Vector Formulas

Vector Formulas

Vectors are mathematical objects that express magnitude and direction. They are widely used in a variety of domains, including physics, engineering, and computer science, to describe physical quantities like as displacement, velocity, force, and acceleration. Vectors play an important role in representing the spatial and temporal behavior of objects and events. Learn more vectors and its related formulas in detail.

What are Vectors?

Vectors are mathematical concepts used to describe quantities with both magnitude and direction. They are commonly used in science, engineering, and mathematics to describe physical quantities like force, velocity, acceleration, displacement, and so on.

Vectors are represented geometrically as directed line segments, with the length representing the magnitude and the direction representing the vector’s direction. Vectors can be represented as arrows in space, with a starting point (origin) and an endpoint that indicate both direction and magnitude.

What are Vector Formulas

The different vector formulas used in Maths are mentioned below

Magnitude of a Vector

The magnitude (\(|\mathbf{v}|\) or \(\|\mathbf{v}\|\)) of a vector \(\mathbf{v} = \langle v_x, v_y, v_z \rangle\) in three-dimensional space is given by:

\[ |\mathbf{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2} \]

Unit Vector

A unit vector in the direction of a vector \(\mathbf{v}\) is obtained by dividing the vector by its magnitude. For a vector \(\mathbf{v}\), the unit vector is:

\[ \hat{\mathbf{v}} = \frac{\mathbf{v}}{|\mathbf{v}|} \]

Addition of Vectors

The addition of two vectors \(\mathbf{v}\) and \(\mathbf{u}\) is performed by adding their corresponding components:

\[ \mathbf{v} + \mathbf{u} = \langle v_x + u_x, v_y + u_y, v_z + u_z \rangle \]

Subtraction of Vectors

The subtraction of two vectors \(\mathbf{v}\) and \(\mathbf{u}\) is performed by subtracting their corresponding components:

\[ \mathbf{v} – \mathbf{u} = \langle v_x – u_x, v_y – u_y, v_z – u_z \rangle \]

Scalar Multiplication

Multiplying a vector \(\mathbf{v}\) by a scalar \(k\) scales the magnitude of the vector:

\[ k \mathbf{v} = \langle k v_x, k v_y, k v_z \rangle \]

Dot Product

The dot product (or scalar product) of two vectors \(\mathbf{v}\) and \(\mathbf{u}\) is a scalar quantity given by the sum of the products of their corresponding components:

\[ \mathbf{v} \cdot \mathbf{u} = v_x u_x + v_y u_y + v_z u_z \]

Cross Product

The cross product (or vector product) of two vectors \(\mathbf{v}\) and \(\mathbf{u}\) yields a new vector that is orthogonal to both original vectors. The magnitude of the cross product is equal to the product of the magnitudes of the original vectors multiplied by the sine of the angle between them:

\[ \mathbf{v} \times \mathbf{u} = \langle v_y u_z – v_z u_y, v_z u_x – v_x u_z, v_x u_y – v_y u_x \rangle \]

Angle between Two Vectors

The angle \(\theta\) between two vectors \(\mathbf{v}\) and \(\mathbf{u}\) can be calculated using the dot product formula and the magnitudes of the vectors. The formula to find the angle between two vectors is given by:

\[ \cos \theta = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}| |\mathbf{u}|} \]

where:

– \( \theta \) is the angle between the two vectors.

– \( \mathbf{v} \cdot \mathbf{u} \) denotes the dot product (or scalar product) of the vectors \( \mathbf{v} \) and \( \mathbf{u} \).

– \( |\mathbf{v}| \) and \( |\mathbf{u}| \) represent the magnitudes (or lengths) of the vectors \( \mathbf{v} \) and \( \mathbf{u} \), respectively.

Once you find \( \cos \theta \) using the dot product formula, you can then use the inverse cosine function (arccos) to find the angle \( \theta \) itself:

\[ \theta = \arccos \left( \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}| |\mathbf{u}|} \right) \]

This formula gives the angle between the two vectors in radians. If you want the angle in degrees, you can convert it by multiplying by \( \frac{180}{\pi} \).

Triangular Law of Additions

The resultant R will be the sum of two vectors if forces Vector A and Vector B are acting in the same direction.

If \(\vec{A}\) and \(\vec{B}\) are two vectors, their resultant \(\vec{R}\)$ is given by:
\[ \vec{R} = \vec{A} + \vec{B} \]

Parallelogram Law of Addition

The diagonal of a parallelogram formed at the same point can be used to represent the resultant of two forces, Vector A and Vector B if they are represented by the neighbouring sides of the parallelogram.

Examples using Vector Formula

Example 1:  Consider two vectors \( \mathbf{v} = \langle 2, -3, 1 \rangle \) and \( \mathbf{u} = \langle -1, 4, 2 \rangle \). Find the dot product of these two vectors.

Solution:

The dot product of two vectors \( \mathbf{v} \) and \( \mathbf{u} \) is calculated by multiplying their corresponding components and then summing up the results. Mathematically, it’s given by:

\[ \mathbf{v} \cdot \mathbf{u} = v_x u_x + v_y u_y + v_z u_z \]

Substituting the given values:

\[ \mathbf{v} \cdot \mathbf{u} = (2)(-1) + (-3)(4) + (1)(2) \]

\[ \mathbf{v} \cdot \mathbf{u} = -2 – 12 + 2 = -12 \]

So, the dot product of \( \mathbf{v} \) and \( \mathbf{u} \) is \( -12 \).

Example 2: Find the cross product of the vectors \( \mathbf{v} = \langle 1, 2, 3 \rangle \) and \( \mathbf{u} = \langle 4, 5, 6 \rangle \).

Solution:

The cross product of two vectors \( \mathbf{v} \) and \( \mathbf{u} \) is calculated using the determinant of a matrix formed by the components of the vectors. The cross product is given by:

\[ \mathbf{v} \times \mathbf{u} = \langle v_y u_z – v_z u_y, v_z u_x – v_x u_z, v_x u_y – v_y u_x \rangle \]

Substituting the given values:

\[ \mathbf{v} \times \mathbf{u} = \langle (2)(6) – (3)(5), (3)(4) – (1)(6), (1)(5) – (2)(4) \rangle \]

\[ \mathbf{v} \times \mathbf{u} = \langle 12 – 15, 12 – 6, 5 – 8 \rangle \]

\[ \mathbf{v} \times \mathbf{u} = \langle -3, 6, -3 \rangle \]

So, the cross product of \( \mathbf{v} \) and \( \mathbf{u} \) is \( \langle -3, 6, -3 \rangle \).

Example 3: Consider two vectors \( \mathbf{v} = \langle 3, -4 \rangle \) and \( \mathbf{u} = \langle -2, 5 \rangle \). Find the angle between these two vectors.

Solution:

First, we’ll calculate the dot product of the two vectors using the formula \( \mathbf{v} \cdot \mathbf{u} = v_x u_x + v_y u_y \):

\[ \mathbf{v} \cdot \mathbf{u} = (3)(-2) + (-4)(5) = -6 – 20 = -26 \]

Next, we’ll find the magnitudes of each vector using the formula \( |\mathbf{v}| = \sqrt{v_x^2 + v_y^2} \) and \( |\mathbf{u}| = \sqrt{u_x^2 + u_y^2} \):

\[ |\mathbf{v}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \]

\[ |\mathbf{u}| = \sqrt{(-2)^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29} \]

Now, we’ll use these values to calculate \( \cos \theta \) using the formula \( \cos \theta = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}| |\mathbf{u}|} \):

\[ \cos \theta = \frac{-26}{(5)(\sqrt{29})} \]

\[ \cos \theta = \frac{-26}{5\sqrt{29}} \]

Finally, we’ll find the angle \( \theta \) using the inverse cosine function \( \arccos \):

\[ \theta = \arccos \left( \frac{-26}{5\sqrt{29}} \right) \]

\[ \theta \approx \arccos \left( -0.903 \right) \]

\[ \theta \approx 2.665 \text{ radians} \]

Example 4: Find the angle between the vectors \( \mathbf{v} = \langle 1, 2, 3 \rangle \) and \( \mathbf{u} = \langle -1, 1, 4 \rangle \).

Solution:

Following the same steps as in Example 1, we calculate the dot product of the vectors:

\[ \mathbf{v} \cdot \mathbf{u} = (1)(-1) + (2)(1) + (3)(4) = -1 + 2 + 12 = 13 \]

Then, we find the magnitudes of the vectors:

\[ |\mathbf{v}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14} \]

\[ |\mathbf{u}| = \sqrt{(-1)^2 + 1^2 + 4^2} = \sqrt{18} \]

Now, we calculate \( \cos \theta \):

\[ \cos \theta = \frac{\mathbf{v} \cdot \mathbf{u}}{|\mathbf{v}| |\mathbf{u}|} = \frac{13}{\sqrt{14} \sqrt{18}} \]

Finally, we find \( \theta \):

\[ \theta = \arccos \left( \frac{13}{\sqrt{14} \sqrt{18}} \right) \]

\[ \theta \approx \arccos \left( 0.725 \right) \]

\[ \theta \approx 0.756 \text{ radians} \]