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Scalar Product, Angle and Projection

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Student-friendly explanation

How to write this in exams

1
Start with the exact idea
The scalar product of two vectors a and b is a · b = |a||b|cos theta. It gives a scalar, not a vector. In component form, if a = a1i + a2j + a3k and b = b1i + b2j + b3k, then a · b = a1b1 + a2b2 + a3b3.
2
Then show how to use it
Compute a · b using components; compute magnitudes if angle or projection is needed; substitute in the correct formula; check whether the expected result is scalar, angle, projection length, or work.
3
Add one concrete example
For a = 2i + j - 2k and b = i - 3j + k, a · b = 2(1) + 1(-3) + (-2)(1) = -3.
4
Avoid this incomplete answer
Using sin theta instead of cos theta for dot product, or writing i, j, k in the final dot product answer.

Definition

Example

For a = 2i + j - 2k and b = i - 3j + k, a · b = 2(1) + 1(-3) + (-2)(1) = -3.

Rule to remember

Key formulas: a · b = |a||b|cos theta; component form a · b = a1b1 + a2b2 + a3b3; cos theta = (a · b)/(|a||b|); scalar projection of a on b = (a · b)/|b|. Conditions: angle formula needs non-zero vectors.

Memory hook

Dot product measures how much one vector goes along another; zero means no along-part for non-zero perpendicular vectors.

Examples and method

Worked example

Find the angle between a = i + j and b = i - j. First, a · b = 1(1) + 1(-1) = 0. Also |a| = sqrt(2), |b| = sqrt(2). Using cos theta = (a · b)/(|a||b|), cos theta = 0/(sqrt(2)sqrt(2)) = 0. Hence theta = 90 degrees, so the vectors are perpendicular.

Method to apply

Compute a · b using components; compute magnitudes if angle or projection is needed; substitute in the correct formula; check whether the expected result is scalar, angle, projection length, or work.

Diagram support

A diagram showing two vectors with included angle theta and projection of one vector on the other helps distinguish dot product, angle, and projection.

How CBSE asks it

Common forms include finding angle between vectors, checking perpendicularity, finding projection, and work done by a force.

Avoid common mistakes

Common confusion

Students sometimes treat the dot product answer as a vector or forget that perpendicularity requires non-zero vectors.

Common wrong answer

Using sin theta instead of cos theta for dot product, or writing i, j, k in the final dot product answer.

Exam tip

Use component dot product first, then compare with |a||b|cos theta if angle is required.

Quick check

If a = i + 2j + 2k and b = 2i - j, find a · b.

a · b = 1(2) + 2(-1) + 2(0) = 0. The non-zero vectors are perpendicular.

Answer writing and exam use

1-mark answer

2-mark answer

The scalar product of two vectors a and b is a · b = |a||b|cos theta. It gives a scalar, not a vector. In component form, if a = a1i + a2j + a3k and b = b1i + b2j + b3k, then a · b = a1b1 + a2b2 + a3b3. Key formulas: a · b = |a||b|cos theta; component form a · b = a1b1 + a2b2 + a3b3; cos theta = (a · b)/(|a||b|); scalar projection of a on b = (a · b)/|b|. Conditions: angle formula needs non-zero vectors. For a = 2i + j - 2k and b = i - 3j + k, a · b = 2(1) + 1(-3) + (-2)(1) = -3.

3-mark answer

Dot product connects algebra with angle. If a · b = 0 and both vectors are non-zero, then the vectors are perpendicular. Projection of a on b is (a · b)/|b| as a scalar component in the direction of b. Key formulas: a · b = |a||b|cos theta; component form a · b = a1b1 + a2b2 + a3b3; cos theta = (a · b)/(|a||b|); scalar projection of a on b = (a · b)/|b|. Conditions: angle formula needs non-zero vectors. Find the angle between a = i + j and b = i - j. First, a · b = 1(1) + 1(-1) = 0. Also |a| = sqrt(2), |b| = sqrt(2). Using cos theta = (a · b)/(|a||b|), cos theta = 0/(sqrt(2)sqrt(2)) = 0. Hence theta = 90 degrees, so the vectors are perpendicular. Common forms include finding angle between vectors, checking perpendicularity, finding projection, and work done by a force. Using sin theta instead of cos theta for dot product, or writing i, j, k in the final dot product answer.

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