In mathematics, a function usually gives one value for each input. But in many cases, one value is not enough to describe a situation. For example, position, velocity, and force have both size and direction. To represent such quantities, we use vector-valued functions. A vector-valued function gives a vector as its output instead of a single number. These functions help us describe motion and paths in space in a simple and clear way.
Vector Vector Valued Function can be written as:
\vec{r}(t) = f(t)\hat{i} + g(t)\hat{j} + h(t)\hat{k}
Where
- t is the input variable (often representing time)
- f(t), g(t), h(t) are the component functions, each of which is a real-valued function.
Examples on Vector-Valued Functions
2D Vector-Valued Function:
\vec{r}(t) = \langle \cos(t),\ \sin(t) \rangle This represents a parametric equation of a circle.
3D Vector-Valued Function:
\vec{r}(t) = \langle t,\ t^{2},\ t^{3} \rangle This represents a space curve in three dimensions.
More generally, a vector-valued function is defined as:
\vec{f}(t) = \langle f_{1}(t),\ f_{2}(t),\ \ldots,\ f_{n}(t) \rangle
where t is the independent variable and
Magnitude of Vectors:
The magnitude (or length) of a vector
f_1(t), f_2(t), ..., f_n(t) in n-dimensional space is given by:
|\vec{v}| = \sqrt{v_{1}^{2} + v_{2}^{2} + \cdots + v_{n}^{2}} For example, for
\vec{v} = \langle 3, 4 \rangle in 2-dimensional space:
|\vec{v}| = \sqrt{3^{2} + 4^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5
Unit Vector:
To find the unit vector
\hat{v} in the direction of a given vector\vec{v} , divide the vector by its magnitude:
\hat{v} = \frac{\vec{v}}{|\vec{v}|}
Zero Vector:
The zero vector is a vector where all components are zero. In n-dimensional space:
\vec{0} = \langle 0, 0, \ldots, 0 \rangle The zero vector has a magnitude of 0 and does not have a specific direction.
Vector Operations:
Addition: The sum of two vectors
\vec{u} = \langle u_{1}, u_{2}, \ldots, u_{n} \rangle and\vec{v} = \langle v_{1}, v_{2}, \ldots, v_{n} \rangle is
\vec{u} + \vec{v} = \langle u_{1} + v_{1},\ u_{2} + v_{2},\ \ldots,\ u_{n} + v_{n} \rangle Subtraction: The difference between two vectors 𝑢 and 𝑣 is:
\vec{u} - \vec{v} = \langle u_{1} - v_{1},\ u_{2} - v_{2},\ \ldots,\ u_{n} - v_{n} \rangle Scalar Multiplication: Multiplying a vector
\vec{v} = \langle v_{1}, v_{2}, \ldots, v_{n} \rangle by a scalar gives:
c\vec{v} = \langle cv_{1},\ cv_{2},\ \ldots,\ cv_{n} \rangle Dot Product: The dot product of two vectors
\vec{u} = \langle u_{1}, u_{2}, \ldots, u_{n} \rangle and\vec{v} = \langle v_{1}, v_{2}, \ldots, v_{n} \rangle is:
\vec{u} \cdot \vec{v} = u_{1}v_{1} + u_{2}v_{2} + \cdots + u_{n}v_{n} Cross Product: The cross product of two vectors
\vec{u} = \langle u_{1}, u_{2}, u_{3} \rangle and \vec{v} = \langle v_{1}, v_{2}, v_{3} \rangle in 3-dimensional space is:
\vec{u} \times \vec{v} = \langle u_{2}v_{3} - u_{3}v_{2},\ u_{3}v_{1} - u_{1}v_{3},\ u_{1}v_{2} - u_{2}v_{1} \rangle
Vector Components:
A vector in 𝑛-dimensional space can be broken down into its components along each axis. For a vector
\vec{v} = \langle v_{1}, v_{2}, \ldots, v_{n} \rangle:
- 𝑣1 is the component along the 𝑥-axis.
- 𝑣2 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 :Find the angle between the vectors v = (3, -4) and u = (5, 2).
Solution:
Step 1: Compute the dot product
\vec{v} \cdot \vec{u}:
\vec{v} \cdot \vec{u} = (3)(5) + (-4)(2) = 15 - 8 = 7 Step 2: Compute the magnitudes
|\vec{v}| = \sqrt{3^{2} + (-4)^{2}} = \sqrt{9 + 16} = \sqrt{25} = 5
|\vec{u}| = \sqrt{5^{2} + 2^{2}} = \sqrt{25 + 4} = \sqrt{29} Step 3: Compute cos(𝜃):
\cos(\theta) = \frac{\vec{v} \cdot \vec{u}}{|\vec{v}|\ |\vec{u}|} = \frac{7}{5\sqrt{29}} \approx \frac{7}{26.93} \approx 0.2601 Step 4: Compute θ:
\theta = \cos^{-1}(0.2601) \approx 74.93° So, the angle between the vectors 𝑣v and 𝑢u is approximately 75.15∘
Practice Problem 2: 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.
Solution:
a) The magnitude of 𝑢:
|\vec{u}| = \sqrt{4^{2} + (-2)^{2} + 1^{2}} = \sqrt{16 + 4 + 1} = \sqrt{21} b) The unit vector in the direction of 𝑣:
|\vec{v}| = \sqrt{(-1)^{2} + 3^{2} + (-2)^{2}} = \sqrt{1 + 9 + 4} = \sqrt{14}
\hat{v} = \frac{1}{\sqrt{14}}(-1, 3, -2) = \left\langle \frac{-1}{\sqrt{14}},\ \frac{3}{\sqrt{14}},\ \frac{-2}{\sqrt{14}} \right\rangle c) The dot product of 𝑣 and 𝑢:
\vec{v} \cdot \vec{u} = (-1)(4) + (3)(-2) + (-2)(1) = -4 - 6 - 2 = -12 d) The angle between 𝑣 and 𝑢:
\cos(\theta) = \frac{\vec{v} \cdot \vec{u}}{|\vec{v}|\ |\vec{u}|} = \frac{-12}{\sqrt{14} \cdot \sqrt{21}} = \frac{-12}{\sqrt{294}} \approx -0.6987
\theta = \cos^{-1}(-0.6987) \approx 134.43°
Practice Problem 4: Find the cross product of the vectors v = (2, 1, -3) and u = (-1, 4, 2).
Solution:
First component:
v_{2}u_{3} - v_{3}u_{2} = (1)(2) - (-3)(4) = 2 + 12 = 14 Second component:
v_{3}u_{1} - v_{1}u_{3} = (-3)(-1) - (2)(2) = 3 - 4 = -1 Third component:
v_{1}u_{2} - v_{2}u_{1} = (2)(4) - (1)(-1) = 8 + 1 = 9
\vec{v} \times \vec{u} = (14,\ -1,\ 9) So, the cross product of the vectors 𝑣 = (2, 1, −3) and 𝑢 = (−1, 4, 2) is (14, −1, 9).
Practice Problem 5: Find a vector that is perpendicular to both v = (3, 1, -2) and u = (2, -1, 4).
Solution:
We compute
\vec{v} \times \vec{u}:
\vec{v} \times \vec{u} = \langle v_{2}u_{3} - v_{3}u_{2},\ v_{3}u_{1} - v_{1}u_{3},\ v_{1}u_{2} - v_{2}u_{1} \rangle First component: 1⋅4 − (−2) ⋅ (−1) = 4 − 2 = 4 − 2 = 2
Second component: (−2)⋅2 − 3⋅4 = − 4 − 12 = − 4 − 12 = − 16
Third component: 3⋅(−1) − 1⋅ 2 = - 5
\vec{v} \times \vec{u} = (2,\ -16,\ -5) So, the vector (2,−16,−5) is perpendicular to both 𝑣 = (3, 1, −2) and 𝑢 = (2, −1, 4).