3.2 Dot Product and Vector Projections

Dot (Scalar) Product — Definition

The dot product (scalar product) of two vectors $\vec{a}$ and $\vec{b}$ is defined as:

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

where $\theta$ is the angle between the two vectors ($0^{\circ }\le \theta \le 180^{\circ }$).

The result is a scalar (a real number), not a vector.

Geometric Interpretation

$\vec{a}\cdot \vec{b}$ equals the magnitude of $\vec{a}$ multiplied by the projection of $\vec{b}$ onto $\vec{a}$ (or vice versa).

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Dot Product in Component Form

If $\vec{u}=u_1\hat{i}+u_2\hat{j}+u_3\hat{k}$ and $\vec{v}=v_1\hat{i}+v_2\hat{j}+v_3\hat{k}$, then:

$\vec{u}\cdot \vec{v}=u_1{v}_1+u_2{v}_2+u_3{v}_3$

This follows from the orthonormality of $\hat{i},\hat{j},\hat{k}$:

• $\hat{i}\cdot \hat{i}=\hat{j}\cdot \hat{j}=\hat{k}\cdot \hat{k}=1$
• $\hat{i}\cdot \hat{j}=\hat{j}\cdot \hat{k}=\hat{k}\cdot \hat{i}=0$

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Angle Between Two Vectors

Rearranging the dot product definition:

$\cos \theta =\frac{\vec{u}\cdot \vec{v}}{|\vec{u}||\vec{v}|}$

$\theta ={\cos }^{-1}\left(\frac{\vec{u}\cdot \vec{v}}{|\vec{u}||\vec{v}|}\right),0^{\circ }\le \theta \le 180^{\circ }$

Example: Find the angle between $\vec{a}=2\hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+2\hat{k}$.

$\vec{a}\cdot \vec{b}=(2)(1)+(1)(-1)+(-1)(2)=2-1-2=-1$ $|\vec{a}|=\sqrt{4+1+1}=\sqrt{6},|\vec{b}|=\sqrt{1+1+4}=\sqrt{6}$ $\cos \theta =\frac{-1}{\sqrt{6}\cdot \sqrt{6}}=\frac{-1}{6}⟹\theta ={\cos }^{-1}\left(-\frac{1}{6}\right)$

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Condition for Orthogonality

Two non-zero vectors $\vec{a}$ and $\vec{b}$ are perpendicular (orthogonal) if and only if:

$\vec{a}\cdot \vec{b}=0$

This is because $\cos 90^{\circ }=0$.

Example: Show that $\vec{a}=2\hat{i}-3\hat{j}+\hat{k}$ and $\vec{b}=\hat{i}+\hat{j}+\hat{k}$ are not orthogonal.

$\vec{a}\cdot \vec{b}=2-3+1=0⟹\text{They ARE orthogonal.}$

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Projection of a Vector Along Another

The scalar projection of $\vec{a}$ along $\vec{b}$ is:

${\text{proj}}_{\vec{b}}\vec{a}=\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}$

The vector projection of $\vec{a}$ onto $\vec{b}$ is:

${\overrightarrow{\text{proj}}}_{\vec{b}}\vec{a}=\frac{\vec{a}\cdot \vec{b}}{|\vec{b}{|}^2}\vec{b}$

Example: Find the projection of $\vec{a}=3\hat{i}+4\hat{j}$ along $\vec{b}=\hat{i}+\hat{j}$.

$\vec{a}\cdot \vec{b}=3+4=7,|\vec{b}|=\sqrt{2}$ $\text{Scalar projection}=\frac{7}{\sqrt{2}}=\frac{7\sqrt{2}}{2}$

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Direction Cosines

For a vector $\mathbf{v}=v_1\hat{i}+v_2\hat{j}+v_3\hat{k}$ with magnitude $|\mathbf{v}|$, the direction cosines are:

$\cos \alpha =\frac{v_1}{|\mathbf{v}|},\cos \beta =\frac{v_2}{|\mathbf{v}|},\cos \gamma =\frac{v_3}{|\mathbf{v}|}$

They are the components of the unit vector $\hat{v}=\frac{\mathbf{v}}{|\mathbf{v}|}$, and satisfy:

${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$

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Work Done by a Constant Force

If a constant force $\vec{F}$ moves an object through displacement $\vec{d}$, the work done is:

$W=\vec{F}\cdot \vec{d}=|\vec{F}||\vec{d}|\cos \theta$

where $\theta$ is the angle between $\vec{F}$ and $\vec{d}$. Work is a scalar quantity measured in joules (J).

Example: A force $\vec{F}=5\hat{i}+3\hat{j}$ N displaces an object by $\vec{d}=4\hat{i}+2\hat{j}$ m.

$W=\vec{F}\cdot \vec{d}=(5)(4)+(3)(2)=20+6=26\text{ J}$

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Key Identity: Sum of Three Vectors Equal to Zero

If $\vec{a}+\vec{b}+\vec{c}=\vec{0}$, squaring both sides:

$|\vec{a}{|}^2+|\vec{b}{|}^2+|\vec{c}{|}^2+2(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})=0$

$\therefore \vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}=-\frac{|\vec{a}{|}^2+|\vec{b}{|}^2+|\vec{c}{|}^2}{2}$