Dot Product & Cross Product

Two ways to multiply vectors

Adding two vectors gives a vector and is unambiguous. Multiplying them is not: there are two distinct, equally legitimate products, and they answer different physical questions. The dot product (also called the scalar product) takes two vectors and returns a single number. The cross product (also called the vector product) takes two vectors and returns a third vector, perpendicular to both.

The distinction is not a notational nicety. Work done by a force is a number — it cannot point anywhere — so it is built from a dot product. Torque produces rotation about an axis and therefore has a direction of its own, so it is built from a cross product. Before solving any product problem, ask one diagnostic question: is the answer a scalar or a vector? The answer decides which product you need.

The dot product (scalar product)

The dot product of two vectors \(\vec A\) and \(\vec B\) separated by angle \(\theta\) is defined as

$$\vec A\cdot\vec B = AB\cos\theta$$

where \(A\) and \(B\) are the magnitudes. Because \(\cos\theta\) runs from \(+1\) at \(\theta=0^\circ\) to \(0\) at \(\theta=90^\circ\) to \(-1\) at \(\theta=180^\circ\), the dot product is maximum when the vectors are parallel, zero when they are perpendicular, and negative when they point in broadly opposite directions. The product carries the units of both vectors but no direction.

Two consequences are worth memorising. The dot product of a vector with itself returns its magnitude squared, so \(A=\sqrt{\vec A\cdot\vec A}\) — this is the vector form of the Pythagorean magnitude. And the unit-vector rules above are what make the component formula (below) so mechanical: every cross term \(\hat i\cdot\hat j\) collapses to zero, leaving only the matched terms.

Geometric meaning — projection

The dot product has a clean geometric reading. Write \(\vec A\cdot\vec B = A\,(B\cos\theta)\). The quantity \(B\cos\theta\) is the projection of \(\vec B\) onto \(\vec A\) — the length of the "shadow" that \(\vec B\) casts along the direction of \(\vec A\). So the dot product is the magnitude of one vector times the amount of the other that lies along it. When the vectors are perpendicular, the shadow has zero length and the dot product vanishes.

This projection picture is the reason work behaves the way it does. Only the part of a force that lies along the displacement does work; a force perpendicular to the motion — the normal reaction on a sliding block, the tension in a string for circular motion — does zero work, because its projection onto the displacement is zero. The dot product encodes that physics automatically.

The cross product (vector product)

The cross product of \(\vec A\) and \(\vec B\) is defined as

$$\vec A\times\vec B = AB\sin\theta\,\hat n$$

where \(\hat n\) is a unit vector perpendicular to the plane containing \(\vec A\) and \(\vec B\), with its sense fixed by the right-hand rule. Because the factor is \(\sin\theta\), the cross product is zero when the vectors are parallel (\(\theta=0^\circ\) or \(180^\circ\)) and maximum when they are perpendicular — the exact opposite behaviour of the dot product.

The cyclic rule is worth drilling: marching forward through \(\hat i\to\hat j\to\hat k\to\hat i\) gives a positive result, while going backwards introduces a minus sign. The magnitude \(|\vec A\times\vec B| = AB\sin\theta\) has a tidy geometric meaning too — it equals the area of the parallelogram with \(\vec A\) and \(\vec B\) as adjacent sides, so half of it is the area of the triangle they span.

The right-hand rule

The magnitude \(AB\sin\theta\) leaves the direction of the cross product to be fixed, and that is the job of the right-hand rule. Point the fingers of your right hand along the first vector \(\vec A\), then curl them toward the second vector \(\vec B\) through the smaller angle. Your outstretched thumb points along \(\vec A\times\vec B\), perpendicular to the plane of \(\vec A\) and \(\vec B\).

This is exactly why \(\vec A\times\vec B = -(\vec B\times\vec A)\). Curling from \(\vec B\) to \(\vec A\) reverses the curl, so the thumb flips to the opposite side of the plane. The magnitudes are identical; only the sign of \(\hat n\) changes. In physics this matters enormously: torque is defined as \(\vec\tau=\vec r\times\vec F\) and never \(\vec F\times\vec r\), because the order fixes whether the body turns clockwise or anticlockwise.

Component and determinant forms

Geometry gives the meaning; components give the arithmetic. Write \(\vec A = A_x\hat i + A_y\hat j + A_z\hat k\) and \(\vec B = B_x\hat i + B_y\hat j + B_z\hat k\). Using \(\hat i\cdot\hat i=1\) and \(\hat i\cdot\hat j=0\), every cross term in the dot product vanishes and only the matched terms survive:

$$\vec A\cdot\vec B = A_xB_x + A_yB_y + A_zB_z$$

The cross product, using the cyclic unit-vector rules, is most reliably written as a \(3\times3\) determinant with the unit vectors in the top row:

$$\vec A\times\vec B = \begin{vmatrix} \hat i & \hat j & \hat k \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix} = (A_yB_z - A_zB_y)\hat i - (A_xB_z - A_zB_x)\hat j + (A_xB_y - A_yB_x)\hat k$$

Notice the sign on the middle (\(\hat j\)) term — it carries a minus, a standard determinant expansion. Two diagnostic checks settle which product a problem needs: is the answer a number or a vector?, and does the answer vanish when the vectors line up or when they cross perpendicularly? One question, and the choice is made.

Dot vs cross — side by side

The two products are complementary in almost every respect. The card below collects the contrasts you should be able to recall instantly in the exam hall.

Worked examples

Three short calculations cover the question types NEET draws from this topic: a dot-and-cross from components, the angle between two vectors, and the area of a parallelogram from the cross-product magnitude.