer

   $\det =1(1-0)-1(0-1)+0=2\ne 0$: independent, so a basis.

   Common mistake: Checking only that no vector is zero.

6. Find the volume of the parallelepiped with edges $(1,2,0)$, $(0,1,3)$, $(2,0,1)$.

   Show answer

   $\det =1(1-0)-2(0-6)+0=13$: volume 13.

   Common mistake: Using the dot product of two vectors only.

7. Find the area of the parallelogram with sides $\vec{u}=(1,2,2)$ and $\vec{v}=(2,1,-2)$.

   Show answer

   $\vec{u}\times \vec{v}=(-4-2,4+2,1-4)=(-6,6,-3)$; area $=\sqrt{81}=9$.

   Common mistake: Halving the result (that would be a triangle).

8. Are $A(1,0,2)$, $B(2,1,3)$, $C(4,3,5)$ collinear?

   Show answer

   $\overrightarrow{AB}=(1,1,1)$, $\overrightarrow{AC}=(3,3,3)=3\overrightarrow{AB}$: yes.

   Common mistake: Comparing only two coordinates.

Matching and drag-and-drop

1. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. What is the dimension of the subspace $\{(x,y,z):x+y+z=0\}$?	a. $90^{\circ }$
   2. Find the coordinates of $(4,1,3)$ in the basis $\{(1,0,0),(0,1,0),(1,1,1)\}$.	b. $(1,-2,3)$
   3. $|\vec{u}|=2$, $|\vec{v}|=3$ and $\vec{u}\cdot \vec{v}=1$. Find $|\vec{u}+\vec{v}|$.	c. $\sqrt{15}$
   4. Find the angle between $(1,2,2)$ and $(2,1,-2)$.	d. $2$

   Show answer

   1 → d ($2$) · 2 → b ($(1,-2,3)$) · 3 → c ($\sqrt{15}$) · 4 → a ($90^{\circ }$)

   • Step 1. $z=-x-y$, so every vector is $x(1,0,-1)+y(0,1,-1)$.
     Step 2. These two vectors are independent, so the dimension is 2.
   • Step 1. $a(1,0,0)+b(0,1,0)+c(1,1,1)=(a+c,b+c,c)$.
     Step 2. $c=3$, $b+3=1\Rightarrow b=-2$, $a+3=4\Rightarrow a=1$.
   • Step 1. $|\vec{u}+\vec{v}{|}^2=|\vec{u}{|}^2+2\vec{u}\cdot \vec{v}+|\vec{v}{|}^2$.
     Step 2. $4+2+9=15$, so $|\vec{u}+\vec{v}|=\sqrt{15}$.
   • Step 1. Dot product: $2+2-4=0$.
     Step 2. A zero dot product (non-zero vectors) means the angle is $90^{\circ }$.

2. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $1$  ·  $\left(\frac{3}{2},\frac{3}{2},0\right)$  ·  $\frac{11}{3}$  ·  $11$
   1. Find the scalar projection of $\vec{u}=(3,4,0)$ on $\vec{v}=(1,2,2)$. →
   2. Find the scalar triple product $\vec{u}\cdot (\vec{v}\times \vec{w})$ for $\vec{u}=(1,2,3)$, $\vec{v}=(0,1,4)$, $\vec{w}=(2,0,1)$. →
   3. Find the vector projection of $(1,2,3)$ on $(1,1,0)$. →
   Show answer

   1: $\frac{11}{3}$ · 2: $11$ · 3: $\left(\frac{3}{2},\frac{3}{2},0\right)$

   • Step 1. $\vec{u}\cdot \vec{v}=3+8+0=11$ and $|\vec{v}|=\sqrt{1+4+4}=3$.
     Step 2. Scalar projection $=\frac{\vec{u}\cdot \vec{v}}{|\vec{v}|}=\frac{11}{3}$.
   • Step 1. It equals $\left|\begin{array}{ccc}1& 2& 3\\ 0& 1& 4\\ 2& 0& 1\end{array}\right|$.
     Step 2. $1(1-0)-2(0-8)+3(0-2)=1+16-6=11$.
   • Step 1. Dot product $=1+2+0=3$; $|\vec{v}{|}^2=2$.
     Step 2. Projection $=\frac{3}{2}(1,1,0)=\left(\frac{3}{2},\frac{3}{2},0\right)$.

3. Matching

   Match each item in column A with the correct answer in column B. Write the letter next to each number.

   Column A	Column B
   1. $\vec{u}=(1,2,-1)$ and $\vec{v}=(3,0,2)$. Find $\vec{u}\cdot \vec{v}$.	a. $\{(x,y,z):x=2y\}$
   2. Which set is a subspace of ${\mathbb{R}}^3$?	b. $\frac{1}{3}$
   3. Find $m$ so that $(2,m,1)$ is perpendicular to $(1,3,-4)$.	c. $1$
   4. Find the volume of the tetrahedron built on $(1,0,0)$, $(1,1,0)$ and $(1,1,2)$.	d. $m=\frac{2}{3}$

   Show answer

   1 → c ($1$) · 2 → a ($\{(x,y,z):x=2y\}$) · 3 → d ($m=\frac{2}{3}$) · 4 → b ($\frac{1}{3}$)

   • Step 1. Multiply matching components: $1\times 3+2\times 0+(-1)\times 2$.
     Step 2. $3+0-2=1$.
   • Step 1. A subspace must contain $(0,0,0)$ and be closed under addition and scalar multiplication.
     Step 2. $x=2y$ is linear with no constant term, so it passes all tests. The others fail ($z=1$ and $x+y=1$ miss the origin; $x=y^2$ is not closed under scaling).
   • Step 1. Perpendicular means dot product 0: $2+3m-4=0$.
     Step 2. $3m=2$, $m=\frac{2}{3}$.
   • Step 1. $\left|\begin{array}{ccc}1& 0& 0\\ 1& 1& 0\\ 1& 1& 2\end{array}\right|=1\times 1\times 2=2$ (triangular).
     Step 2. Tetrahedron volume $=\frac{1}{6}|\det |=\frac{2}{6}=\frac{1}{3}$.

4. Drag and drop

   Choose the correct word or phrase from the box for each gap. One word in the box is not needed.

   $(-2,-3,1)$  ·  $0$  ·  $a=4,b=1,c=0$  ·  $a=5,b=4,c=1$
   1. For any vectors $\vec{u}$ and $\vec{v}$ in ${\mathbb{R}}^3$, $(\vec{u}\times \vec{v})\cdot \vec{u}=$ →
   2. Write $(5,4,1)$ as $a(1,1,0)+b(1,0,1)+c(0,1,1)$. →
   3. Find $(1,0,2)\times (0,1,3)$. →
   Show answer

   1: $0$ · 2: $a=4,b=1,c=0$ · 3: $(-2,-3,1)$

   • Step 1. $\vec{u}\times \vec{v}$ is perpendicular to $\vec{u}$.
     Step 2. The dot product of perpendicular vectors is 0.
   • Step 1. Components: $a+b=5$, $a+c=4$, $b+c=1$.
     Step 2. Adding: $2(a+b+c)=10$, so $a+b+c=5$.
     Step 3. $c=5-5=0$, $b=5-4=1$, $a=5-1=4$.
   • Step 1. $x$: $0\cdot 3-2\cdot 1=-2$; $y$: $2\cdot 0-1\cdot 3=-3$.
     Step 2. $z$: $1\cdot 1-0\cdot 0=1$.

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