What are Vector Valued Functions?

A vector-valued function can be written as:

?(?)=⟨?1(?),?2(?),…,??(?)⟩r(t)=⟨f1​(t),f2​(t),…,fn​(t)⟩

where t is the input variable (often representing time) and ?1(?),?2(?),…,??(?)f1​(t),f2​(t),…,fn​(t) are the component functions, each of which is a real-valued function.

Examples on Vector-Valued Functions

1. 2D Vector-Valued Function:

?(?)=⟨cos⁡(?),sin⁡(?)⟩r(t)=⟨cos(t),sin(t)⟩

This represents a parametric equation of a circle.

1. 3D Vector-Valued Function:

?(?)=⟨?,?2,?3⟩r(t)=⟨t,t2,t3⟩

This represents a space curve in three dimensions.

Vector-valued functions are functions that assign vectors to each value of an independent variable. Mathematically, a vector-valued function can be defined as f(t) = ⟨f₁(t), f₂(t), …, fₙ(t)⟩, where t is the independent variable and f₁(t), f₂(t), …, fₙ(t) are scalar functions determining the components of the vector function at each value of t.

This function describes the trajectory of the particle in three-dimensional space as it moves along a helical path.

Magnitude of Vectors:

The magnitude (or length) of a vector ?=⟨?1,?2,…,??⟩v=⟨v1​,v2​,…,vn​⟩ in ?n-dimensional space is given by:

∣?∣=?12+?22+…+??2∣v∣=v12​+v22​+…+vn2​​

For example, for a vector ?=⟨3,4⟩v=⟨3,4⟩ in 2-dimensional space:

∣?∣=32+42=9+16=25=5∣v∣=32+42​=9+16​=25​=5

Unit Vector:

A unit vector is a vector with a magnitude of 1. It indicates direction but has no units of magnitude. To find a unit vector ?^v^ in the direction of a given vector ?v, divide the vector by its magnitude:

?^=?∣?∣v^=∣v∣v​

For example, for ?=⟨3,4⟩v=⟨3,4⟩:

∣?∣=5∣v∣=5

?^=⟨3,4⟩5=⟨35,45⟩v^=5⟨3,4⟩​=⟨53​,54​⟩

Zero Vector:

The zero vector (denoted as 00) is a vector where all components are zero. In ?n-dimensional space, the zero vector is:

0=⟨0,0,…,0⟩0=⟨0,0,…,0⟩

The zero vector has a magnitude of 0 and does not have a specific direction.

Vector Operations:

1. Addition: The sum of two vectors ?=⟨?1,?2,…,??⟩u=⟨u1​,u2​,…,un​⟩ and ?=⟨?1,?2,…,??⟩v=⟨v1​,v2​,…,vn​⟩ is:?+?=⟨?1+?1,?2+?2,…,??+??⟩u+v=⟨u1​+v1​,u2​+v2​,…,un​+vn​⟩
2. Subtraction: The difference between two vectors ?u and ?v is:?−?=⟨?1−?1,?2−?2,…,??−??⟩u−v=⟨u1​−v1​,u2​−v2​,…,un​−vn​⟩
3. Scalar Multiplication: Multiplying a vector ?=⟨?1,?2,…,??⟩v=⟨v1​,v2​,…,vn​⟩ by a scalar ?c is:??=⟨??1,??2,…,???⟩cv=⟨cv1​,cv2​,…,cvn​⟩
4. Dot Product: The dot product of two vectors ?=⟨?1,?2,…,??⟩u=⟨u1​,u2​,…,un​⟩ and ?=⟨?1,?2,…,??⟩v=⟨v1​,v2​,…,vn​⟩ is:?⋅?=?1?1+?2?2+…+????u⋅v=u1​v1​+u2​v2​+…+un​vn​
5. Cross Product: The cross product of two vectors in 3-dimensional space ?=⟨?1,?2,?3⟩u=⟨u1​,u2​,u3​⟩ and ?=⟨?1,?2,?3⟩v=⟨v1​,v2​,v3​⟩ is:?×?=⟨?2?3−?3?2,?3?1−?1?3,?1?2−?2?1⟩u×v=⟨u2​v3​−u3​v2​,u3​v1​−u1​v3​,u1​v2​−u2​v1​⟩

Vector Components:

A vector in ?n-dimensional space can be broken down into its components along each axis. For a vector ?=⟨?1,?2,…,??⟩v=⟨v1​,v2​,…,vn​⟩:

• ?1v1​ is the component along the ?x-axis.
• ?2v2​ is the component along the ?y-axis.
• ??vn​ is the component along the ?n-th axis.

These components represent the projections of the vector onto the respective axes. The original vector is the sum of these component vectors.

Practice Problems on Vector Valued Functions

Practice Problem 1:

Let v = (2, -3, 5) and u = (-1, 4, 2). Compute:

a) The magnitude of v.

b) The unit vector in the direction of u.

c) The dot product of v and u.

d) The cross product of v and u.

Given:

?=(2,−3,5)v=(2,−3,5)?=(−1,4,2)u=(−1,4,2)

a) The magnitude of ?v:

The magnitude of a vector ?=(?1,?2,?3)v=(v1​,v2​,v3​) is given by:

∣?∣=?12+?22+?32∣v∣=v12​+v22​+v32​​

For ?=(2,−3,5)v=(2,−3,5):

∣?∣=22+(−3)2+52∣v∣=22+(−3)2+52​∣?∣=4+9+25∣v∣=4+9+25​∣?∣=38∣v∣=38​

So, the magnitude of ?v is 3838​.

b) The unit vector in the direction of ?u:

The unit vector ?^u^ in the direction of ?=(?1,?2,?3)u=(u1​,u2​,u3​) is given by:

?^=?∣?∣u^=∣u∣u​

First, we find the magnitude of ?u:

∣?∣=(−1)2+42+22∣u∣=(−1)2+42+22​∣?∣=1+16+4∣u∣=1+16+4​∣?∣=21∣u∣=21​

Now, the unit vector ?^u^ is:

?^=121(−1,4,2)u^=21​1​(−1,4,2)?^=(−121,421,221)u^=(−21​1​,21​4​,21​2​)

So, the unit vector in the direction of ?u is (−121,421,221)(−21​1​,21​4​,21​2​).

c) The dot product of ?v and ?u:

The dot product of two vectors ?=(?1,?2,?3)v=(v1​,v2​,v3​) and ?=(?1,?2,?3)u=(u1​,u2​,u3​) is given by:

?⋅?=?1?1+?2?2+?3?3v⋅u=v1​u1​+v2​u2​+v3​u3​

For ?=(2,−3,5)v=(2,−3,5) and ?=(−1,4,2)u=(−1,4,2):

?⋅?=2⋅(−1)+(−3)⋅4+5⋅2v⋅u=2⋅(−1)+(−3)⋅4+5⋅2?⋅?=−2−12+10v⋅u=−2−12+10?⋅?=−4v⋅u=−4

So, the dot product of ?v and ?u is −4−4.

d) The cross product of ?v and ?u:

The cross product of two vectors ?=(?1,?2,?3)v=(v1​,v2​,v3​) and ?=(?1,?2,?3)u=(u1​,u2​,u3​) is given by:

?×?=(?2?3−?3?2,?3?1−?1?3,?1?2−?2?1)v×u=(v2​u3​−v3​u2​,v3​u1​−v1​u3​,v1​u2​−v2​u1​)

For ?=(2,−3,5)v=(2,−3,5) and ?=(−1,4,2)u=(−1,4,2):

?×?=((−3)⋅2−5⋅4,5⋅(−1)−2⋅2,2⋅4−(−3)⋅(−1))v×u=((−3)⋅2−5⋅4,5⋅(−1)−2⋅2,2⋅4−(−3)⋅(−1))?×?=(−6−20,−5−4,8−3)v×u=(−6−20,−5−4,8−3)?×?=(−26,−9,5)v×u=(−26,−9,5)

So, the cross product of ?v and ?u is (−26,−9,5)(−26,−9,5).

Practice Problem 2:

Find the angle between the vectors v = (3, -4) and u = (5, 2).

Given:

?=(3,−4)v=(3,−4)?=(5,2)u=(5,2)

Step 1: Compute the dot product ?⋅?v⋅u:

?⋅?=3⋅5+(−4)⋅2v⋅u=3⋅5+(−4)⋅2?⋅?=15−8v⋅u=15−8?⋅?=7v⋅u=7

Step 2: Compute the magnitudes of ?v and ?u:

For ?=(3,−4)v=(3,−4):

∣?∣=32+(−4)2∣v∣=32+(−4)2​∣?∣=9+16∣v∣=9+16​∣?∣=25∣v∣=25​∣?∣=5∣v∣=5

For ?=(5,2)u=(5,2):

∣?∣=52+22∣u∣=52+22​∣?∣=25+4∣u∣=25+4​∣?∣=29∣u∣=29​

Step 3: Compute cos⁡(?)cos(θ):

cos⁡(?)=?⋅?∣?∣∣?∣cos(θ)=∣v∣∣u∣v⋅u​cos⁡(?)=75⋅29cos(θ)=5⋅29​7​cos⁡(?)=7529cos(θ)=529​7​

Step 4: Compute ?θ:

?=cos⁡−1(7529)θ=cos−1(529​7​)

To get the exact angle, we can use a calculator:

?≈cos⁡−1(7529)θ≈cos−1(529​7​)?≈cos⁡−1(726.870)θ≈cos−1(26.8707​)?≈cos⁡−1(0.2606)θ≈cos−1(0.2606)?≈75.15∘θ≈75.15∘

So, the angle between the vectors ?v and ?u is approximately 75.15∘75.15∘

Practice Problem 3:

Let v = (-1, 3, -2) and u = (4, -2, 1). Compute:

a) The magnitude of u.

b) The unit vector in the direction of v.

c) The dot product of v and u.

d) The angle between v and u.

Given:

?=(−1,3,−2)v=(−1,3,−2)?=(4,−2,1)u=(4,−2,1)

a) The magnitude of ?u:

The magnitude of a vector ?=(?1,?2,?3)u=(u1​,u2​,u3​) is given by:

∣?∣=?12+?22+?32∣u∣=u12​+u22​+u32​​

For ?=(4,−2,1)u=(4,−2,1):

∣?∣=42+(−2)2+12∣u∣=42+(−2)2+12​∣?∣=16+4+1∣u∣=16+4+1​∣?∣=21∣u∣=21​

So, the magnitude of ?u is 2121​.

b) The unit vector in the direction of ?v:

The unit vector ?^v^ in the direction of ?=(?1,?2,?3)v=(v1​,v2​,v3​) is given by:

?^=?∣?∣v^=∣v∣v​

First, we find the magnitude of ?v:

∣?∣=(−1)2+32+(−2)2∣v∣=(−1)2+32+(−2)2​∣?∣=1+9+4∣v∣=1+9+4​∣?∣=14∣v∣=14​

Now, the unit vector ?^v^ is:

?^=114(−1,3,−2)v^=14​1​(−1,3,−2)?^=(−114,314,−214)v^=(−14​1​,14​3​,−14​2​)

So, the unit vector in the direction of ?v is (−114,314,−214)(−14​1​,14​3​,−14​2​).

c) The dot product of ?v and ?u:

The dot product of two vectors ?=(?1,?2,?3)v=(v1​,v2​,v3​) and ?=(?1,?2,?3)u=(u1​,u2​,u3​) is given by:

?⋅?=?1?1+?2?2+?3?3v⋅u=v1​u1​+v2​u2​+v3​u3​

For ?=(−1,3,−2)v=(−1,3,−2) and ?=(4,−2,1)u=(4,−2,1):

?⋅?=(−1)⋅4+3⋅(−2)+(−2)⋅1v⋅u=(−1)⋅4+3⋅(−2)+(−2)⋅1?⋅?=−4−6−2v⋅u=−4−6−2?⋅?=−12v⋅u=−12

So, the dot product of ?v and ?u is −12−12.

d) The angle between ?v and ?u:

The angle ?θ between two vectors ?v and ?u can be found using the dot product formula:

cos⁡(?)=?⋅?∣?∣∣?∣cos(θ)=∣v∣∣u∣v⋅u​

We already have: ?⋅?=−12v⋅u=−12∣?∣=14∣v∣=14​∣?∣=21∣u∣=21​

Now, calculate cos⁡(?)cos(θ):

cos⁡(?)=−1214⋅21cos(θ)=14​⋅21​−12​cos⁡(?)=−12294cos(θ)=294​−12​cos⁡(?)=−12294≈−0.6987cos(θ)=294​−12​≈−0.6987

Now, find ?θ using the inverse cosine function:

?=cos⁡−1(−0.6987)θ=cos−1(−0.6987)

Using a calculator:

?≈134.43∘θ≈134.43∘

So, the angle between the vectors ?v and ?u is approximately 134.43∘134.43∘

Practice Problem 4:

Find the cross product of the vectors v = (2, 1, -3) and u = (-1, 4, 2).

Given: ?=(2,1,−3)v=(2,1,−3)?=(−1,4,2)u=(−1,4,2)

Compute each component of the cross product:

1. First component: ?2?3−?3?2=1⋅2−(−3)⋅4v2​u3​−v3​u2​=1⋅2−(−3)⋅4=2+12=2+12=14=14
2. Second component: ?3?1−?1?3=(−3)⋅(−1)−2⋅2v3​u1​−v1​u3​=(−3)⋅(−1)−2⋅2=3−4=3−4=−1=−1
3. Third component: ?1?2−?2?1=2⋅4−1⋅(−1)v1​u2​−v2​u1​=2⋅4−1⋅(−1)=8+1=8+1=9=9

Therefore, the cross product ?×?v×u is:

?×?=(14,−1,9)v×u=(14,−1,9)

So, the cross product of the vectors ?=(2,1,−3)v=(2,1,−3) and ?=(−1,4,2)u=(−1,4,2) is (14,−1,9)(14,−1,9).

Practice Problem 5:

Find a vector that is perpendicular to both v = (3, 1, -2) and u = (2, -1, 4).

Given: ?=(3,1,−2)v=(3,1,−2)?=(2,−1,4)u=(2,−1,4)

The cross product ?×?v×u is calculated as follows:

?×?=(?2?3−?3?2,?3?1−?1?3,?1?2−?2?1)v×u=(v2​u3​−v3​u2​,v3​u1​−v1​u3​,v1​u2​−v2​u1​)

Let’s compute each component:

1. First component: ?2?3−?3?2=1⋅4−(−2)⋅(−1)v2​u3​−v3​u2​=1⋅4−(−2)⋅(−1)=4−2=4−2=2=2
2. Second component: ?3?1−?1?3=(−2)⋅2−3⋅4v3​u1​−v1​u3​=(−2)⋅2−3⋅4=−4−12=−4−12=−16=−16
3. Third component: ?1?2−?2?1=3⋅(−1)−1⋅2v1​u2​−v2​u1​=3⋅(−1)−1⋅2=−3−2=−3−2=−5=−5

Therefore, the cross product ?×?v×u is:

?×?=(2,−16,−5)v×u=(2,−16,−5)

So, the vector (2,−16,−5)(2,−16,−5) is perpendicular to both ?=(3,1,−2)v=(3,1,−2) and ?=(2,−1,4)u=(2,−1,4).