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Dot product Knowledge-Based Study

Physics · Introduction to Physics · Physics · Math Tools for Physics

Physics topic explanations on Dot product.

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  1. What is the Dot Product?

    In physics, vectors can be multiplied in two distinct ways depending on whether the result is a scalar or another vector. The dot product (also known as the scalar product) takes two vectors and produces a single scalar value (a real number with magnitude and sign, but no spatial direction).

    Geometric Formulation

    Given two vectors A⃗\vec{A} and B⃗\vec{B} placed tail-to-tail with an angle θ\theta between them (where 0∘≤θ≤180∘0^\circ \le \theta \le 180^\circ), their dot product is defined as:

    A⃗⋅B⃗=∣A⃗∣∣B⃗∣cos⁡θ=ABcos⁡θ\vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}|\cos\theta = AB\cos\theta

    Here, AA and BB represent the magnitudes (lengths) of vectors A⃗\vec{A} and B⃗\vec{B}, and θ\theta is the smallest angle between their directions.

    Physical Interpretation: Scalar Projection

    The term Bcos⁡θB\cos\theta represents the component or projection of vector B⃗\vec{B} along the line of vector A⃗\vec{A}. Similarly, Acos⁡θA\cos\theta is the projection of A⃗\vec{A} along B⃗\vec{B}. Thus, the dot product measures the degree to which two vectors point in the same direction.

    Evaluating the Dot Product Using Components

    When vectors are expressed in Cartesian coordinate form using the standard orthonormal unit vectors i^\hat{i}, j^\hat{j}, and k^\hat{k}, we can compute the dot product directly from their scalar components.

    Dot Products of Unit Vectors

    Because the unit vectors each have a magnitude of 1 and are mutually perpendicular:

    • Parallel unit vectors: i^⋅i^=j^⋅j^=k^⋅k^=(1)(1)cos⁡(0∘)=1\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = (1)(1)\cos(0^\circ) = 1
    • Orthogonal unit vectors: i^⋅j^=j^⋅k^=i^⋅k^=(1)(1)cos⁡(90∘)=0\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{i} \cdot \hat{k} = (1)(1)\cos(90^\circ) = 0

    Algebraic Formula

    Let A⃗=Axi^+Ayj^+Azk^\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k} and B⃗=Bxi^+Byj^+Bzk^\vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}. Distributing the dot product over the components yields:

    A⃗⋅B⃗=AxBx+AyBy+AzBz\vec{A} \cdot \vec{B} = A_x B_x + A_y B_y + A_z B_z

    This algebraic form enables direct calculation of the scalar product without knowing the angle between the vectors beforehand.

    Mathematical Properties

    The dot product satisfies several algebraic properties essential for vector manipulation in physics:

    • Commutative Property: A⃗⋅B⃗=B⃗⋅A⃗\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}
    • Distributive Property: A⃗⋅(B⃗+C⃗)=A⃗⋅B⃗+A⃗⋅C⃗\vec{A} \cdot (\vec{B} + \vec{C}) = \vec{A} \cdot \vec{B} + \vec{A} \cdot \vec{C}
    • Multiplication by a Scalar: c(A⃗⋅B⃗)=(cA⃗)⋅B⃗=A⃗⋅(cB⃗)c(\vec{A} \cdot \vec{B}) = (c\vec{A}) \cdot \vec{B} = \vec{A} \cdot (c\vec{B})

    Sign and Orientation

    The sign of A⃗⋅B⃗\vec{A} \cdot \vec{B} depends entirely on cos⁡θ\cos\theta:

    • Acute Angle (0∘≤θ<90∘0^\circ \le \theta < 90^\circ): cos⁡θ>0\cos\theta > 0, so A⃗⋅B⃗>0\vec{A} \cdot \vec{B} > 0.
    • Orthogonal Vectors (θ=90∘\theta = 90^\circ): cos⁡(90∘)=0\cos(90^\circ) = 0, so A⃗⋅B⃗=0\vec{A} \cdot \vec{B} = 0. This provides an absolute test for perpendicularity between two non-zero vectors.
    • Obtuse Angle (90∘<θ≤180∘90^\circ < \theta \le 180^\circ): cos⁡θ<0\cos\theta < 0, so A⃗⋅B⃗<0\vec{A} \cdot \vec{B} < 0.
    • Self Dot Product: A⃗⋅A⃗=AAcos⁡(0∘)=A2=∣A⃗∣2\vec{A} \cdot \vec{A} = A A \cos(0^\circ) = A^2 = |\vec{A}|^2. Thus, the magnitude of a vector is ∣A⃗∣=A⃗⋅A⃗|\vec{A}| = \sqrt{\vec{A} \cdot \vec{A}}.

    Work Done by a Constant Force

    In classical mechanics, the primary application of the dot product is finding the mechanical work WW performed by a constant force F⃗\vec{F} acting on an object that undergoes a displacement d⃗\vec{d}:

    W=F⃗⋅d⃗=Fdcos⁡θW = \vec{F} \cdot \vec{d} = F d \cos\theta

    Only the component of force parallel to the displacement contributes to the work done. A force acting strictly perpendicular to the displacement (such as the normal force on a horizontal surface) does zero work because cos⁡(90∘)=0\cos(90^\circ) = 0.

    Finding the Angle Between Two Vectors

    Combining the geometric and component definitions allows us to determine the angle θ\theta between any two known vectors:

    cos⁡θ=A⃗⋅B⃗AB=AxBx+AyBy+AzBzAx2+Ay2+Az2Bx2+By2+Bz2\cos\theta = \frac{\vec{A} \cdot \vec{B}}{AB} = \frac{A_x B_x + A_y B_y + A_z B_z}{\sqrt{A_x^2 + A_y^2 + A_z^2}\sqrt{B_x^2 + B_y^2 + B_z^2}}

    Example

    Consider A⃗=3i^+4j^\vec{A} = 3\hat{i} + 4\hat{j} and B⃗=4i^−3j^\vec{B} = 4\hat{i} - 3\hat{j}:

    A⃗⋅B⃗=(3)(4)+(4)(−3)=12−12=0\vec{A} \cdot \vec{B} = (3)(4) + (4)(-3) = 12 - 12 = 0

    Since their dot product is zero, vectors A⃗\vec{A} and B⃗\vec{B} are mutually perpendicular (θ=90∘\theta = 90^\circ).