Scalar product of two vectors

Two non-zero vectors $\bar a,\bar b$, drawn from a common initial point, make an angle $\theta$ with

$0\le\theta\le\pi$; this angle is written $\bar a\wedge\bar b$. Collinear vectors make an angle of $0$ if they

point the same way and $\pi$ if opposite.

Definition (scalar/dot product). $\bar a\cdot\bar b=|\bar a||\bar b|\cos\theta$. Since this is a real

number, the dot product is also called the scalar product.

Notes.

1. If $\bar a=\bar 0$ or $\bar b=\bar 0$, $\theta$ is undefined and $\bar a\cdot\bar b$ is defined to be 0.
2. If $\theta=0$: $\bar a\cdot\bar b=|\bar a||\bar b|$; in particular $\bar a\cdot\bar a=|\bar a|^2$.
3. If $\theta=\pi$: $\bar a\cdot\bar b=-|\bar a||\bar b|$.
4. $\bar a\perp\bar b\iff\bar a\cdot\bar b=0$ (given both are non-zero), and conversely $\bar a\cdot\bar b=0\implies\bar a=\bar 0$ or $\bar b=\bar 0$ or $\theta=\pi/2$. Also $\bar a\cdot\bar b=\bar b\cdot\bar a$ (commutative).
5. Distributive over addition: $\bar a\cdot(\bar b+\bar c)=\bar a\cdot\bar b+\bar a\cdot\bar c$.
6. For scalars $m,n$: $(m\bar a)\cdot(n\bar b)=mn(\bar a\cdot\bar b)$ and $(m\bar a)\cdot\bar b=m(\bar a\cdot\bar b)=\bar a\cdot(m\bar b)$.
7. Cauchy-Schwarz inequality: $|\bar a\cdot\bar b|\le|\bar a||\bar b|$.

For the standard basis, $\hat\imath\cdot\hat\imath=\hat\jmath\cdot\hat\jmath=\hat k\cdot\hat k=1$ and

$\hat\imath\cdot\hat\jmath=\hat\jmath\cdot\hat k=\hat k\cdot\hat\imath=0$; consequently, for $\bar a=a_1\hat\imath+a_2\hat\jmath+a_3\hat k$ and $\bar b=b_1\hat\imath+b_2\hat\jmath+b_3\hat k$,

Worked examples.

• $\bar a\cdot\bar b=|\bar a||\bar b|\cos45^\circ=3\cdot6\cdot\tfrac{\sqrt2}{2}=9\sqrt2$ for $|\bar a|=3,|\bar b|=6$.
• For $\bar a=3\hat\imath+4\hat\jmath-5\hat k,\ \bar b=3\hat\imath-4\hat\jmath-5\hat k$: (i) $\bar a\cdot\bar b=9-16+25=18$; (ii) $|\bar a|=|\bar b|=\sqrt{50}$, so $\cos\theta=18/50$, giving $\theta=\cos^{-1}(18/50)$; (iii) scalar projection of $\bar a$ on $\bar b$ is $(\bar a\cdot\bar b)/|\bar b|=18/\sqrt{50}$; (iv) vector projection of $\bar b$ along $\bar a$ is $\dfrac{\bar a\cdot\bar b}{|\bar a|^2}\bar a=\dfrac{18}{50}(3 \hat\imath+4\hat\jmath-5\hat k)$.
• To find $\lambda$ making $3\hat\imath+2\hat\jmath+9\hat k$ and $\hat\imath+\lambda\hat\jmath+3\hat k$ perpendicular: set the dot product $3+2\lambda+27=0\Rightarrow\lambda=-15$; to make them parallel, match ratios of like components, $3/1=2/\lambda=9/3\Rightarrow\lambda=2/3$.
• Given $\bar a=\hat\imath+2\hat\jmath-3\hat k,\bar b=3\hat\imath-\hat\jmath+2\hat k$, the angle between $2\bar a+\bar b$ and $\bar a+2\bar b$ is found by first simplifying each combination into a single vector, then applying $\cos\theta=(\bar m\cdot\bar n)/(|\bar m||\bar n|)$ to the two simplified vectors.
• A line making angles $90^\circ,60^\circ,30^\circ$ with the X,Y,Z axes has direction cosines $0,\tfrac12, \tfrac{\sqrt3}2$.
• To find the vector projection of $\overrightarrow{PQ}$ on $\overrightarrow{AB}$ from four given points: form both vectors from the point coordinates, then apply the vector-projection formula $\dfrac{\overrightarrow{PQ}\cdot\overrightarrow{AB}}{|\overrightarrow{AB}|^2}\,\overrightarrow{AB}$.
• To find $\lambda$ making the angle between $\hat\imath+\lambda^2\hat\jmath+4\hat k$ and $2\hat\imath-\lambda\hat\jmath+2\lambda\hat k$ obtuse: the angle is obtuse exactly when the dot product is negative, so simplify the dot product to an inequality in $\lambda$ and solve it.
• Direction cosines of $2\hat\imath+2\hat\jmath-\hat k$: divide by its magnitude $3$ to get $(\tfrac23,\tfrac23,-\tfrac13)$.
• If a line is inclined at $45^\circ$ to the X-axis and $60^\circ$ to the Y-axis, the third direction cosine $n$ is found from $l^2+m^2+n^2=1$, giving $n=\pm\tfrac12$; multiplying the direction cosines by the given total length $12$ gives the position vector of the far endpoint.
• Direction ratios from two given points, with the requirement that the angle $\alpha$ with the X-axis be …