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Vectors: Magnitude, Direction and Unit Vector

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Student-friendly explanation

How to write this in exams

1
Start with the exact idea
A vector is a quantity having both magnitude and direction. If a = a1i + a2j + a3k, then its magnitude is |a| = sqrt(a1^2 + a2^2 + a3^2), and a unit vector in its direction is a/|a| when a is not the zero vector.
2
Then show how to use it
Write the vector in component form; square and add components for magnitude; divide each component by magnitude for unit vector; for direction cosines, identify component divided by magnitude.
3
Add one concrete example
For a = 2i - 3j + 6k, |a| = sqrt(4 + 9 + 36) = 7. A unit vector along a is (2/7)i - (3/7)j + (6/7)k.
4
Avoid this incomplete answer
Dividing by the sum of components instead of the magnitude, for example using 2 + 4 - 1 instead of sqrt(21).

Definition

Example

For a = 2i - 3j + 6k, |a| = sqrt(4 + 9 + 36) = 7. A unit vector along a is (2/7)i - (3/7)j + (6/7)k.

Rule to remember

Key rules: |a1i + a2j + a3k| = sqrt(a1^2 + a2^2 + a3^2); unit vector along non-zero a is a/|a|; direction cosines satisfy l^2 + m^2 + n^2 = 1. The unit vector formula is not defined for the zero vector.

Memory hook

Magnitude is the length; unit vector is the same direction with length 1.

Examples and method

Worked example

Find the unit vector along AB where A(1, 2, 3) and B(3, 6, 2). AB = (3 - 1)i + (6 - 2)j + (2 - 3)k = 2i + 4j - k. |AB| = sqrt(4 + 16 + 1) = sqrt(21). Unit vector = (2i + 4j - k)/sqrt(21).

Method to apply

Write the vector in component form; square and add components for magnitude; divide each component by magnitude for unit vector; for direction cosines, identify component divided by magnitude.

Diagram support

A diagram showing a directed line segment from O to P(x, y, z), with projections on coordinate axes, helps connect position vector and direction cosines.

How CBSE asks it

Questions usually ask for magnitude, unit vector, position vector of a point, or direction cosines of a given vector.

Avoid common mistakes

Common confusion

Students often write only the magnitude and forget that a vector answer also needs direction or component form.

Common wrong answer

Dividing by the sum of components instead of the magnitude, for example using 2 + 4 - 1 instead of sqrt(21).

Exam tip

Before finalising an answer, check whether the question asks for magnitude, vector, unit vector, position vector, or direction cosines.

Quick check

Find the magnitude of a = 3i + 4j - 12k.

|a| = sqrt(3^2 + 4^2 + (-12)^2) = sqrt(169) = 13.

Answer writing and exam use

1-mark answer

2-mark answer

A vector is a quantity having both magnitude and direction. If a = a1i + a2j + a3k, then its magnitude is |a| = sqrt(a1^2 + a2^2 + a3^2), and a unit vector in its direction is a/|a| when a is not the zero vector. Key rules: |a1i + a2j + a3k| = sqrt(a1^2 + a2^2 + a3^2); unit vector along non-zero a is a/|a|; direction cosines satisfy l^2 + m^2 + n^2 = 1. The unit vector formula is not defined for the zero vector. For a = 2i - 3j + 6k, |a| = sqrt(4 + 9 + 36) = 7. A unit vector along a is (2/7)i - (3/7)j + (6/7)k.

3-mark answer

A position vector locates a point with respect to the origin. For point P(x, y, z), OP = xi + yj + zk. Direction cosines are the cosines of the angles made by a vector with the positive x-, y-, and z-axes. If l, m, n are direction cosines, then l^2 + m^2 + n^2 = 1. Key rules: |a1i + a2j + a3k| = sqrt(a1^2 + a2^2 + a3^2); unit vector along non-zero a is a/|a|; direction cosines satisfy l^2 + m^2 + n^2 = 1. The unit vector formula is not defined for the zero vector. Find the unit vector along AB where A(1, 2, 3) and B(3, 6, 2). AB = (3 - 1)i + (6 - 2)j + (2 - 3)k = 2i + 4j - k. |AB| = sqrt(4 + 16 + 1) = sqrt(21). Unit vector = (2i + 4j - k)/sqrt(21). Questions usually ask for magnitude, unit vector, position vector of a point, or direction cosines of a given vector. Dividing by the sum of components instead of the magnitude, for example using 2 + 4 - 1 instead of sqrt(21).

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