Wize High School Grade 12 Calculus Textbook > Vectors

Vector Operations

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Vector Scalar Multiplication
Given the vector u⃗=[u1, u2]\vec{u}=\left[u_1,\ u_2\right]u=[u1​, u2​] in R2R^2R2, we can multiply u⃗\vec{u}u by any scalar (number) kkk.

Albegraically
ku⃗=k[u1, u2]=[ku1, ku2]k\vec{u}=k\left[u_1,\ u_2\right]=\left[ku_1,\ ku_2\right]ku=k[u1​, u2​]=[ku1​, ku2​]

Geometrically
The scalar multiple of a vector is a vector that is parallel to the original vector
If the scalar kkk is positive, we will get a vector that is in the same direction as the original vector
If the scalar kkk is negative, we will get a vector that is in the opposite direction as the original vector
If the scalar k>1k>1k>1 or k<−1k<-1k<−1, we get a vector that is stretched
If the scalar −1<k<1-1<k<1−1<k<1, we get a vector that is compressed (shrunk)

Example
If v⃗=[−1, 4]\vec{v}=\left[-1,\ 4\right]v=[−1, 4], find −3v⃗-3\vec{v}−3v and 12v⃗\frac{1}{2}\vec{v}21​v.

−3v⃗=−3[−1,  4]=[−3, −12]-3\vec{v}=-3\left[-1,\ \ 4\right]=\left[-3,\ -12\right]−3v=−3[−1,  4]=[−3, −12]
12v⃗=12[−1, 4]=[−12, 2]\frac{1}{2}\vec{v}=\frac{1}{2}\left[-1,\ 4\right]=\left[-\frac{1}{2},\ 2\right]21​v=21​[−1, 4]=[−21​, 2]

Wize Tip
Scalar multiplication in R3R^3R3 works the same way as R2R^2R2!

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Vector Addition
Given the vectors u⃗=[u1, u2]\vec{u}=\left[u_1,\ u_2\right]u=[u1​, u2​] and v⃗=[v1, v2]\vec{v}=\left[v_1,\ v_2\right]v=[v1​, v2​] in R2R^2R2, we can add the two vectors.

Adding Algebraically
Add component by component
u⃗+v⃗=[u1+v1, u2+v2]\vec{u}+\vec{v}=\left[u_1+v_1,\ u_2+v_2\right]u+v=[u1​+v1​, u2​+v2​]

Adding Geometrically (Triangle Rule)
We line up the vectors from "tip to tail" and find the resultant vector from the starting point to the ending point

Wize Tip
When adding multiple vectors, find the "short-cut" from the starting point to the ending point.

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Example
If u1⃗=[4, 4]\vec{u_1}=\left[4,\ 4\right]u1​​=[4, 4], u2⃗=[−5, 0]\vec{u_2}=\left[-5,\ 0\right]u2​​=[−5, 0], u3⃗=[3, 12]\vec{u_3}=\left[3,\ 12\right]u3​​=[3, 12], u4⃗=[4, −17]\vec{u_4}=\left[4,\ -17\right]u4​​=[4, −17], find u1⃗+u2⃗+u3⃗+u4⃗\vec{u_1}+\vec{u_2}+\vec{u_3}+\vec{u_4}u1​​+u2​​+u3​​+u4​​.

[4+(−5)+3+4, 4+0+12+(−17)]=[6, −1]\left[4+\left(-5\right)+3+4,\ 4+0+12+\left(-17\right)\right]=\left[6,\ -1\right][4+(−5)+3+4, 4+0+12+(−17)]=[6, −1]

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Vector Subtraction
Given the vectors u⃗=[u1, u2]\vec{u}=\left[u_1,\ u_2\right]u=[u1​, u2​] and v⃗=[v1, v2]\vec{v}=\left[v_1,\ v_2\right]v=[v1​, v2​] in R2R^2R2, we can subtract the two vectors.

Subtracting Algebraically
Add the negative of the vector
u⃗+(−v⃗)=[u1+(−v1), u2+(−v2)]\vec{u}+\left(-\vec{v}\right)=\left[u_1+\left(-v_1\right),\ u_2+\left(-v_2\right)\right]u+(−v)=[u1​+(−v1​), u2​+(−v2​)]

or

Subtract component by component
u⃗−v⃗=[u1−v1, u2−v2]\vec{u}-\vec{v}=\left[u_1-v_1,\ u_2-v_2\right]u−v=[u1​−v1​, u2​−v2​]

Subtracting Geometrically
We line up the vectors from "tail to tail" and find the resultant vector from the tip of the second vector to the tip of the first vector

Wize Tip
Vector addition and subtraction are done the same way in R3R^3R3 as it is in R2R^2R2!


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Position Vector
The position vector in R2R^2R2 (or R3R^3R3) that points from point A to point B is AB→=B→−A→\overrightarrow{AB}=\overrightarrow{B}-\overrightarrow{A}AB=B−A.

Practice: Vector Operations
If u⃗=[1, 0, −1]\vec{u}=\left[1,\ 0,\ -1\right]u=[1, 0, −1], v⃗=[−2, 4, 1]\vec{v}=\left[-2,\ 4,\ 1\right]v=[−2, 4, 1], and w⃗=[3, 9, −12]\vec{w}=\left[3,\ 9,\ -12\right]w=[3, 9, −12], determine 5u⃗−12v⃗+23w⃗5\vec{u}-\frac{1}{2}\vec{v}+\frac{2}{3}\vec{w}5u−21​v+32​w.

[6, 8, −252]\left[6,\ 8,\ -\frac{25}{2}\right][6, 8, −225​]

[6, 8, −272]\left[6,\ 8,\ -\frac{27}{2}\right][6, 8, −227​]

[8, 4, −272]\left[8,\ 4,\ -\frac{27}{2}\right][8, 4, −227​]

[8, 4, −252]\left[8,\ 4,\ -\frac{25}{2}\right][8, 4, −225​]

I don't know

Practice: Position Vectors
Given the points A(1, 2, 3)A\left(1,\ 2,\ 3\right)A(1, 2, 3), B(0, −4, 1)B\left(0,\ -4,\ 1\right)B(0, −4, 1), and C(2, 8, 5)C\left(2,\ 8,\ 5\right)C(2, 8, 5), select all of the following statements that is/are true.

AB→\overrightarrow{AB}AB is opposite to BA→\overrightarrow{BA}BA

AB→\overrightarrow{AB}AB is opposite to −BA→-\overrightarrow{BA}−BA

2AB→=−2[1, 6, 2]2\overrightarrow{AB}=-2\left[1,\ 6,\ 2\right]2AB=−2[1, 6, 2]

AB→\overrightarrow{AB}AB and BC→\overrightarrow{BC}BC are parallel

AB→\overrightarrow{AB}AB and BC→\overrightarrow{BC}BC are opposite

I don't know

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Parallelogram Rule
The parallelogram rule is another way to represent vector addition and subtraction.

If we draw a parallelogram with sides u⃗\vec{u}u and v⃗\vec{v}v, u⃗+v⃗\vec{u}+\vec{v}u+v and u⃗−v⃗\vec{u}-\vec{v}u−v form the diagonals of the parallelogram.


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Finding Magnitudes
We can use the Cosine Law to calculate magnitudes: c2=a2+b2−2abcos⁡θc^2=a^2+b^2-2ab\cos\theta c2=a2+b2−2abcosθ
If we line up the vectors from tip to tail, suppose the angle between u⃗\vec{u}u and v⃗\vec{v}v is θ\thetaθ, then ∣∣u⃗+v⃗∣∣2=∣∣u⃗∣∣2+∣∣v⃗∣∣2−2∣∣u⃗∣∣∣∣v⃗∣∣cos⁡θ\left|\left|\vec{u}+\vec{v}\right|\right|^2=\left|\left|\vec{u}\right|\right|^2+\left|\left|\vec{v}\right|\right|^2-2\left|\left|\vec{u}\right|\right|\left|\left|\vec{v}\right|\right|\cos\theta∣∣u+v∣∣2=∣∣u∣∣2+∣∣v∣∣2−2∣∣u∣∣∣∣v∣∣cosθ

If we line up the vectors from tail to tail, suppose the angle between u⃗\vec{u}u and v⃗\vec{v}v is ϕ\phiϕ, then ∣∣u⃗−v⃗∣∣2=∣∣u⃗∣∣2+∣∣v⃗∣∣2−2∣∣u⃗∣∣∣∣v⃗∣∣cos⁡ϕ\left|\left|\vec{u}-\vec{v}\right|\right|^2=\left|\left|\vec{u}\right|\right|^2+\left|\left|\vec{v}\right|\right|^2-2\left|\left|\vec{u}\right|\right|\left|\left|\vec{v}\right|\right|\cos\phi∣∣u−v∣∣2=∣∣u∣∣2+∣∣v∣∣2−2∣∣u∣∣∣∣v∣∣cosϕ

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Finding Angles
We can use the Sine Law to calculate the angles: sin⁡Aa=sin⁡Bb=sin⁡Cc\displaystyle\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}asinA​=bsinB​=csinC​

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Example: Finding Magnitude & Direction
Suppose that u⃗\vec uu and v⃗\vec vv are unit vectors that make an angle of 40°40\degree40° with each other.
a) Compute the magnitude of 2u⃗+3v⃗2\vec u+3\vec v2u+3v
b) Compute the angle of 2u⃗+3v⃗2\vec u+3\vec v2u+3v relative to u⃗\vec uu

Part a)
Since u⃗\vec uu and v⃗\vec vv are unit vectors, their magnitudes are 1:

The diagram below shows 2u⃗+3v⃗2\vec u+3\vec v2u+3v. Once we line up the vectors from tip to tail, we see that the angle between the two vectors is 180°−40°=140°180\degree-40\degree=140\degree180°−40°=140°


Using the Cosine law, we see that 
∣∣u⃗+v⃗∣∣2=∣∣u⃗∣∣2+∣∣v⃗∣∣2−2∣∣u⃗∣∣∣∣v⃗∣∣cos⁡140°\left|\left|\vec{u}+\vec{v}\right|\right|^2=\left|\left|\vec{u}\right|\right|^2+\left|\left|\vec{v}\right|\right|^2-2\left|\left|\vec{u}\right|\right|\left|\left|\vec{v}\right|\right|\cos140\degree∣∣u+v∣∣2=∣∣u∣∣2+∣∣v∣∣2−2∣∣u∣∣∣∣v∣∣cos140°:
∣∣u⃗+v⃗∣∣2=22+32−2(2)(3)cos⁡140°\left|\left|\vec{u}+\vec{v}\right|\right|^2=2^2+3^2-2\left(2\right)\left(3\right)\cos140\degree∣∣u+v∣∣2=22+32−2(2)(3)cos140°
∣∣u⃗+v⃗∣∣2≈4+9−12(−0.766)\left|\left|\vec{u}+\vec{v}\right|\right|^2\approx4+9-12\left(-0.766\right)∣∣u+v∣∣2≈4+9−12(−0.766)
∣∣u⃗+v⃗∣∣2≈22.192\left|\left|\vec{u}+\vec{v}\right|\right|^2\approx22.192∣∣u+v∣∣2≈22.192
∣∣u⃗+v⃗∣∣≈22.192≈4.71\left|\left|\vec{u}+\vec{v}\right|\right|\approx\sqrt{22.192}\approx4.71∣∣u+v∣∣≈22.192​≈4.71
Therefore, the magnitude of 2u⃗+3v⃗≈22.192\vec{u}+3\vec{v}\approx22.192u+3v≈22.19

Part b)
Using the Sine law, we see that
sin⁡θ∣∣v⃗∣∣=sin⁡140°∣∣u⃗+v⃗∣∣\displaystyle\frac{\sin\theta}{\left|\left|\vec{v}\right|\right|}=\frac{\sin140\degree}{\left|\left|\vec{u}+\vec{v}\right|\right|}∣∣v∣∣sinθ​=∣∣u+v∣∣sin140°​
sin⁡θ=sin⁡140°∣∣u⃗+v⃗∣∣∣∣v⃗∣∣\displaystyle\sin\theta=\frac{\sin140\degree}{\left|\left|\vec{u}+\vec{v}\right|\right|}\left|\left|\vec{v}\right|\right|sinθ=∣∣u+v∣∣sin140°​∣∣v∣∣
θ=sin⁡−1(sin⁡140°∣∣u⃗+v⃗∣∣∣∣v⃗∣∣)\displaystyle\theta=\sin^{-1}\left(\frac{\sin140\degree}{\left|\left|\vec{u}+\vec{v}\right|\right|}\left|\left|\vec{v}\right|\right|\right)θ=sin−1(∣∣u+v∣∣sin140°​∣∣v∣∣)
θ≈24.17°\theta\approx24.17\degreeθ≈24.17°
Therefore,  2u⃗+3v⃗2\vec u+3\vec v2u+3v is 24.17°24.17\degree 24.17° measured counter clockwise from u⃗\vec uu

Practice: Finding Magnitudes
Suppose that ∣∣u⃗∣∣=3||\vec u||=3∣∣u∣∣=3, ∣∣v⃗∣∣=4||\vec v||=4∣∣v∣∣=4, and the angle between the two vectors when placed tail to tail is 60°60\degree60°.
a) Compute ∣∣u⃗+v⃗∣∣||\vec u+\vec v||∣∣u+v∣∣
b) Computer ∣∣u⃗−v⃗∣∣||\vec u-\vec v||∣∣u−v∣∣
c) Computer ∣∣v⃗−u⃗∣∣||\vec v-\vec u||∣∣v−u∣∣

[Enter your answers as exact values, without rounding)

||\vec u+\vec v||=

||\vec u-\vec v||=

||\vec v-\vec u||=

I don't know

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Vector Properties
Given that u⃗, v⃗, w⃗\vec{u},\ \vec{v},\ \vec wu, v, w are vectors both in R2R^2R2 or R3R^3R3, a, ba,\ ba, b are scalars.
Justify the following properties.
Commutative Property for Addition: u⃗+v⃗=v⃗+u⃗\vec{u}+\vec{v}=\vec{v}+\vec{u}u+v=v+u
Associative Property for Addition: (u⃗+v⃗)+w⃗=u⃗+(v⃗+w⃗)\left(\vec{u}+\vec{v}\right)+\vec{w}=\vec{u}+\left(\vec{v}+\vec{w}\right)(u+v)+w=u+(v+w)
Distributive Property for Addition: a(u⃗+v⃗)=au⃗+av⃗a\left(\vec{u}+\vec{v}\right)=a\vec{u}+a\vec{v}a(u+v)=au+av
Associative Law for Scalars: (ab)u⃗=a(bu)=b(au⃗)\left(ab\right)\vec{u}=a\left(bu\right)=b\left(a\vec{u}\right)(ab)u=a(bu)=b(au)
Distrobutive Law for Scalars: (a+b)u⃗=au⃗+bu⃗\left(a+b\right)\vec{u}=a\vec{u}+b\vec{u}(a+b)u=au+bu
There exists a zero vector 0⃗\vec{0}0 such that:
v⃗+0⃗=v⃗\vec{v}+\vec{0}=\vec{v}v+0=v
v⃗+(−v⃗)=0⃗\vec{v}+\left(-\vec{v}\right)=\vec{0}v+(−v)=0
Magnitude Properties
∣∣av⃗∣∣=∣a∣∣∣v⃗∣∣\left|\left|a\vec{v}\right|\right|=\left|a\right|\left|\left|\vec{v}\right|\right|∣∣av∣∣=∣a∣∣∣v∣∣
∣∣u⃗+v⃗∣∣≤∣∣u⃗∣∣+∣∣v⃗∣∣\left|\left|\vec{u}+\vec{v}\right|\right|\le\left|\left|\vec{u}\right|\right|+\left|\left|\vec{v}\right|\right|∣∣u+v∣∣≤∣∣u∣∣+∣∣v∣∣
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Can You Answer These?
❓ Can we add a vector in R2R^2R2 and a vector in R3R^3R3? 

No

❓ If u⃗=2i+j−k\vec{u}=2i+j-ku=2i+j−k and v⃗=i−j\vec{v}=i-jv=i−j, what is u⃗+v⃗\vec{u}+\vec{v}u+v?

[2, 1, −1]+[1, −1, 0]=[3, 0, −1]\left[2,\ 1,\ -1\right]+\left[1,\ -1,\ 0\right]=\left[3,\ 0,\ -1\right][2, 1, −1]+[1, −1, 0]=[3, 0, −1]

❓ When will ∣∣u⃗+v⃗∣∣=∣∣u⃗∣∣+∣∣v⃗∣∣\left|\left|\vec{u}+\vec{v}\right|\right|=\left|\left|\vec{u}\right|\right|+\left|\left|\vec{v}\right|\right|∣∣u+v∣∣=∣∣u∣∣+∣∣v∣∣?

When u⃗\vec{u}u and v⃗\vec{v}v are parallel.

Extra Practice

Vector Operations

Practice: Collinear Vectors
Given  u⃗=⟨2,−5,6⟩,  v⃗=⟨−4,10,−12⟩,\vec{u}=\lang 2,−5,6\rang ,\ \ \vec{v}=\lang−4,10,−12\rang ,u=⟨2,−5,6⟩,  v=⟨−4,10,−12⟩, and  𝑤⃗=⟨−2,5,−6⟩𝑤⃗=\lang −2,5,−6 \rangw⃗=⟨−2,5,−6⟩, select all statements that are true.