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Components of Vector in Algebra

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Estimated time: 7 minutes
CBSE: Class 12

Introduction

A vector in three-dimensional geometry can be written in terms of its components along the x-, y-, and z-axes using the unit vectors \[\vec{i}, \vec{j}, \vec{k}\]. This component form makes it easier to find magnitude, compare vectors, and perform operations like addition and subtraction.

CBSE: Class 12

Definition: Component Form of a Vector

If P(x, y, z) is a point, then its position vector is

\[\vec{OP} = x\hat{i} + y\hat{j} + z\hat{k}\]

This is called the component form of a vector.

CBSE: Class 12

Properties of Vectors in Component Form

  • Magnitude
    \[|\vec{r}| = \sqrt{x^2 + y^2 + z^2}\]
  • Addition
    If \[\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\] and
    \[\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\]
    then \[\vec{a} + \vec{b} = (a_1 + b_1)\hat{i} + (a_2 + b_2)\hat{j} + (a_3 + b_3)\hat{k}\]
  • Subtraction
    \[\vec{a} - \vec{b} = (a_1 - b_1)\hat{i} + (a_2 - b_2)\hat{j} + (a_3 - b_3)\hat{k}\]
  • Scalar Multiplication
    \[\lambda\vec{a} = (\lambda a_1)\hat{i} + (\lambda a_2)\hat{j} + (\lambda a_3)\hat{k}\]
  • Equality of Vectors
    Two vectors are equal if their corresponding components are equal.
  • Collinearity of Vectors

    Two vectors are collinear if one is a scalar multiple of the other, i.e. \[\vec{b} = \lambda\vec{a}\]

  • Equality of Vectors
    Two vectors are equal if their corresponding components are equal.
CBSE: Class 12

Example 1

Find the values of \[x, y\] and \[z\] so that the vectors \[\vec{a} = x\hat{i} + 2\hat{j} + z\hat{k}\]and \[\vec{b} = 2\hat{i} + y\hat{j} + \hat{k}\]are equal.

Solution: Note that two vectors are equal if and only if their corresponding components are equal. Thus, the given vectors \[\vec{a}\] and \[\vec{b}\] will be equal if and only if

\[x = 2, \quad y = 2, \quad z = 1\]
CBSE: Class 12

Example 2

Question: Find a vector of magnitude 7 units in the direction of:

\[\vec{a} = \vec{i} - 2\vec{j}\]

Solution:

Step 1: Find the magnitude of \[\vec{a}\].

\[|\vec{a}| = \sqrt{1^2 + (-2)^2} = \sqrt{5}\]

Step 2: Find the unit vector in the same direction.

\[\hat{a} = \frac{1}{\sqrt{5}}(\vec{i} - 2\vec{j})\]

Step 3: Multiply by 7.

\[\vec{v} = 7\hat{a} = \frac{7}{\sqrt{5}}(\vec{i} - 2\vec{j})\]
\[\vec{v} = \frac{7}{\sqrt{5}}\vec{i} - \frac{14}{\sqrt{5}}\vec{j}\]

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