The magnitude of a vector is its length, found using the Pythagorean theorem. For a 2D vector, square the two components, add them, and take the square root. For a 3D vector, do the same thing with all three components.
Notes
Practice Problems
\(\textbf{1)}\) Find \(|\vec{a}|\) where \(\vec{a}=3\vec{i}-4\vec{j}\)
The answer is \(|\vec{a}|=5 \)
\(\,\,\,|\vec{a}|=\sqrt{3^2+(-4)^2}\)
\(\,\,\,|\vec{a}|=\sqrt{9+16}\)
\(\,\,\,|\vec{a}|=\sqrt{25}\)
\(\,\,\,|\vec{a}|=5\)
\(\textbf{2)}\) Find \(|\vec{b}|\) where \(\vec{b}=3\vec{i}-4\vec{j}+6\vec{k}\)
The answer is \(|\vec{b}|=\sqrt{61} \)
\(\,\,\,|\vec{b}|=\sqrt{3^2+(-4)^2+6^2}\)
\(\,\,\,|\vec{b}|=\sqrt{9+16+36}\)
\(\,\,\,|\vec{b}|=\sqrt{61}\)
\(\textbf{3)}\) Find \(|\vec{v}|\) where \(\vec{v}=\langle -4,5 \rangle\)
The answer is \(|\vec{v}|=\sqrt{41} \)
\(\,\,\,|\vec{v}|=\sqrt{(-4)^2+5^2}\)
\(\,\,\,|\vec{v}|=\sqrt{16+25}\)
\(\,\,\,|\vec{v}|=\sqrt{41}\)
\(\textbf{4)}\) Find \(|\vec{m}|\) where \(\vec{m}=\langle 1,7,-3 \rangle\)
The answer is \(|\vec{m}|=\sqrt{59} \)
\(\,\,\,|\vec{m}|=\sqrt{1^2+7^2+(-3)^2}\)
\(\,\,\,|\vec{m}|=\sqrt{1+49+9}\)
\(\,\,\,|\vec{m}|=\sqrt{59}\)
\(\textbf{5)}\) Find \(|\vec{u}|\) where \(\vec{u}=\langle 5,12 \rangle\)
The answer is \(|\vec{u}|=13 \)
\(\,\,\,|\vec{u}|=\sqrt{5^2+12^2}\)
\(\,\,\,|\vec{u}|=\sqrt{25+144}\)
\(\,\,\,|\vec{u}|=\sqrt{169}\)
\(\,\,\,|\vec{u}|=13\)
\(\textbf{6)}\) Find \(|\vec{w}|\) where \(\vec{w}=-6\vec{i}+8\vec{j}\)
The answer is \(|\vec{w}|=10 \)
\(\,\,\,|\vec{w}|=\sqrt{(-6)^2+8^2}\)
\(\,\,\,|\vec{w}|=\sqrt{36+64}\)
\(\,\,\,|\vec{w}|=\sqrt{100}\)
\(\,\,\,|\vec{w}|=10\)
\(\textbf{7)}\) Find \(|\vec{p}|\) where \(\vec{p}=2\vec{i}-3\vec{j}+6\vec{k}\)
The answer is \(|\vec{p}|=7 \)
\(\,\,\,|\vec{p}|=\sqrt{2^2+(-3)^2+6^2}\)
\(\,\,\,|\vec{p}|=\sqrt{4+9+36}\)
\(\,\,\,|\vec{p}|=\sqrt{49}\)
\(\,\,\,|\vec{p}|=7\)
\(\textbf{8)}\) Find \(|\vec{r}|\) where \(\vec{r}=8\vec{i}-15\vec{j}\)
The answer is \(|\vec{r}|=17 \)
\(\,\,\,|\vec{r}|=\sqrt{8^2+(-15)^2}\)
\(\,\,\,|\vec{r}|=\sqrt{64+225}\)
\(\,\,\,|\vec{r}|=\sqrt{289}\)
\(\,\,\,|\vec{r}|=17\)
\(\textbf{9)}\) Find \(|\vec{q}|\) where \(\vec{q}=\langle -9,-40 \rangle\)
The answer is \(|\vec{q}|=41 \)
\(\,\,\,|\vec{q}|=\sqrt{(-9)^2+(-40)^2}\)
\(\,\,\,|\vec{q}|=\sqrt{81+1600}\)
\(\,\,\,|\vec{q}|=\sqrt{1681}\)
\(\,\,\,|\vec{q}|=41\)
\(\textbf{10)}\) Find \(|\vec{t}|\) where \(\vec{t}=\langle 1,2,2 \rangle\)
The answer is \(|\vec{t}|=3 \)
\(\,\,\,|\vec{t}|=\sqrt{1^2+2^2+2^2}\)
\(\,\,\,|\vec{t}|=\sqrt{1+4+4}\)
\(\,\,\,|\vec{t}|=\sqrt{9}\)
\(\,\,\,|\vec{t}|=3\)
\(\textbf{11)}\) Find \(|\vec{d}|\) where \(\vec{d}=\langle 4,0,-3 \rangle\)
The answer is \(|\vec{d}|=5 \)
\(\,\,\,|\vec{d}|=\sqrt{4^2+0^2+(-3)^2}\)
\(\,\,\,|\vec{d}|=\sqrt{16+0+9}\)
\(\,\,\,|\vec{d}|=\sqrt{25}\)
\(\,\,\,|\vec{d}|=5\)
\(\textbf{12)}\) Find \(|\vec{x}|\) where \(\vec{x}=6\vec{i}+2\vec{j}+3\vec{k}\)
The answer is \(|\vec{x}|=7 \)
\(\,\,\,|\vec{x}|=\sqrt{6^2+2^2+3^2}\)
\(\,\,\,|\vec{x}|=\sqrt{36+4+9}\)
\(\,\,\,|\vec{x}|=\sqrt{49}\)
\(\,\,\,|\vec{x}|=7\)
\(\textbf{13)}\) Find \(|\vec{y}|\) where \(\vec{y}=-2\vec{i}+6\vec{j}-9\vec{k}\)
The answer is \(|\vec{y}|=11 \)
\(\,\,\,|\vec{y}|=\sqrt{(-2)^2+6^2+(-9)^2}\)
\(\,\,\,|\vec{y}|=\sqrt{4+36+81}\)
\(\,\,\,|\vec{y}|=\sqrt{121}\)
\(\,\,\,|\vec{y}|=11\)
\(\textbf{14)}\) Find \(|\vec{z}|\) where \(\vec{z}=\langle 0,5,-12 \rangle\)
The answer is \(|\vec{z}|=13 \)
\(\,\,\,|\vec{z}|=\sqrt{0^2+5^2+(-12)^2}\)
\(\,\,\,|\vec{z}|=\sqrt{0+25+144}\)
\(\,\,\,|\vec{z}|=\sqrt{169}\)
\(\,\,\,|\vec{z}|=13\)
\(\textbf{15)}\) Find \(|\vec{s}|\) where \(\vec{s}=\langle -7,24 \rangle\)
The answer is \(|\vec{s}|=25 \)
\(\,\,\,|\vec{s}|=\sqrt{(-7)^2+24^2}\)
\(\,\,\,|\vec{s}|=\sqrt{49+576}\)
\(\,\,\,|\vec{s}|=\sqrt{625}\)
\(\,\,\,|\vec{s}|=25\)
Challenge Problems
\(\textbf{16)}\) Find the distance from \(A(1,2)\) to \(B(7,10)\) using the magnitude of a vector.
The answer is \(10\)
\(\,\,\,\overrightarrow{AB}=\langle7-1,10-2\rangle\)
\(\,\,\,\overrightarrow{AB}=\langle6,8\rangle\)
\(\,\,\,|\overrightarrow{AB}|=\sqrt{6^2+8^2}\)
\(\,\,\,|\overrightarrow{AB}|=\sqrt{36+64}\)
\(\,\,\,|\overrightarrow{AB}|=10\)
\(\textbf{17)}\) Find the distance from \(A(-2,5)\) to \(B(10,0)\) using the magnitude of a vector.
The answer is \(13\)
\(\,\,\,\overrightarrow{AB}=\langle10-(-2),0-5\rangle\)
\(\,\,\,\overrightarrow{AB}=\langle12,-5\rangle\)
\(\,\,\,|\overrightarrow{AB}|=\sqrt{12^2+(-5)^2}\)
\(\,\,\,|\overrightarrow{AB}|=\sqrt{144+25}\)
\(\,\,\,|\overrightarrow{AB}|=13\)
\(\textbf{18)}\) Find the distance from \(A(1,-1,2)\) to \(B(5,2,14)\) using the magnitude of a vector.
The answer is \(13\)
\(\,\,\,\overrightarrow{AB}=\langle5-1,2-(-1),14-2\rangle\)
\(\,\,\,\overrightarrow{AB}=\langle4,3,12\rangle\)
\(\,\,\,|\overrightarrow{AB}|=\sqrt{4^2+3^2+12^2}\)
\(\,\,\,|\overrightarrow{AB}|=\sqrt{16+9+144}\)
\(\,\,\,|\overrightarrow{AB}|=13\)
\(\textbf{19)}\) Find \(k\) if \(\vec{v}=\langle 3,k\rangle\) and \(|\vec{v}|=5\).
The answer is \(k=\pm4\)
\(\,\,\,|\vec{v}|=\sqrt{3^2+k^2}\)
\(\,\,\,5=\sqrt{9+k^2}\)
\(\,\,\,25=9+k^2\)
\(\,\,\,16=k^2\)
\(\,\,\,k=\pm4\)
\(\textbf{20)}\) Find \(k\) if \(\vec{v}=\langle 2,-3,k\rangle\) and \(|\vec{v}|=7\).
The answer is \(k=\pm6\)
\(\,\,\,|\vec{v}|=\sqrt{2^2+(-3)^2+k^2}\)
\(\,\,\,7=\sqrt{4+9+k^2}\)
\(\,\,\,49=13+k^2\)
\(\,\,\,36=k^2\)
\(\,\,\,k=\pm6\)
See Related Pages\(\)
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