Components of a Vector and Linear Combinations

Position Vector of a Point and a Vector joining Two Points

Vectors can be used to represent the position of a point in space relative to a fixed reference point (origin) and to represent the displacement or translation from one point to another.

Position Vector of a Point

Consider a three-dimensional space with a fixed origin $O$. In a Cartesian coordinate system, the origin is typically taken as $O(0, 0, 0)$. Let P be any point in this space with coordinates $(x, y, z)$.

The position vector of the point P with respect to the origin O is the vector $\vec{OP}$. This vector is a directed line segment starting from the origin O and ending at the point P. It uniquely determines the location of point P in space relative to the origin.

Using the standard unit vectors $\hat{i}, \hat{j}, \hat{k}$ along the positive X, Y, and Z axes respectively, the position vector $\vec{OP}$ can be expressed in component form. If P has coordinates $(x, y, z)$, then its position vector $\vec{OP}$ is given by:

Here, $x, y, z$ are the scalar components of $\vec{OP}$ along the x, y, and z axes, respectively.

The magnitude of the position vector $\vec{OP}$ is the distance of the point P from the origin O. Using the distance formula (or Pythagorean theorem in 3D), the magnitude is:

Vector Joining Two Points

A vector connecting two points represents the displacement from the initial point to the terminal point. Let $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ be two points in space. We want to find the vector $\vec{AB}$, which is the vector directed from point A to point B.

We can use the concept of position vectors to find the vector $\vec{AB}$. Let O be the origin. The position vector of point A is $\vec{OA}$, and the position vector of point B is $\vec{OB}$.

Consider the triangle OAB. According to the Triangle Law of Vector Addition, if we go from O to A ($\vec{OA}$) and then from A to B ($\vec{AB}$), the resultant displacement is from O to B ($\vec{OB}$).

To find the vector $\vec{AB}$, we can rearrange this equation:

Substituting the component forms of $\vec{OB}$ and $\vec{OA}$:

This is the component form of the vector joining point A to point B. The scalar components of $\vec{AB}$ are $(x_2 - x_1)$, $(y_2 - y_1)$, and $(z_2 - z_1)$. This formula highlights a crucial concept: the vector from A to B is obtained by subtracting the coordinates of the initial point A from the coordinates of the terminal point B.

The magnitude of the vector $\vec{AB}$ is the distance between points A and B. Using the distance formula (which is the magnitude of the component form of $\vec{AB}$):

Answer:

Components of a Vector in a Plane (2D) and Magnitude

To effectively perform algebraic operations with vectors, it is often useful to represent them in terms of their components along the coordinate axes. We begin with vectors in a two-dimensional plane.

Component Form of a Vector in 2D

Consider a two-dimensional Cartesian coordinate system with a fixed origin O and perpendicular axes, the x-axis and the y-axis. Let $\hat{i}$ be the unit vector along the positive x-axis and $\hat{j}$ be the unit vector along the positive y-axis. These two unit vectors are orthogonal (perpendicular) and form a basis for the 2D plane.

Let $\vec{r}$ be any vector in the 2D plane originating from the origin O. Let P be the terminal point of $\vec{r}$, with coordinates $(x, y)$. Thus, $\vec{r} = \vec{OP}$.

Draw a perpendicular from point P to the x-axis, meeting it at point M. The coordinates of M are $(x, 0)$. The vector $\vec{OM}$ lies along the x-axis and has magnitude $|x|$. Since $\vec{OM}$ points along the x-axis, it can be written as $x\hat{i}$.

Draw a perpendicular from point P to the y-axis, meeting it at point N. The coordinates of N are $(0, y)$. The vector $\vec{ON}$ lies along the y-axis and has magnitude $|y|$. Since $\vec{ON}$ points along the y-axis, it can be written as $y\hat{j}$. Note that $\vec{MP} = \vec{ON}$.

Applying the Triangle Law of Vector Addition to $\triangle OMP$ (or the Parallelogram Law using $\vec{OM}$ and $\vec{ON}$ which form a rectangle OMPN), the position vector $\vec{OP}$ is the sum of vectors $\vec{OM}$ and $\vec{MP}$ (or $\vec{ON}$):

Substituting the component forms of $\vec{OM}$ and $\vec{MP}$ (where $\vec{MP} = \vec{ON} = y\hat{j}$):

This is the component form (or rectangular form) of the vector $\vec{r}$ in 2D. In this form:

• The scalars $x$ and $y$ are called the scalar components or rectangular components of the vector $\vec{r}$ along the x and y axes, respectively.
• The vectors $x\hat{i}$ and $y\hat{j}$ are called the vector components of $\vec{r}$ along the x and y axes, respectively.

Any vector $\vec{a}$ in a plane can be represented in the form $\vec{a} = a_1\hat{i} + a_2\hat{j}$, where $a_1$ and $a_2$ are its scalar components.

Magnitude of a Vector in 2D Component Form

If a vector $\vec{r} = x\hat{i} + y\hat{j}$ starts from the origin and ends at $P(x,y)$, its magnitude is the distance OP. Applying the Pythagorean theorem to the right-angled triangle OMP:

Taking the square root, we get the magnitude:

For a general vector $\vec{a} = a_1\hat{i} + a_2\hat{j}$, its magnitude is $|\vec{a}| = \sqrt{a_1^2 + a_2^2}$.

If the vector does not start from the origin but joins point $A(x_1, y_1)$ to point $B(x_2, y_2)$, its component form is $\vec{AB} = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j}$. The magnitude is then $|\vec{AB}| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$, which is the distance formula in 2D.

Answer:

Components of a Vector in Space (3D) and Magnitude

The concept of vector components extends naturally from two dimensions to three dimensions. In 3D space, a vector can be resolved into components along the three perpendicular coordinate axes.

Component Form of a Vector in 3D

Consider a three-dimensional Cartesian coordinate system with origin O and the three mutually perpendicular axes X, Y, and Z. Let $\hat{i}, \hat{j}, \hat{k}$ be the standard unit vectors along the positive X, Y, and Z axes, respectively. These three unit vectors are orthogonal and form a basis for 3D space.

Let $\vec{r}$ be any vector in 3D space originating from the origin O, with its terminal point at P$(x, y, z)$. So, $\vec{r} = \vec{OP}$.

By extending the method used in 2D (applying vector addition laws or considering projections), the position vector $\vec{OP}$ can be uniquely expressed as the sum of three vectors, each lying along one of the coordinate axes. If we project the point P onto the x, y, and z axes, we get points corresponding to the scalar components $x, y, z$. The vector components along the axes will be $x\hat{i}$ (along X), $y\hat{j}$ (along Y), and $z\hat{k}$ (along Z).

The position vector $\vec{r}$ can be written as the sum of these vector components:

This is the component form (or rectangular form) of the vector $\vec{r}$ in 3D space. In this form:

• The scalars $x, y, z$ are called the scalar components or rectangular components of the vector $\vec{r}$ along the x, y, and z axes, respectively.
• The vectors $x\hat{i}, y\hat{j}, z\hat{k}$ are called the vector components of $\vec{r}$ along the x, y, and z axes, respectively.

Any vector $\vec{a}$ in 3D space can be represented in the form $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, where $a_1, a_2, a_3$ are its scalar components.

Magnitude of a Vector in 3D Component Form

If a vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ starts from the origin O and ends at $P(x,y,z)$, its magnitude is the distance OP. This can be found by applying the Pythagorean theorem twice or using the 3D distance formula, which is derived from it.

The magnitude of $\vec{r}$ is:

For a general vector $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, its magnitude is $|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$.

If the vector does not start from the origin but joins point $A(x_1, y_1, z_1)$ to point $B(x_2, y_2, z_2)$, its component form is $\vec{AB} = (x_2-x_1)\hat{i} + (y_2-y_1)\hat{j} + (z_2-z_1)\hat{k}$. The magnitude is then $|\vec{AB}| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^1}$, which is the distance formula in 3D, matching what we found earlier.

Answer:

Linear Combination of Vectors

Vector operations like addition and scalar multiplication can be combined to form new vectors. A fundamental concept that arises from these operations is the linear combination of vectors.

Definition of a Linear Combination

A vector $\vec{r}$ is said to be a linear combination of a set of vectors $\vec{v_1}, \vec{v_2}, ..., \vec{v_n}$ if $\vec{r}$ can be expressed as the sum of scalar multiples of these vectors.

Mathematically, $\vec{r}$ is a linear combination of $\vec{v_1}, \vec{v_2}, ..., \vec{v_n}$ if there exist scalars $c_1, c_2, ..., c_n$ such that:

The scalars $c_1, c_2, ..., c_n$ are called the coefficients of the linear combination. These coefficients are real numbers.

For example, $2\vec{a} + 3\vec{b}$ is a linear combination of vectors $\vec{a}$ and $\vec{b}$ with coefficients $c_1 = 2$ and $c_2 = 3$. Similarly, $\vec{a} - \vec{b} = 1\cdot\vec{a} + (-1)\cdot\vec{b}$ is a linear combination of $\vec{a}$ and $\vec{b}$ with coefficients $c_1 = 1$ and $c_2 = -1$.

Calculation of a Linear Combination in Component Form

When vectors are given in component form, calculating a linear combination is straightforward. We apply the rules for scalar multiplication and vector addition component-wise.

For instance, consider two vectors in 3D space in component form:

Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$

Let $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$

To find the linear combination $x\vec{a} + y\vec{b}$, where $x$ and $y$ are scalars, we first perform the scalar multiplications and then the vector addition:

Now, add the resulting vectors:

$$ x\vec{a} + y\vec{b} = ((xa_1) + (yb_1))\hat{i} + ((xa_2) + (yb_2))\hat{j} + ((xa_3) + (yb_3))\hat{k} $$

Thus, the scalar components of the linear combination $x\vec{a} + y\vec{b}$ are $(xa_1 + yb_1)$, $(xa_2 + yb_2)$, and $(xa_3 + yb_3)$. This principle extends to any number of vectors in any dimension.

Answer:

Collinearity and Coplanarity of Vectors using Components

The component form of vectors provides powerful algebraic tools to determine important geometric relationships between vectors, specifically collinearity (lying on the same line) and coplanarity (lying on the same plane).

Condition for Collinearity using Components

Two non-zero vectors $\vec{a}$ and $\vec{b}$ are collinear if and only if they are parallel, meaning one is a scalar multiple of the other. That is, $\vec{a} = \lambda \vec{b}$ for some non-zero scalar $\lambda$.

Let the vectors be given in component form:

$\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$

$\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$

If $\vec{a} = \lambda \vec{b}$, then by the definition of scalar multiplication in component form:

Since two vectors are equal if and only if their corresponding components are equal, we have:

Assuming none of $b_1, b_2, b_3$ are zero, we can write:

Thus, two non-zero vectors are collinear if and only if their corresponding scalar components are proportional.

Important Note: If a component of $\vec{b}$ is zero (say $b_i = 0$), then for $\vec{a} = \lambda \vec{b}$ to hold for some non-zero $\lambda$, the corresponding component of $\vec{a}$ ($a_i$) must also be zero. If $a_i = 0$ and $b_i = 0$, the ratio $\frac{a_i}{b_i}$ is indeterminate, but the condition $a_i = \lambda b_i$ (which becomes $0 = \lambda \cdot 0$) is satisfied for any $\lambda$. So, the rule of proportionality should be applied carefully: $\vec{a}$ and $\vec{b}$ are collinear if and only if $a_i = \lambda b_i$ for all $i=1, 2, 3$ for the same non-zero scalar $\lambda$. Practically, check the ratios of corresponding non-zero components. If $b_i=0$, check if $a_i$ is also 0.

Another condition for collinearity of two non-zero vectors $\vec{a}$ and $\vec{b}$, which we will see later, is that their cross product is the zero vector: $\vec{a} \times \vec{b} = \vec{0}$.

Answer:

Condition for Coplanarity using Components

Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if they lie in the same plane when their initial points are made to coincide.

Algebraically, three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if and only if one of them can be expressed as a linear combination of the other two (provided the other two are non-collinear). More generally, three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if and only if there exist scalars $x, y, z$, not all zero, such that $x\vec{a} + y\vec{b} + z\vec{c} = \vec{0}$.

A convenient and standard method to check for coplanarity using components is by evaluating the scalar triple product of the three vectors. The scalar triple product of $\vec{a}, \vec{b}, \vec{c}$ is denoted by $[\vec{a} \vec{b} \vec{c}]$ or $(\vec{a} \times \vec{b}) \cdot \vec{c}$. Geometrically, the absolute value of the scalar triple product represents the volume of the parallelepiped formed by the three vectors originating from the same point. If the vectors are coplanar, they cannot form a parallelepiped with a non-zero volume, so the volume must be zero.

Let the vectors be given in component form:

$\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$

$\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$

$\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$

The scalar triple product $(\vec{a} \times \vec{b}) \cdot \vec{c}$ can be calculated as the determinant of the matrix formed by the scalar components of the three vectors:

The vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if and only if their scalar triple product is zero.

Answer:

Theorem on Two Non-Zero Non-Collinear Vectors (Basis in 2D)

This theorem is fundamental to understanding how vectors in a plane can be represented and highlights the concept of a basis.

Theorem Statement

If $\vec{a}$ and $\vec{b}$ are any two non-zero and non-collinear vectors in a plane, then any vector $\vec{r}$ in that same plane can be expressed uniquely as a linear combination of $\vec{a}$ and $\vec{b}$.

That is, for any vector $\vec{r}$ coplanar with $\vec{a}$ and $\vec{b}$, there exist a unique pair of scalars $x$ and $y$ such that:

Explanation and Proof of Uniqueness

Let $\vec{a}$ and $\vec{b}$ be two non-zero, non-collinear vectors in a plane. Let $\vec{r}$ be any vector in the same plane.

Consider the vectors $\vec{a}, \vec{b}, \vec{r}$ originating from the same point O. Since $\vec{a}, \vec{b}, \vec{r}$ are all in the same plane, they are coplanar.

Since $\vec{a}$ and $\vec{b}$ are non-zero and non-collinear, the vector $x\vec{a} + y\vec{b}$ can represent any vector in the plane containing $\vec{a}$ and $\vec{b}$ by varying the scalars $x$ and $y$. This can be visualized using the parallelogram law: by scaling $\vec{a}$ and $\vec{b}$ and adding them, we can reach any point in the plane relative to the origin O. Thus, $\vec{r}$ can be expressed as a linear combination of $\vec{a}$ and $\vec{b}$.

Now, we need to prove that this representation is unique. Assume that $\vec{r}$ can be expressed as a linear combination of $\vec{a}$ and $\vec{b}$ in two different ways:

where $x_1, y_1, x_2, y_2$ are scalars.

Equating the two expressions for $\vec{r}$:

Rearranging the terms:

Using the distributive property of scalar multiplication over scalar addition:

Let $k_1 = x_1 - x_2$ and $k_2 = y_1 - y_2$. The equation becomes:

We need to show that the only solution to this equation is $k_1 = 0$ and $k_2 = 0$.

Suppose $k_1 \neq 0$. Then we can write:

This equation shows that $\vec{a}$ is a scalar multiple of $\vec{b}$ (where the scalar is $-\frac{k_2}{k_1}$). This implies that $\vec{a}$ and $\vec{b}$ are collinear. However, our initial condition states that $\vec{a}$ and $\vec{b}$ are non-collinear. This is a contradiction.

Therefore, our assumption that $k_1 \neq 0$ must be false. The only possibility is $k_1 = 0$.

Substituting $k_1=0$ back into the equation $k_1\vec{a} + k_2\vec{b} = \vec{0}$, we get:

Since $\vec{b}$ is a non-zero vector by hypothesis, this equation can only hold if the scalar $k_2$ is zero.

Thus, we have shown that $x_1 = x_2$ and $y_1 = y_2$. This proves that there is only one unique pair of scalars $x$ and $y$ for which $\vec{r} = x\vec{a} + y\vec{b}$.

Concept of Basis

A set of vectors $\{\vec{v_1}, \vec{v_2}, ..., \vec{v_n}\}$ is called a basis for a vector space if every vector in that space can be expressed as a unique linear combination of these vectors.

The theorem above shows that any two non-zero, non-collinear vectors $\{\vec{a}, \vec{b}\}$ in a plane form a basis for the set of all vectors in that plane (which constitutes a 2-dimensional vector space, often denoted $\mathbb{R}^2$).

The most common and convenient basis for the 2D plane is the set of standard unit vectors $\{\hat{i}, \hat{j}\}$. Any vector $\vec{r}$ in the plane can be uniquely written as $\vec{r} = x\hat{i} + y\hat{j}$, where $x$ and $y$ are its scalar components. Here, the coefficients of the linear combination are simply the components of the vector.

Extension to 3D Space

A similar theorem holds for three-dimensional space: Any vector $\vec{r}$ in 3D space can be expressed uniquely as a linear combination of three non-zero, non-coplanar vectors $\vec{a}, \vec{b}, \vec{c}$. That is, there exist unique scalars $x, y, z$ such that $\vec{r} = x\vec{a} + y\vec{b} + z\vec{c}$. These three vectors $\{\vec{a}, \vec{b}, \vec{c}\}$ form a basis for 3D space ($\mathbb{R}^3$).

The standard basis for 3D space is the set of orthogonal unit vectors $\{\hat{i}, \hat{j}, \hat{k}\}$. Any vector $\vec{r}$ in space can be uniquely written as $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$, where $x, y, z$ are its scalar components.