Dot and Cross Products

Two vectors can be multiplied in two physically meaningful ways: the dot product yields a scalar (work, projection), the cross product yields a vector (torque, area, angular momentum). Their algebraic and geometric properties drive most JEE vector questions.

Concept

Dot product: $\vec a \cdot \vec b = a b \cos\theta = a_x b_x + a_y b_y + a_z b_z.$ Commutative; gives projection $a\cos\theta = \vec a \cdot \hat b$. Zero $\Leftrightarrow$ perpendicular.

Cross product: $\vec a \times \vec b = a b \sin\theta\, \hat n,$ where $\hat n$ is normal to the plane of $\vec a, \vec b$ given by the right-hand rule. Anti-commutative: $\vec b \times \vec a = -\vec a \times \vec b$. Zero $\Leftrightarrow$ parallel.

In components, $\vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{vmatrix}.$

Magnitude $|\vec a \times \vec b|$ equals the area of the parallelogram spanned.

Derivation

Define $\vec a \cdot \vec b$ via projections. Write $\vec b = b\cos\theta\,\hat a + \vec b_\perp$ where $\vec b_\perp \perp \vec a$. Then $\vec a \cdot \vec b = a\,(b\cos\theta)$ since $\vec a \cdot \vec b_\perp = 0$. Distributivity over $\hat i, \hat j, \hat k$ with $\hat i \cdot \hat i = 1$, $\hat i \cdot \hat j = 0$ gives the component formula.

For the cross product, set $\hat i \times \hat j = \hat k$ (cyclic), expand bilinearly, and the determinant form follows.

JEE Worked Example

Q. A force $\vec F = 3\hat i + 4\hat j\,\text{N}$ acts at the point $\vec r = 2\hat i + \hat j\,\text{m}$. Find the torque $\vec\tau = \vec r \times \vec F$ about the origin and the work done while displacing along $\vec d = \hat i + \hat j\,\text{m}$.

Solution. $\vec\tau = (2\hat i + \hat j) \times (3\hat i + 4\hat j) = 2\cdot 4\,(\hat i \times \hat j) + 1\cdot 3\,(\hat j \times \hat i) = 8\hat k - 3\hat k = 5\hat k\,\text{N m}$.

Work $W = \vec F \cdot \vec d = 3(1) + 4(1) = 7\,\text{J}$.

Traps

• Treating cross product as commutative.
• Forgetting that the dot product can be negative (obtuse angle).
• Reading off $\sin\theta$ instead of $\cos\theta$ when both vectors are tail-to-tail.
• Sign errors when expanding the determinant — minus on the $\hat j$ row.

Key Takeaways

• Dot product: scalar, commutative, $a b \cos\theta$.
• Cross product: vector, anti-commutative, $a b \sin\theta$.
• Use components for messy 3D problems; geometric form for clean angles.