Wize High School Grade 12 Calculus Textbook > Vectors

Magnitude of a Vector

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Magnitude of a 2D Vector
We can calculate the magnitude (a.k.a. length or size) of a vector in 2D using Pythagorean's Theorem:

Write it Down
The length of the vector v⃗=[v1, v2]\vec{v}=\left[v_1,\ v_2\right]v=[v1​, v2​] is
∣∣v⃗∣∣=∣∣[v1, v2]∣∣=v1 2+v2 2||\vec v||=||[v_1,\ v_2]||=\sqrt{v_1^{\ 2}+v_2^{\ 2}}∣∣v∣∣=∣∣[v1​, v2​]∣∣=v1 2​+v2 2​​


Example
Find the lengths of the vectors v⃗=[−3, 4]\vec{v}=[-3,\ 4]v=[−3, 4] and u⃗=[0, −1]\vec{u}=\left[0,\ -1\right]u=[0, −1]

∣∣v⃗∣∣=(−3)2+(4)2=25=5\left|\left|\vec{v}\right|\right|=\sqrt{\left(-3\right)^2+\left(4\right)^2}=\sqrt{25}=5∣∣v∣∣=(−3)2+(4)2​=25​=5
∣∣u⃗∣∣=02+(−1)2=1=1\left|\left|\vec{u}\right|\right|=\sqrt{0^2+\left(-1\right)^2}=\sqrt{1}=1∣∣u∣∣=02+(−1)2​=1​=1

Magnitude of a 3D Vector
We can calculate the magnitude of a vector in 3D using an extension of this formula.

Write it Down
The length of the vector u⃗=[u1, u2, u3]\vec{u}=\left[u_1,\ u_2,\ u_3\right]u=[u1​, u2​, u3​] is
∣∣u⃗∣∣=∣∣[u1, u2, u3]∣∣=u1 2+u2 2+u3 2\left|\left|\vec{u}\right|\right|=\left|\left|\left[u_1,\ u_2,\ u_3\right]\right|\right|=\sqrt{u_1^{\ 2}+u_2^{\ 2}+u_3^{\ 2}}∣∣u∣∣=∣∣[u1​, u2​, u3​]∣∣=u1 2​+u2 2​+u3 2​​

Practice: Magnitude of a Vector
An aircraft's location is 200 miles from base, at a direction of 30° W of S, at an  altitude of 35,000 ft. Find the distance between the aircraft and the base. 

Note: 1 mile = 5280 feet

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Unit Vector
A unit vector is a vector that has a magnitude of 1.

Example
Show that the following vectors are all unit vectors
u⃗=[1, 0]\vec{u}=\left[1,\ 0\right]u=[1, 0] is a unit vector

∣∣u⃗∣∣=12+02=1\left|\left|\vec{u}\right|\right|=\sqrt{1^2+0^2}=1∣∣u∣∣=12+02​=1

v⃗=[12, −12]\vec{v}=\left[\frac{1}{\sqrt{2}},\ -\frac{1}{\sqrt{2}}\right]v=[2​1​, −2​1​] is a unit vector

∣∣v⃗∣∣=(12)2+(−12)2=12+12=1=1\left|\left|\vec{v}\right|\right|=\sqrt{\left(\frac{1}{\sqrt{2}}\right)^2+\left(-\frac{1}{\sqrt{2}}\right)^2}=\sqrt{\frac{1}{2}+\frac{1}{2}}=\sqrt{1}=1∣∣v∣∣=(2​1​)2+(−2​1​)2​=21​+21​​=1​=1

w⃗=[0, 35, −45]\vec{w}=\left[0,\ \frac{3}{5},\ -\frac{4}{5}\right]w=[0, 53​, −54​] is a unit vector

∣∣w⃗∣∣=02+(35)2+(−45)2=925+1625=1=1\left|\left|\vec{w}\right|\right|=\sqrt{0^2+\left(\frac{3}{5}\right)^2+\left(-\frac{4}{5}\right)^2}=\sqrt{\frac{9}{25}+\frac{16}{25}}=\sqrt{1}=1∣∣w∣∣=02+(53​)2+(−54​)2​=259​+2516​​=1​=1

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Finding a Unit Vector in a Certain Direction
Given a vector v⃗\vec vv, the unit vector in the same direction as v⃗\vec vv is v^=v⃗∣∣v⃗∣∣\displaystyle \hat v=\frac{\vec v}{||\vec v||}v^=∣∣v∣∣v​
Example
Find the unit vector in the same direction as u⃗=[1, 2, −3]\vec u=\left[1,\ 2,\ -3\right]u=[1, 2, −3]

u^=[1, 2, −3]∣∣ [1, 2, −3] ∣∣\displaystyle \hat u=\frac{[1,\ 2,\ -3]}{||\ [1,\ 2,\ -3]\ ||}u^=∣∣ [1, 2, −3] ∣∣[1, 2, −3]​
u^=[1, 2, −3]12+22+(−3)2\displaystyle \hat u=\frac{[1,\ 2,\ -3]}{\sqrt {1^2+2^2+(-3)^2}}u^=12+22+(−3)2​[1, 2, −3]​
u^=[1, 2, −3]14\displaystyle \hat u=\frac{[1,\ 2,\ -3]}{\sqrt {14}}u^=14​[1, 2, −3]​ or [114, 214, −314]\left[\frac{1}{\sqrt{14}},\ \frac{2}{\sqrt{14}},\ -\frac{3}{\sqrt{14}}\right][14​1​, 14​2​, −14​3​]

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Special Unit Vectors
In 2D, we have the following special unit vectors:
i⃗=[1, 0]\vec i=[1,\ 0]i=[1, 0] → unit vector along the positive xxx-axis
j⃗=[0, 1]\vec j=[0,\ 1]j​=[0, 1] → unit vector along the positive yyy-axis
In 3D, we have the following special unit vectors:
i⃗=[1, 0, 0]\vec i=[1,\ 0,\ 0]i=[1, 0, 0] → unit vector along the positive xxx-axis
j⃗=[0, 1, 0]\vec j=[0,\ 1,\ 0]j​=[0, 1, 0] → unit vector along the positive yyy-axis
k⃗=[0, 0, 1]\vec k=[0,\ 0,\ 1]k=[0, 0, 1] → unit vector along the positive zzz-axis

Example
2i⃗−3j⃗+k⃗=2\vec i-3\vec j+\vec k=2i−3j​+k= [2, -3, 1]

Practice: Unit Vector
Find the unit vector in the same direction as i⃗−5k⃗\vec i-5\vec ki−5k.

[124, 0, −524]\left[\frac{1}{\sqrt{24}},\ 0,\ -\frac{5}{\sqrt{24}}\right][24​1​, 0, −24​5​]

[126, 0, −526]\left[\frac{1}{\sqrt{26}},\ 0,\ -\frac{5}{\sqrt{26}}\right][26​1​, 0, −26​5​]

[127, 0, −527]\left[\frac{1}{\sqrt{27}},\ 0,\ -\frac{5}{\sqrt{27}}\right][27​1​, 0, −27​5​]

[126, 126, −526]\left[\frac{1}{\sqrt{26}},\ \frac{1}{\sqrt{26}},\ -\frac{5}{\sqrt{26}}\right][26​1​, 26​1​, −26​5​]

[124, 124, −524]\left[\frac{1}{\sqrt{24}},\ \frac{1}{\sqrt{24}},\ -\frac{5}{\sqrt{24}}\right][24​1​, 24​1​, −24​5​]

I don't know

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Directional Cosine for 3D Vectors
Given any vector v⃗=[v1, v2, v3]\vec v=[v_1,\ v_2,\ v_3]v=[v1​, v2​, v3​] in 3D, we can define the following angles:
α\alphaα is the angle v⃗\vec{v}v forms with the xxx- axis
β\betaβ is the angle v⃗\vec{v}v forms with the yyy- axis
γ\gammaγ is the angle v⃗\vec{v}v forms with the zzz- axis

If v⃗\vec{v}v is a unit vector, then v⃗=[cos⁡α, cos⁡β, cos⁡γ]\vec{v}=\left[\cos\alpha,\ \cos\beta,\ \cos\gamma\right]v=[cosα, cosβ, cosγ]  ⟶  these 3 components are called the directional cosines of the vector v⃗\vec{v}v.

General Directional Cosines
Given any vector v⃗=[v1, v2, v3]\vec{v}=\left[v_1,\ v_2,\ v_3\right]v=[v1​, v2​, v3​], 
cos⁡α=v1∣∣v⃗∣∣\displaystyle\cos\alpha=\frac{v_1}{\left|\left|\vec{v}\right|\right|}cosα=∣∣v∣∣v1​​
cos⁡β=v2∣∣v⃗∣∣\displaystyle\cos\beta=\frac{v_2}{\left|\left|\vec{v}\right|\right|}cosβ=∣∣v∣∣v2​​
cos⁡γ=v3∣∣v⃗∣∣\displaystyle\cos\gamma=\frac{v_3}{\left|\left|\vec{v}\right|\right|}cosγ=∣∣v∣∣v3​​