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Vector Product, Direction and Area

The vector product of two vectors a and b is a x b = |a||b|sin theta n, where n is a unit vector perpendicular to the plane of a and b in the direction given by the right-hand rule.

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Student-friendly explanation

The cross product gives a vector. Its magnitude equals the area of the parallelogram formed by the two vectors. If a and b are parallel non-zero vectors, then a x b = 0 because sin theta = 0.

How to write this in exams

  1. 1

    Start with the exact idea

    The vector product of two vectors a and b is a x b = |a||b|sin theta n, where n is a unit vector perpendicular to the plane of a and b in the direction given by the right-hand rule.

  2. 2

    Then show how to use it

    Write vectors in component form; set up the 3 by 3 determinant with i, j, k in the first row; expand carefully with minus sign in the j term; simplify components; use magnitude if area is required; apply right-hand rule for direction.

  3. 3

    Add one concrete example

    i x j = k, j x k = i, k x i = j, while j x i = -k. Order matters in cross product.

  4. 4

    Avoid this incomplete answer

    Forgetting the negative sign before the j component during determinant expansion.

Definition

The vector product of two vectors a and b is a x b = |a||b|sin theta n, where n is a unit vector perpendicular to the plane of a and b in the direction given by the right-hand rule.

Example

i x j = k, j x k = i, k x i = j, while j x i = -k. Order matters in cross product.

Rule to remember

Key formulas: a x b = |a||b|sin theta n; |a x b| = area of parallelogram; area of triangle = (1/2)|a x b|. Component determinant: a x b = | i j k; a1 a2 a3; b1 b2 b3 |. Conditions: direction follows right-hand rule; zero result occurs for parallel vectors or when one vector is zero.

Memory hook

Cross product points across the plane and its length gives area.

Examples and method

Worked example

Find a x b for a = i + 2j + 3k and b = 2i - j + k. Using determinant, a x b = i(2(1) - 3(-1)) - j(1(1) - 3(2)) + k(1(-1) - 2(2)) = i(5) - j(-5) + k(-5) = 5i + 5j - 5k.

Method to apply

Write vectors in component form; set up the 3 by 3 determinant with i, j, k in the first row; expand carefully with minus sign in the j term; simplify components; use magnitude if area is required; apply right-hand rule for direction.

Diagram support

Use a diagram showing two vectors forming a parallelogram, the perpendicular direction n, and right-hand rule orientation.

How CBSE asks it

Questions ask for cross product, area of parallelogram or triangle, direction perpendicular to two vectors, and checking whether vectors are parallel.

Avoid common mistakes

Common confusion

Students often use a x b = b x a, but cross product is anti-commutative: b x a = -(a x b).

Common wrong answer

Forgetting the negative sign before the j component during determinant expansion.

Exam tip

When calculating by determinant, keep the signs of i, j, k components carefully, especially the middle component.

Quick check

What is the magnitude of a x b if |a| = 5, |b| = 4, and the angle between them is 30 degrees?

|a x b| = |a||b|sin 30 degrees = 5 x 4 x 1/2 = 10.

Answer writing and exam use

1-mark answer

The vector product of two vectors a and b is a x b = |a||b|sin theta n, where n is a unit vector perpendicular to the plane of a and b in the direction given by the right-hand rule.

2-mark answer

The vector product of two vectors a and b is a x b = |a||b|sin theta n, where n is a unit vector perpendicular to the plane of a and b in the direction given by the right-hand rule. Key formulas: a x b = |a||b|sin theta n; |a x b| = area of parallelogram; area of triangle = (1/2)|a x b|. Component determinant: a x b = | i j k; a1 a2 a3; b1 b2 b3 |. Conditions: direction follows right-hand rule; zero result occurs for parallel vectors or when one vector is zero. i x j = k, j x k = i, k x i = j, while j x i = -k. Order matters in cross product.

3-mark answer

The cross product gives a vector. Its magnitude equals the area of the parallelogram formed by the two vectors. If a and b are parallel non-zero vectors, then a x b = 0 because sin theta = 0. Key formulas: a x b = |a||b|sin theta n; |a x b| = area of parallelogram; area of triangle = (1/2)|a x b|. Component determinant: a x b = | i j k; a1 a2 a3; b1 b2 b3 |. Conditions: direction follows right-hand rule; zero result occurs for parallel vectors or when one vector is zero. Find a x b for a = i + 2j + 3k and b = 2i - j + k. Using determinant, a x b = i(2(1) - 3(-1)) - j(1(1) - 3(2)) + k(1(-1) - 2(2)) = i(5) - j(-5) + k(-5) = 5i + 5j - 5k. Questions ask for cross product, area of parallelogram or triangle, direction perpendicular to two vectors, and checking whether vectors are parallel. Forgetting the negative sign before the j component during determinant expansion.
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