Class 12 — Mathematics (NCERT)

Vector Algebra: Dot Product, Cross Product and Projections

Class 12

✓By the end of this lesson students will be able to define and compute the dot product (scalar product) of two vectors.
✓By the end of this lesson students will be able to define and compute the cross product (vector product) of two vectors.
✓By the end of this lesson students will be able to apply the dot and cross products to find the angle between vectors and the area of geometric figures.
✓By the end of this lesson students will be able to calculate the scalar and vector projections of one vector onto another.
✓By the end of this lesson students will be able to understand the geometric interpretations and properties of dot and cross products.

Key concepts

Dot Product (Scalar Product)

The dot product of two non-zero vectors $\vec{a}$ and $\vec{b}$ , denoted by $\vec{a} \cdot \vec{b}$ , is a scalar quantity defined as the product of their magnitudes and the cosine of the angle $\theta$ between them (where $0 \le \theta \le \pi$ ). 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 their dot product is $a_1 b_1 + a_2 b_2 + a_3 b_3$ . The dot product is commutative, i.e., $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$ . If $\vec{a} \cdot \vec{b} = 0$ , then the vectors $\vec{a}$ and $\vec{b}$ are perpendicular (orthogonal), provided they are non-zero vectors.

\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \quad \text{or} \quad \vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3

Cross Product (Vector Product)

The cross product of two non-zero vectors $\vec{a}$ and $\vec{b}$ , denoted by $\vec{a} \times \vec{b}$ , is a vector quantity defined as $|\vec{a}| |\vec{b}| \sin \theta \hat{n}$ , where $\theta$ is the angle between them ( $0 \le \theta \le \pi$ ) and $\hat{n}$ is a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ , in the direction given by the right-hand rule. 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 their cross product can be found using a determinant. The cross product is anti-commutative, i.e., $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$ . If $\vec{a} \times \vec{b} = \vec{0}$ , then the vectors $\vec{a}$ and $\vec{b}$ are parallel (collinear), provided they are non-zero vectors. The magnitude of the cross product, $|\vec{a} \times \vec{b}|$ , represents the area of the parallelogram formed by adjacent sides $\vec{a}$ and $\vec{b}$ .

\vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin \theta \hat{n} \quad \text{or} \quad \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}

Scalar Projection of a Vector on Another Vector

The scalar projection of vector $\vec{a}$ on vector $\vec{b}$ is the length of the component of $\vec{a}$ along the direction of $\vec{b}$ . It is a scalar value and can be positive, negative, or zero depending on the angle between the vectors. If the angle is acute, the projection is positive; if obtuse, it is negative; if right, it is zero.

\text{Projection of } \vec{a} \text{ on } \vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}

Vector Projection of a Vector on Another Vector

The vector projection of vector $\vec{a}$ on vector $\vec{b}$ is a vector quantity that represents the component of $\vec{a}$ that lies along the direction of $\vec{b}$ . Its magnitude is the absolute value of the scalar projection, and its direction is the same as $\vec{b}$ (or opposite if the scalar projection is negative).

\text{Projection of } \vec{a} \text{ on } \vec{b} = \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \vec{b}

Key facts to remember

1Dot Product: $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta = a_1 b_1 + a_2 b_2 + a_3 b_3$ .
2Angle between vectors: $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$ .
3Condition for perpendicular vectors: $\vec{a} \cdot \vec{b} = 0$ (for non-zero vectors).
4Cross Product: $\vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin \theta \hat{n}$ .
5Algebraic Cross Product: $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$ .
6Condition for parallel vectors: $\vec{a} \times \vec{b} = \vec{0}$ (for non-zero vectors).
7Area of parallelogram: $|\vec{a} \times \vec{b}|$ . Area of triangle: $\frac{1}{2} |\vec{a} \times \vec{b}|$ .
8Scalar projection of $\vec{a}$ on $\vec{b}$ : $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$ .
9Vector projection of $\vec{a}$ on $\vec{b}$ : $\left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \vec{b}$ .

Worked examples

Example 1

Find the dot product of vectors $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}$ . Also, find the angle between them.

IGiven vectors:

\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}

and

\vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}

IIFirst, calculate the dot product

\vec{a} \cdot \vec{b}

III

\vec{a} \cdot \vec{b} = (2)(6) + (2)(-3) + (-1)(2)

\vec{a} \cdot \vec{b} = 12 - 6 - 2 = 4

VNext, calculate the magnitudes of the vectors:

|\vec{a}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3

VII

|\vec{b}| = \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36 + 9 + 4} = \sqrt{49} = 7

VIIINow, use the formula

\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}

to find the angle

\theta

\cos \theta = \frac{4}{(3)(7)} = \frac{4}{21}

\theta = \cos^{-1}\left(\frac{4}{21}\right)

Answer

The dot product $\vec{a} \cdot \vec{b} = 4$ . The angle between the vectors is $\theta = \cos^{-1}\left(\frac{4}{21}\right)$ .

Example 2

Find the cross product of vectors $\vec{p} = 3\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{q} = \hat{i} - 3\hat{j} + \hat{k}$ . Hence, find the area of the parallelogram whose adjacent sides are $\vec{p}$ and $\vec{q}$ .

IGiven vectors:

\vec{p} = 3\hat{i} + \hat{j} - 2\hat{k}

and

\vec{q} = \hat{i} - 3\hat{j} + \hat{k}

IICalculate the cross product

\vec{p} \times \vec{q}

using the determinant form:

III

\vec{p} \times \vec{q} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -2 \\ 1 & -3 & 1 \end{vmatrix}

= \hat{i}((1)(1) - (-2)(-3)) - \hat{j}((3)(1) - (-2)(1)) + \hat{k}((3)(-3) - (1)(1))

= \hat{i}(1 - 6) - \hat{j}(3 + 2) + \hat{k}(-9 - 1)

= -5\hat{i} - 5\hat{j} - 10\hat{k}

VIIThe area of the parallelogram is given by

|\vec{p} \times \vec{q}|

VIII

|\vec{p} \times \vec{q}| = \sqrt{(-5)^2 + (-5)^2 + (-10)^2}

= \sqrt{25 + 25 + 100} = \sqrt{150}

\sqrt{150} = \sqrt{25 \times 6} = 5\sqrt{6}

Answer

The cross product $\vec{p} \times \vec{q} = -5\hat{i} - 5\hat{j} - 10\hat{k}$ . The area of the parallelogram is $5\sqrt{6}$ square units.

Remember that the area is always a positive scalar quantity.

Example 3

Find the scalar projection and vector projection of the vector $\vec{a} = \hat{i} + 3\hat{j} + 7\hat{k}$ on the vector $\vec{b} = 7\hat{i} - \hat{j} + 8\hat{k}$ .

IGiven vectors:

\vec{a} = \hat{i} + 3\hat{j} + 7\hat{k}

and

\vec{b} = 7\hat{i} - \hat{j} + 8\hat{k}

IIFirst, calculate the dot product

\vec{a} \cdot \vec{b}

III

\vec{a} \cdot \vec{b} = (1)(7) + (3)(-1) + (7)(8)

= 7 - 3 + 56 = 60

VNext, calculate the magnitude of vector

\vec{b}

|\vec{b}| = \sqrt{7^2 + (-1)^2 + 8^2} = \sqrt{49 + 1 + 64} = \sqrt{114}

VIINow, find the scalar projection of

\vec{a}

\vec{b}

VIIIScalar projection

= \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{60}{\sqrt{114}}

9Finally, find the vector projection of

\vec{a}

\vec{b}

10Vector projection

= \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \vec{b}

= \left( \frac{60}{(\sqrt{114})^2} \right) (7\hat{i} - \hat{j} + 8\hat{k})

= \frac{60}{114} (7\hat{i} - \hat{j} + 8\hat{k})

= \frac{10}{19} (7\hat{i} - \hat{j} + 8\hat{k})

= \frac{70}{19}\hat{i} - \frac{10}{19}\hat{j} + \frac{80}{19}\hat{k}

Answer

Scalar projection of $\vec{a}$ on $\vec{b}$ is $\frac{60}{\sqrt{114}}$ . Vector projection of $\vec{a}$ on $\vec{b}$ is $\frac{70}{19}\hat{i} - \frac{10}{19}\hat{j} + \frac{80}{19}\hat{k}$ .

Scalar projection is a number, vector projection is a vector.

Common mistakes

✗Confusing dot product (scalar) with cross product (vector).
✗Incorrectly applying the right-hand rule for the direction of the cross product.
✗Forgetting to divide by the magnitude of the *second* vector when calculating projection (e.g., projection of $\vec{a}$ on $\vec{b}$ uses $|\vec{b}|$ in the denominator).
✗Using the magnitude of the cross product for the area of a triangle without dividing by 2.
✗Errors in determinant calculation for the cross product, especially sign errors for the $\hat{j}$ component.

Exam tips

★Memorise all formulas for dot product, cross product, and projections. Understand the conditions for perpendicular and parallel vectors.
★Practice determinant calculations for the cross product thoroughly to avoid computational errors.
★Always specify if the answer is a scalar or a vector, especially for projection questions.
★For geometric applications (area), ensure the final answer is a positive scalar quantity with appropriate units (e.g., square units).

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