Namrata Das Exams Prep Master
Exams Prep Master
The magnitude of a vector is the length of a vector or the distance between the initial and the final position of a vector. Whereas, the vector is the one that has both the magnitude as well as the direction and follows the triangle law of addition. Some important and common examples of vectors are displacements, velocity, acceleration, and force. The magnitude of a vector formula can be calculated in two ways. Firstly, the magnitude is calculated for a vector when its final point is at origin (0,0) while in the other instance, the initial and the final point of the vector is at definite points (x1, y1) and (x2, y2) respectively. Moreover, to calculate the magnitude of a vector, we need to find the length of the given vector quantity or the total distance from the initial point to the final position. It can be easily evaluated with the help of certain formulas that we will be discussing here.
| Table of Content |
Key takeaways: Vector quantity, the magnitude of a two-dimensional vector, the magnitude of a three-dimensional vector, displacements, velocity, acceleration, force
Also read: Isosceles Triangle Theorems
Vector Quantity
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“A vector quantity is a quantity that has both, the magnitude as well as the direction.” It is said to follow the triangle law of addition or the parallelogram law of addition. A vector is generally represented using bold alphabets or alphabets with a bar above them, say v.
Magnitude of a Vector
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The magnitude of a vector quantity is the length of the vector, which is represented as |a| (using a modulus operator) for a vector ‘a’. The magnitude of a vector is also called its absolute value. Moreover, the magnitude of a vector can never be negative, it can either be positive or zero. For any given vector with its direction ratios along the x-axis, y-axis, and z-axis, its magnitude will be equal to the square root of the sum of the square of its particular direction axis.
Magnitude of a Vector Formula
| Magnitude Formula for a Vector when the initial points are (x1, y1) and final points are (x2, y2) | |v| = √(x2 + x1)2 + (y2 + y1)2 |
| Magnitude Formula for a Vector when final point is origin | |v| = √x2 + y2 |
Let us know about the above formulas in detail.
- Let there be a vector,
\(\bar{A} = x \hat{i} + y \hat{j} + z \hat{k}\)
Where, x, y, and z are the components of A along x, y, and z-axis, or we can say they are the coordinates of the object.
And its magnitude will be calculated as:
\(|A| = \sqrt {x^2_1 + y^2_1 + z^2_1}\)
- Let there be a vector whose initial points are ( x, y) and the endpoints are at the origin (0,0), in that case, the magnitude will be calculated as:
\(| \vec{V} | = \sqrt {x^2 + y^2}\)
- Let there be a vector with its initial point as (x1,y1) and its final points as (x2,y2), its magnitude will be calculated as:
Also Read:
| Related Articles | ||
|---|---|---|
| Vector Algebra Ncert Solutions | Scalar triple product of vectors Important Questions | Scalar triple product of vectors |
| Direction of a vector formula | Adjacency Matrix | Cross Product |
How to Calculate the Magnitude of a Vector Quantity?
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Following are the steps included calculating the magnitude of a two-dimensional vector from its coordinates,
Step-1 Firstly, identity the coordinates of the vector.
Step-2 Then calculate the sum of the square of each of its components.
Step-3 Find the square root of the obtained sum.
Hence, the formula thus obtained is used to calculate the magnitude of any two-dimensional vector (say v) with its coordinates as (0,0) or (x1,y1).
Note: The formula is obtained using the Pythagoras theorem or using the distance formula.
The formula is:
\(| \vec{V} | = \sqrt {x^2 + y^2}\)
The steps included for calculating the magnitude of a three-dimensional vector from its coordinates:
Step-1 Firstly, identify the coordinates of the vector.
Step-2 Then calculate the sum of the square of each of its components.
Step-3 Find the square root of the sum obtained.
Hence, the formula therefore attained is used to calculate the magnitude of any three-dimensional vector (say V) with its coordinates as (x1, y1, z1).
The formula is:
\(|V| = \sqrt {x^2_1 + y^2_1 + z^2_1}\)
Things to Remember
- The magnitude of a given vector is represented using a modulus operator, always. Illustration: |v|
- The magnitude of a vector can never be expressed as negative. It can either be positive or zero.
- The reason is the modulus operation, which converts all the negative values to positive values.
- For a two-dimensional vector with (x, y) as its coordinates, we use the formula:
- For a three-dimensional vector with(x, y, z) as its coordinates, we use the formula:
- The formula for the calculation of the magnitude of a vector is taken from the distance formula or the Pythagoras theorem.
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Sample Questions
Ques 1: Calculate the magnitude of vector a with (3,4) as its initial point and (0,0) as its endpoints. (3 marks)
Ans: Given, Initial point = (3,4), where x1 = 3 and y1 = 4
Final point = (0,0), both x2 and y2 are 0.
In such a case,
The formula for calculating the magnitude will be,
|a| = √x2 + y2
|a| = √(- 3)2 + (- 4)2
= √9+16
= √ 25
|a| = 5.
Hence, the magnitude of vector a is 5.
Ques 2: Vector U has an initial point (4,2) and a final point (7,6). Find the value of |U|. (3 marks)
Ans: Given, Initial points (X1, y1) = (4,2), where x1 = 4 and y1 = 2
Final points(X2, y2) = (7,6) where x2 = 7 and y2 = 6
So, |U| = √(x2 – x1) 2 + (y2 – y1)2
|U| = √(7 – 4)2 + (6 – 2)2
= √(3)2 + (4)2
= √ 9 + 16
= √ 25
|U| = 5.
Hence, the magnitude of the vector U is 5.
Ques 3: Calculate the magnitude of the vector A = 5i + 6j + 7z. (4 marks)
Ans: Given, the components of the vector A as,
X = 5, y = 6 and z = 7.
The formula to be used here is,
|A| = √x2 + y2 + z2
So, |A| = √ (5)2 + (6)2 + (7)2
= √ 25 + 36 + 49
= √ 110
|A| = 10.48.
Hence, the magnitude of the vector |A| is 10.48.
Ques 4: Calculate the magnitude of the vector V = 2i + 7j + 5 z. (4 marks)
Ans: Given, The components of the vector V as,
X= 2, y = 7 and z = 5 .
The formula to be used here is,
|V| = √x2 + y2 + z2
So, |V| = √(2)2 + (7)2 + (5)2
= √4 + 49 + 25
=√78
|V| = 8.83.
Hence, the magnitude of the vector |V| is 8.83.
Ques 5: Find the magnitude of the given vectors: |A| = -2i + 5j (3 marks)
Ans: As given in the above question,
X=(-2) and y= 5
The formula to be used is:
|A| = √x2 + y2
= √(-2)2 + (5)2
= √4 + 25
= √ 29
|A| = 5.38.
Hence, the magnitude of the vector |A| is 5.38.
Ques 6: If vector |a| = 2, |b| = 7 and (a x b) = (3i+2j+6k), what is the angle between a and b. (3 marks)
Ans: Given: |\(\overrightarrow{a}\)| 2 |\(\overrightarrow{b}\)|2 = |\(\overrightarrow{a}\)x\(\overrightarrow{b}\)|2 + |\(\overrightarrow{a}\).\(\overrightarrow{b}\)|2
⇒ \(\overrightarrow{a}\)x\(\overrightarrow{b}\) = |\(\overrightarrow{a}\)||\(\overrightarrow{b}\)| sinθ
⇒ |\(\overrightarrow{a}\)x\(\overrightarrow{b}\)| = \(\sqrt{3^2 + 2^2 + 6^2} = 1\)
⇒ 7 = 7 x 2sinθ
⇒ sinθ = \(\frac{1}{2}\)
⇒θ = sin – 1\(\frac{1}{2}\)
Ques 7: Given a = (1/7)(2i + 3j + 6k), b = (1/7)(3i - 6j + 2k), c = (1/7)(6i + 2j - 3k), i, j, k being a right handed orthogonal system of unit vectors in space, find the vectors a, b, c is also another system. (5 marks)
Ans: In order to show that the vectors a, b, c is right handed orthogonal system of unit vectors, we need to prove:
(a) |\(\overrightarrow{a}\)| = |b| = |\(\overrightarrow{c}\)| = 1
(b) \(\overrightarrow{a}\) x \(\overrightarrow{b}\) = \(\overrightarrow{c}\)
(c) \(\overrightarrow{b}\) x \(\overrightarrow{c}\) = \(\overrightarrow{a}\)
(d) \(\overrightarrow{c}\) x \(\overrightarrow{a}\) = \(\overrightarrow{b}\)
Let us assume each of these one at a time.
(a) Recall the magnitude of the vector \(x \hat{i} + y \hat{j} + z \hat{k}\) is,
|\(x \hat{i} + y \hat{j} + z \hat{k}\)| = \(\sqrt{x^2 + y^2 + z^2}\)
First, we will find |\(\overrightarrow{a}\)|.
|\(\overrightarrow{a}\)| = \(\frac{1}{7}\)\(\sqrt{2^2 + 3^2 + 6^2}\)
⇒ |\(\overrightarrow{a}\)| = \(\frac{1}{7}\)\(\sqrt{4 + 9 + 36}\)
⇒ |\(\overrightarrow{a}\)| = \(\frac{1}{7}\)\(\sqrt{49}\) = \(\frac{1}{7}\) x 7
∴ |\(\overrightarrow{a}\)| = 1
Now, we will find |\(\overrightarrow{a}\)|.
|\(\overrightarrow{b}\)| = \(\frac{1}{7}\)\(\sqrt{3^2 + (- 6)^2 + 2^2}\)
⇒ |\(\overrightarrow{b}\)| = \(\frac{1}{7}\)\(\sqrt{9 + 36 + 4}\)
⇒ |\(\overrightarrow{b}\)| = \(\frac{1}{7}\)\(\sqrt{49}\)= \(\frac{1}{7}\) x 7
∴ |\(\overrightarrow{b}\)| = 1
And finally we will find the vector c,
|\(\overrightarrow{c}\)| = \(\frac{1}{7}\)\(\sqrt{6^2 + 2^2 + (-3)^2}\)
⇒ |\(\overrightarrow{c}\)| = \(\frac{1}{7}\)\(\sqrt{36 + 4 + 9}\)
⇒ |\(\overrightarrow{c}\)| = \(\frac{1}{7}\)\(\sqrt{49}\) = \(\frac{1}{7}\) x 7
∴ |\(\overrightarrow{c}\)| = 1
(b) Now, we will evaluate the vector vector a x b.
The cross product of two vectors:
\(\overrightarrow{a}\) = a1\(\hat{i}\) + a2\(\hat{j}\) + a3\(\hat{k}\) and \(\overrightarrow{b}\) = b1\(\hat{i}\) + b2\(\hat{j}\) + b3\(\hat{k}\) is
\(\overrightarrow{a}\) x \(\overrightarrow{b}\) = \(\begin{bmatrix}\hat{i} & \hat{j} & \hat{k} \\[0.3em]a_1 & a_2 & a_3 \\[0.3em]b_1 & b_2 & b_3 \\[0.3em] \end{bmatrix}\)
⇒ \(\overrightarrow{a}\) x \(\overrightarrow{b}\) = \(\frac{1}{7}\) x \(\frac{1}{7}\) \(\begin{bmatrix}\hat{i} & \hat{j} & \hat{k} \\[0.3em]2 & 3 & 6 \\[0.3em]3 & -6 & 2 \\[0.3em] \end{bmatrix}\)
⇒ \(\overrightarrow{a}\) x \(\overrightarrow{b}\)
= \(\frac{1}{49}\)(\(\hat{i}\)[(3)(2) – (-6)(6)] – \(\hat{j}\)[(2)(2) – (3)(6)] + \(\hat{k}\)[(2)(-6) – (3)(3)])
⇒ \(\overrightarrow{a}\) x \(\overrightarrow{b}\) =\(\frac{1}{49}\)( \(\hat{i}\)[6 + 36] – [4 – 18] + \(\hat{k}\)[ – 12 – 9])
⇒ \(\overrightarrow{a}\) x \(\overrightarrow{b}\) = \(\frac{1}{49}\) (42\(\hat{i}\) + 14\(\hat{j}\) – 21\(\hat{k}\))
∴ \(\overrightarrow{a}\) x \(\overrightarrow{b}\) = \(\frac{1}{7}\)(6\(\hat{i}\) + 2\(\hat{j}\) – 3\(\hat{k}\)) = \(\overrightarrow{c}\)
Hence, we have \(\overrightarrow{a}\) x \(\overrightarrow{b}\) = \(\overrightarrow{c}\)
(c) Now, we will evaluate the vector b x vector c,
(d) Now, we will evaluate the vector c x vector a,
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