Magnitude of a Vector: Formula and Direction

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Vectors are objects that share the properties of magnitude and direction. It is necessary to calculate the vector length before we can calculate its magnitude. A vector quantity is, for example, a velocity, displacement, force, momentum, etc. Scalar quantities, however, include speed, mass, distance, volume, temperature, etc. Vectors have both magnitude and direction, while scalars have only magnitude. When calculating the length of a vector (say v), the magnitude of the vector formula is used. In a nutshell, the vector length is the distance from the initial point to the endpoint. Distances are calculated by using the formula for distance.

Table of Content

Magnitude of a Vector Formula
Direction of a Vector
Things to Remember
Sample Questions

Keywords- Magnitude of vector, Vector, Magnitude, Formula, Direction, Distance, Root, Coordinates

Also read: Isosceles Triangle Theorems

Magnitude of a Vector Formula

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Magnitude is a way to summarize a vector's numeric value. The magnitude describes the magnitude of a vector. It is impossible to have a negative magnitude for a vector, which means its magnitude is always positive or zero.

In the case of AB, we will assume it has both magnitude and direction. The distance between the initial point A and endpoint B is required to calculate the magnitude of the vector \(\overrightarrow{AB}\). Consider the XY plane, where A possesses coordinates (x₀, y₀) and B possesses coordinates (x₁, y₁). As a result, the magnitude of the vector \(\overrightarrow{AB}\) can be calculated by distance formula can be written as:

\(\overrightarrow{AB}\) = \(\sqrt{(x_1-x_0)^2+ (y_1-y_0)^2}\)

Using the example of starting at (x, y) and ending at the origin, the magnitude of the vector formula is as follows: \(\mid \overrightarrow{AB} \mid\) = \(\sqrt{x^2 + y^2}\)

From the coordinates of a two-dimensional vector, we can determine its magnitude

Identification of its components is the first step.
Identify their sums by finding the squares of their perimeters.
Take the square root of the result.

The formula for calculating a vector's magnitude follows v = (x₁, y₁) is: \(\mid \overrightarrow{V} \mid\) = \(\sqrt{x^2 + y^2}\)

This formula is derived from the Pythagorean theorem.

A three-dimensional vector's magnitude can be calculated from its coordinates by using the formula below.

Identification of its components is the first step.
Secondly, sum the squares of each component.
Thirdly, take the square root of the sum.

Thus, the formula to determine the magnitude of a vector

V = (x₁+ y₁+ z₁) is \(\mid {V} \mid\)= \(\sqrt{x_1^2+y_1^2+z_1^2}\)

Direction of a Vector

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An angle with the horizontal line is what determines the direction of a vector. Methods that are used to determine the vector's direction \(\overrightarrow{AB}\) is:

tan α = y/x; zero is the endpoint.

In mathematical terms, Y is the change in a vertical line and X is the change in a horizontal line

Or tan α = y₁– y₀ / x₁ – x₀ ; where (x₀ , y₀) is initial point and (x₁ , y₁) is the endpoint.

Also read: Difference between Sequence and Series

Things to Remember

To determine the magnitude of a vector, its length must be determined.
Whenever a vector has a magnitude, it is called ?a?.
For a two-dimensional vector a, where a = (a₁, a₂ ), ||a|| = √a¹₁+a²₂
For a three-dimensional vector a, where a = (a₁, a₂, a₃), ||a|| = √a²₁+a²₂+a²₃
It is always possible to convert a vector to arbitrary dimensions by using the formula for magnitude.
Let's take a look of such a fourth-dimensional vector: a, where a = (a₁, a₂, a₃, a₄),

|a|= √a²₁+a²₂+a²₃+a²₄

Also read: First Order Differential Equation

Sample Questions

Ques. Find a vector of magnitude 11 in the direction opposite to that of \(\overrightarrow{PQ}\), where P and Q are the points (1,3,2) and (-1,0,8), respectively. (4 marks)

Ans. The vector with initial point P(1,3,2) and terminal point Q (-1,0,8) is given by

Ques. Find the position vector of a point R which divides the line joining the two points P and Q with position vectors \(\overrightarrow{OP}\) = \(\overrightarrow{2a} + \overrightarrow{b}\) and \(\overrightarrow{OQ}\)= \(\overrightarrow{a} - \overrightarrow{2b}\), respectively, in the ratio 1:2, (i) internally and (ii) externally. (4 marks)

Ans. (i) The position vector of the point R dividing the join of P and Q internally in the ratio 1:2 is given by

(ii) The position vector of the point R' dividing the join of P and Q in the ratio 1:2 externally is given by

Ques. If the points (-1,-1,2),(2,m,5) and (3,11,6) are collinear, find the value of m. (4 marks)

Ans. Let the given points be A (-1,-1,2),B(2,m,5) and C (3,11,6). Then

Ques. Using vectors, prove that cos(A-B)= cosAcosB + sinAsinB. (4 marks)

Ans. Let \(\hat{OP}\) and \(\hat{OQ}\) be unit vectors making angles A and B, respectively, with positive direction of x-axis. Then ∠QOP=A-B [Fig. 10.1]

Ques. Find a vector r of magnitude 3√2 units which makes an angle of \(\frac{\pi}{4}\) and \(\frac{I}{2}\) with y and z - axes, respectively. (4 marks)

Ans. Here m=cos\(\frac{\pi}{4}\)= \(\frac{1}{\sqrt{2}}\) and n = cos\(\frac{\pi}{2}\)=0.

Therefore, 1²+ m²+ n²= 1 gives

Therefore, 12 + m2 + n2= 1 gives — Therefore, 1²+ m²+ n²= 1

Ques. If \(\overrightarrow{a}\) = 2iˆ-jˆ+kˆ,\(\overrightarrow{b}\) = iˆ+jˆ-2kˆ and \(\overrightarrow{c}\) = iˆ+3jˆ-kˆ, find such that \(\overrightarrow{a}\) is perpendicular to λ\(\overrightarrow{b}\) + \(\overrightarrow{c}\). (4 marks)

Ans. We have

Ques. Find all vectors of magnitude 103 that are perpendicular to the plane of iˆ+2jˆ+kˆ and iˆ+3jˆ+4kˆ. (4 marks)

Ans. Let \(\overrightarrow{a}\)=iˆ+2jˆ+kˆ and \(\overrightarrow{b}\)=iˆ+3jˆ+4kˆ. Then

\(\overrightarrow{a}\)*\(\overrightarrow{b}\)= \(\mid\)iˆ jˆ kˆ 1 2 1 -1 3 4 \(\mid\)=iˆ(8-3) – jˆ(4+1) + kˆ(3+2)

= 5iˆ-5jˆ+5kˆ |\(\overrightarrow{a}\)*\(\overrightarrow{b}\)|= \(\sqrt{(5)^2+(-5)^2+(5)^2} = \sqrt{3(5)^2} = 5 \sqrt{3}\)

Therefore, unit vector perpendicular to the plane of \(\overrightarrow{a}\) and \(\overrightarrow{b}\) is given by

unit vector perpendicular to the plane of a and b

Ques. Find the unit vector in the direction of the sum of the vectors \(\overrightarrow{a}\)=2iˆ-jˆ+2kˆ and \(\overrightarrow{b}\)=-iˆ+jˆ+3kˆ. (4 marks)

Ans. Let \(\overrightarrow{c}\) denote the sum of \(\overrightarrow{a}\) and \(\overrightarrow{b}\). We have

Let c denote the sum of a and b. We have — Let c denote the sum of a **and b)**.

Ques. Prove that in a ΔABC,\(\frac{sinA}{a} = \frac{sinB}{b} = \frac{sinC}{c}\), where a,b,c represent the magnitudes of the sides opposite to vertices A, B, C, respectively. (4 marks)

Ans. Let the three sides of the triangle BC,CA and AB be represented by \(\overrightarrow{a}\),\(\overrightarrow{b}\) and \(\overrightarrow{c}\), respectively [Fig. 10.2].

Let the three sides of the triangle BC,CA and AB be represented by a,b and c, respectively — Let the three sides of the triangle BC,CA and AB be represented by a, b and c

Mathematics Related Links:

Horizontal and Vertical Lines	Equation of a Line
Calculus Formula	Rolle's Theorem
Trigonometric Functions	Real Valued Function
Arithmetic Progression	Sets
Differentiation and integration	Permutation and Combinations

Magnitude of a Vector: Formula and Direction

Magnitude of a Vector Formula

Direction of a Vector

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.
Find:

The principal value of \[ \sec^{-1}(\sqrt{2})+2\csc^{-1}(-2) \] is:

2.
Find:

The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

3.
Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).

4.
Find:

If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

5.
If \( xy = e^{x - y} \), then find \( \frac{dy}{dx} \).

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Magnitude of a Vector: Formula and Direction

Magnitude of a Vector Formula

Direction of a Vector

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1. Find: The principal value of \[ \sec^{-1}(\sqrt{2})+2\csc^{-1}(-2) \] is:

2. Find: The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

3.Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).

4. Find: If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

5.If \( xy = e^{x - y} \), then find \( \frac{dy}{dx} \).

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
Find:

The principal value of \[ \sec^{-1}(\sqrt{2})+2\csc^{-1}(-2) \] is:

2.
Find:

The shortest distance between the lines: \[ \vec{r}=(4+\lambda)\hat{i}+(2\lambda-1)\hat{j}-3\lambda\hat{k} \] and \[ \vec{r}=(1+2\mu)\hat{i}+(4\mu-1)\hat{j}+(2-5\mu)\hat{k} \]

3.
Using integration, find the area of the region bounded by the curve \( y = x|x| \), the x-axis, and the vertical lines \( x = -2 \) and \( x = 2 \).

4.
Find:

If the function \[ f(x)= \begin{cases} \frac{\sin x}{x}+\cos x, & x\neq0\\ k, & x=0 \end{cases} \] is continuous at \(x=0\), then find the value of \(k\).

5.
If \( xy = e^{x - y} \), then find \( \frac{dy}{dx} \).