Operations on Vectors

Adding and scaling vectors

Vectors add component by component, and a scalar multiplies every component: for $\vec{a} = (a_1,a_2,a_3)$ and $\vec{b} = (b_1,b_2,b_3)$ , $\vec{a}+\vec{b} = (a_1+b_1, a_2+b_2, a_3+b_3)$ and $k\vec{a} = (ka_1, ka_2, ka_3)$ .

The negative of $\vec{a}$ , written $-\vec{a}$ , has the same magnitude but points the opposite way. If $\vec{a} = \overrightarrow{AB}$ , then $-\vec{a} = \overrightarrow{BA}$ .

Dot and cross products

\vec{a}\cdot\vec{b} = a_1b_1 + a_2b_2 + a_3b_3 = |\vec{a}||\vec{b}|\cos\theta

\vec{a}\times\vec{b} = (a_2b_3-a_3b_2,\; a_3b_1-a_1b_3,\; a_1b_2-a_2b_1)

The dot product is a number; the cross product is a vector perpendicular to both $\vec{a}$ and $\vec{b}$ .

Worked example: all four operations

Let $\vec{a} = 2\hat{i}-\hat{j}+3\hat{k}$ and $\vec{b} = \hat{i}+4\hat{j}-2\hat{k}$ .

$\vec{a}+\vec{b} = 3\hat{i}+3\hat{j}+\hat{k}$ and $3\vec{a}-2\vec{b} = 4\hat{i}-11\hat{j}+13\hat{k}$ .
$\vec{a}\cdot\vec{b} = 2 - 4 - 6 = -8$ . With $|\vec{a}| = \sqrt{14}$ and $|\vec{b}| = \sqrt{21}$ , $\cos\theta = \dfrac{-8}{\sqrt{294}} \approx -0.467$ , so $\theta \approx 117.8^\circ$ .
$\vec{a}\times\vec{b} = ((-1)(-2)-(3)(4),\; (3)(1)-(2)(-2),\; (2)(4)-(-1)(1)) = (-10, 7, 9)$ .
Check: $(-10,7,9)\cdot\vec{a} = -20-7+27 = 0$ and $(-10,7,9)\cdot\vec{b} = -10+28-18 = 0$ , so the cross product is perpendicular to both.

Worked example: a scalar triple product identity

Prove that $[\vec{a}+\vec{b}\;\;\vec{b}+\vec{c}\;\;\vec{c}+\vec{a}] = 2[\vec{a}\;\;\vec{b}\;\;\vec{c}]$ , where $[\vec{x}\;\vec{y}\;\vec{z}] = \vec{x}\cdot(\vec{y}\times\vec{z})$ .

Solution: $(\vec{b}+\vec{c})\times(\vec{c}+\vec{a}) = \vec{b}\times\vec{c} + \vec{b}\times\vec{a} + \vec{c}\times\vec{a}$ , since $\vec{c}\times\vec{c} = \vec{0}$ . Dotting with $\vec{a}+\vec{b}$ , every term where a vector meets a cross product containing itself vanishes, leaving $\vec{a}\cdot(\vec{b}\times\vec{c}) + \vec{b}\cdot(\vec{c}\times\vec{a})$ . The scalar triple product is cyclic, so both terms equal $[\vec{a}\;\vec{b}\;\vec{c}]$ , and the sum is twice it.

Worked example: unit vectors summing to zero

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors with $\vec{a}+\vec{b}+\vec{c}=\vec{0}$ , find $\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}$ .

|\vec{a}+\vec{b}+\vec{c}|^2 = 3 + 2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) = 0 \;\Rightarrow\; \vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a} = -\dfrac{3}{2}

Common mistakes

Treating the dot product as a vector. It's a single number.
Reversing the cross product. $\vec{b}\times\vec{a} = -(\vec{a}\times\vec{b})$ .
Losing a sign in the middle component of the cross product: it's $a_3b_1 - a_1b_3$ .
Skipping the perpendicularity check. Dotting the cross product with each vector should give 0.

Where this comes up in exams

The 2025-26 CBSE Class XII syllabus lists addition of vectors, multiplication by a scalar, the negative of a vector, and the scalar (dot) and vector (cross) products. JEE Main 2026 lists the addition of vectors and scalar and vector products, and JEE Advanced 2026 lists dot and cross products and scalar and vector triple products.

Checked against CBSE Mathematics curriculum, Classes XI-XII (2025-26), JEE (Main) 2026 syllabus (NTA) and JEE (Advanced) 2026 syllabus. Syllabi can change from year to year, so confirm with the latest official notification for your exam.

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