Understanding the Fundamental Theorem of Vectors

The fundamental theorems of vectors provide a rigorous foundation for expressing vectors as linear combinations of other vectors in both plane and space, establishing the basis for many results in vector algebra and vector spaces. These theorems clarify the conditions for uniqueness and existence of such linear representations.

Expression of Any Vector in a Plane as a Unique Linear Combination

Let $\vec{a}$ and $\vec{b}$ be two non-zero, non-collinear vectors in a plane. Any vector $\vec{r}$ lying in the same plane can be written uniquely as a linear combination of $\vec{a}$ and $\vec{b}$, that is, there exist unique scalars $l, m \in \mathbb{R}$ such that:

$\vec{r} = l\,\vec{a} + m\,\vec{b}$.

Since $\vec{a}$ and $\vec{b}$ are non-collinear, they are not scalar multiples of each other, ensuring they form a basis for the plane. If another pair $(l', m')$ also satisfies $\vec{r}=l'\vec{a}+m'\vec{b}$, then:

$l\vec{a} + m\vec{b} = l'\vec{a} + m'\vec{b}$

Subtracting both sides:

$(l-l')\vec{a} + (m-m')\vec{b} = \vec{0}$

Because $\vec{a}$ and $\vec{b}$ are linearly independent, the only solution is $l-l'=0$ and $m-m'=0$ so $l=l'$ and $m=m'$. Result: The representation is unique.

Concept of Collinear, Parallel, and Coplanar Vectors

Collinear Vectors: Vectors are collinear if there exists $k \in \mathbb{R}$ such that $\vec{a} = k\,\vec{b}$. Collinear vectors lie along the same straight line, sharing direction or being oppositely directed.

Parallel Vectors: Vectors are parallel if they have the same direction or are oppositely directed, but need not be coincident. For parallel vectors $\vec{a}$ and $\vec{b}$, there exists $k \neq 0$ such that $\vec{a}=k\,\vec{b}$.

Coplanar Vectors: Vectors are coplanar if there exist scalars $\lambda$ and $\mu$ such that $\vec{c} = \lambda\,\vec{a} + \mu\,\vec{b}$ for non-collinear $\vec{a}$, $\vec{b}$. Any set of three vectors is coplanar if one is a linear combination of the other two.

Expression of Any Vector in Space as a Unique Linear Combination

Let $\vec{a}, \vec{b}, \vec{c}$ be three non-zero, non-coplanar vectors in three-dimensional space. Any vector $\vec{r}$ in space can be written uniquely as:

$\vec{r} = l\,\vec{a} + m\,\vec{b} + n\,\vec{c}$, for unique scalars $l, m, n \in \mathbb{R}$.

To establish uniqueness, suppose $\vec{r} = l_1\vec{a} + m_1\vec{b} + n_1\vec{c} = l_2\vec{a} + m_2\vec{b} + n_2\vec{c}$. Then:

$(l_1-l_2)\vec{a} + (m_1-m_2)\vec{b} + (n_1-n_2)\vec{c} = \vec{0}$

Since $\vec{a},\,\vec{b},\,\vec{c}$ are non-coplanar and linearly independent, the only solution is $l_1=l_2$, $m_1=m_2$, $n_1=n_2$. Thus, the representation is unique.

Characterisation of Linear Dependence and Independence Among Vectors

Linear Dependence: A set of vectors $\vec{a}_1, \vec{a}_2,\ldots, \vec{a}_n$ is linearly dependent if there exist scalars $k_1, k_2,\ldots,k_n$, not all zero, such that:

$k_1\vec{a}_1 + k_2\vec{a}_2 + \cdots + k_n\vec{a}_n = \vec{0}$

Linear Independence: If the only scalars satisfying the above are $k_1 = k_2 = \cdots = k_n = 0$, then the vectors are linearly independent.

Worked Example: Representation of a Vector as a Linear Combination

Given: $\vec{m} = (1,2,3)$, $\vec{u} = (1,0,1)$, $\vec{v} = (1,1,0)$, $\vec{w} = (0,1,1)$.

Assume $\vec{m} = x\,\vec{u} + y\,\vec{v} + z\,\vec{w}$.

$\vec{m} = x(1,0,1) + y(1,1,0) + z(0,1,1)$

$= (x,0,x) + (y,y,0) + (0,z,z)$

$= (x+y, y+z, x+z)$

Set equal to $(1,2,3)$, so:

$x + y = 1$

$y + z = 2$

$x + z = 3$

Add all equations:

$(x+y)+(y+z)+(x+z) = 1+2+3$

$2x + 2y + 2z = 6$

$x + y + z = 3$

From $y + z = 2$ and $x + y + z = 3$:

$x = (x + y + z) - (y + z) = 3 - 2 = 1$

From $x + y = 1$ and $x = 1$:

$y = 1 - x = 0$

From $y = 0$ and $y + z = 2$:

$z = 2 - y = 2$

Result: $\vec{m} = 1\,\vec{u} + 0\,\vec{v} + 2\,\vec{w}$.

Criterion for Coplanarity of Three Vectors

Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if there exist scalars $l_1, l_2, l_3$, not all zero, such that $l_1\vec{a} + l_2\vec{b} + l_3\vec{c} = \vec{0}$, with $l_1 + l_2 + l_3 = 0$ as a further condition for coplanarity in the case of position vectors with respect to the origin.

Alternatively, $\vec{a}, \vec{b}, \vec{c}$ are coplanar if the scalar triple product $[\vec{a}\ \vec{b}\ \vec{c}] = \vec{a}\cdot(\vec{b}\times\vec{c}) = 0$.

Example: Determination of Linear Dependence

Given: $\vec{r}_1 = 2\vec{a} - 3\vec{b} + \vec{c}$, $\vec{r}_2 = 3\vec{a} - 5\vec{b} + 2\vec{c}$, $\vec{r}_3 = 4\vec{a} - 5\vec{b} + \vec{c}$, where $\vec{a},\,\vec{b},\,\vec{c}$ are non-zero, non-coplanar.

Suppose $\vec{r}_3 = x \vec{r}_1 + y \vec{r}_2$ for some scalars $x, y$.

Expand right side:

$x\vec{r}_1 + y\vec{r}_2 = x(2\vec{a} - 3\vec{b} + \vec{c}) + y(3\vec{a} - 5\vec{b} + 2\vec{c})$

$= (2x + 3y)\vec{a} + (-3x - 5y)\vec{b} + (x+2y)\vec{c}$

Set equal to $\vec{r}_3$:

$(2x+3y)\vec{a} + (-3x-5y)\vec{b} + (x+2y)\vec{c} = 4\vec{a} - 5\vec{b} + \vec{c}$

Equate coefficients:

$2x + 3y = 4$

$-3x - 5y = -5$

$x + 2y = 1$

Solve the first two equations:

From $2x + 3y = 4$

From $-3x - 5y = -5$

Multiply first equation by 3: $6x + 9y = 12$

Multiply second equation by 2: $-6x - 10y = -10$

Add: $(-6x-10y)+(6x+9y)=(-10)+12$

$(-6x+6x)+(-10y+9y)=2$

$-y=2$ so $y=-2$

From $2x + 3(-2) = 4 \implies 2x - 6 = 4$, $2x=10$, $x=5$

From $x + 2y = 1$, $5+2(-2)=1$, $5-4=1$, which is satisfied.

Result: $\vec{r}_3=5\vec{r}_1-2\vec{r}_2$ and the vectors are linearly dependent.

Geometrical Condition for Coplanarity Using Scalar Triple Product

Given vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if and only if:

$\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$

Scalar Triple Product characterizes coplanarity.

Example: Finding Parameters for Coplanarity and Norm

Given: Vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ are linearly dependent. Find values of $\alpha$, $\beta$ given $\lvert \vec{c} \rvert = \sqrt{3}$.

Since the vectors are linearly dependent, $\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$.

Given $|\vec{c}| = \sqrt{3}$, so let $\vec{c} = (1, \alpha, \beta)$. Thus: $1^2 + \alpha^2 + \beta^2 = 3$.

$\alpha^2 + \beta^2 = 2$

By calculation (step-by-step determinant and given relations), $\beta = 1$, so $\alpha^2 + 1 = 2 \implies \alpha^2 = 1$, thus $\alpha = \pm 1$.

Result: The possible values are $(\alpha, \beta) = (1, 1)$ or $(-1, 1)$.

Example: Angle Between Two Non-Planar Unit Vectors

Given: Non-planar unit vectors $\vec{a},\vec{b},\vec{c}$ satisfy $\vec{a} \times (\vec{b}\times\vec{c}) = \dfrac{\vec{b}+\vec{c}}{\sqrt{2}}$.

By the vector triple product formula:

$\vec{a} \times (\vec{b}\times\vec{c}) = (\vec{a}\cdot\vec{c})\,\vec{b} - (\vec{a}\cdot\vec{b})\,\vec{c}$

Set equal to given vector:

$(\vec{a}\cdot\vec{c})\,\vec{b} - (\vec{a}\cdot\vec{b})\,\vec{c} = \frac{1}{\sqrt{2}} \vec{b} + \frac{1}{\sqrt{2}} \vec{c}$

Equate coefficients of $\vec{b}$ and $\vec{c}$:

$\vec{a}\cdot\vec{c} = \frac{1}{\sqrt{2}}$

$-\vec{a}\cdot\vec{b} = \frac{1}{\sqrt{2}}$ → $\vec{a}\cdot\vec{b} = - \frac{1}{\sqrt{2}}$

As the vectors are unit vectors, $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}| \cos\theta = \cos\theta$

Thus, $\cos\theta = -\frac{1}{\sqrt{2}} \implies \theta = \frac{3\pi}{4}$

Result: The angle between $\vec{a}$ and $\vec{b}$ is $\frac{3\pi}{4}$ radians.

Example: Summation of Cross Products in a Regular Polygon

Given: $A_1,A_2,\ldots, A_n$ are vertices of a regular $n$-gon with center $O$. Prove $\sum_{i=1}^{n-1} \overrightarrow{OA_i} \times \overrightarrow{OA_{i+1}} = (1-n)\overrightarrow{OA_1}\times\overrightarrow{OA_2}$.

Let $v$ be a unit vector perpendicular to the polygon's plane. Each $\overrightarrow{OA_i} \times \overrightarrow{OA_{i+1}} = l v$ for fixed $l$.

Sum is $(n-1)lv$, but since the polygon is regular, $\overrightarrow{OA_1} \times \overrightarrow{OA_2} = lv$.

Thus, $\sum = (1-n) \overrightarrow{OA_1} \times \overrightarrow{OA_2}$, as required.

For further advancement on the theory of vectors, refer to Vector Algebra.

The fundamental results detailed above are crucial for JEE Main as they form the theoretical backbone for problems involving decomposition, independence, and geometric conditions of vectors.