Scalar (Dot) Product
Slope and inclination of lines, the dot product of vectors in the plane and in space, perpendicularity, and equations of lines and planes.
- A vector \(\vec{u}\) in the plane has coordinates \((u_1, u_2)\); in space, \((u_1, u_2, u_3)\).
- Its norm (length) is \(\|\vec{u}\| = \sqrt{u_1^2 + u_2^2}\) (or with three terms in space).
- Vectors add componentwise and scale by a real number.
2.1 Slope and inclination of a line
The inclination of a line is the angle \(\alpha \in [0^\circ, 180^\circ[\) it makes with the positive \(x\)-axis. Its slope is
\[ m = \tan\alpha = \frac{y_2 - y_1}{x_2 - x_1}. \]A line through point \((x_0, y_0)\) with slope \(m\) has equation \(y - y_0 = m(x - x_0)\).
The line through \(A(1,2)\) and \(B(4,8)\) has slope \(m = \dfrac{8-2}{4-1} = 2\), so its inclination is \(\alpha = \tan^{-1}(2) \approx 63.4^\circ\).
2.2 The scalar product of two vectors
For vectors \(\vec{u}\) and \(\vec{v}\) with angle \(\theta\) between them,
\[ \vec{u}\cdot\vec{v} = \|\vec{u}\|\,\|\vec{v}\|\cos\theta. \]The result is a scalar (a number), not a vector. Rearranging gives the angle:
\[ \cos\theta = \frac{\vec{u}\cdot\vec{v}}{\|\vec{u}\|\,\|\vec{v}\|}. \]2.3 Properties of the scalar product
- Commutative: \(\vec{u}\cdot\vec{v} = \vec{v}\cdot\vec{u}\).
- Distributive: \(\vec{u}\cdot(\vec{v}+\vec{w}) = \vec{u}\cdot\vec{v} + \vec{u}\cdot\vec{w}\).
- Scalar factor: \((k\vec{u})\cdot\vec{v} = k(\vec{u}\cdot\vec{v})\).
- \(\vec{u}\cdot\vec{u} = \|\vec{u}\|^2 \ge 0\).
2.4 Coordinate form and perpendicularity
In the plane: \(\vec{u}\cdot\vec{v} = u_1 v_1 + u_2 v_2\).
In space: \(\vec{u}\cdot\vec{v} = u_1 v_1 + u_2 v_2 + u_3 v_3\).
Two non-zero vectors are perpendicular exactly when their dot product is zero:
\[ \vec{u} \perp \vec{v} \iff \vec{u}\cdot\vec{v} = 0. \]For lines, two lines with slopes \(m_1\) and \(m_2\) are perpendicular iff \(m_1 \cdot m_2 = -1\), and parallel iff \(m_1 = m_2\).
Are \(\vec{u} = (2, -1, 3)\) and \(\vec{v} = (1, 5, 1)\) perpendicular? \(\vec{u}\cdot\vec{v} = 2\cdot1 + (-1)\cdot5 + 3\cdot1 = 2 - 5 + 3 = 0\). Yes, they are perpendicular.
2.5 Cartesian equations of planes in space
A plane with normal vector \(\vec{n} = (a, b, c)\) passing through \(P_0(x_0, y_0, z_0)\) consists of all points \(P(x,y,z)\) with \(\vec{n}\cdot\overrightarrow{P_0P} = 0\), i.e.
\[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \;\Longleftrightarrow\; ax + by + cz + d = 0. \]Plane through \((1,0,2)\) with normal \((3,-1,2)\): \(3(x-1) - (y-0) + 2(z-2) = 0 \Rightarrow 3x - y + 2z - 7 = 0.\)
2.6 Relative positions of lines and planes
- Two planes are parallel if their normals are parallel; perpendicular if their normals are perpendicular (\(\vec{n_1}\cdot\vec{n_2}=0\)).
- A line and a plane: the line is parallel to the plane if its direction vector is perpendicular to the plane's normal; perpendicular to the plane if its direction is parallel to the normal.
- Two lines in space may be parallel, intersecting, or skew (non-coplanar, non-intersecting).
Practice problems
- Find the angle between \(\vec{u}=(1,0)\) and \(\vec{v}=(1,1)\).
Show solution
\(\vec{u}\cdot\vec{v} = 1\). \(\|\vec{u}\|=1,\ \|\vec{v}\|=\sqrt2\). \(\cos\theta = \tfrac{1}{\sqrt2}\Rightarrow \theta = 45^\circ\). - Determine \(k\) so that \((2, k)\) and \((3, -6)\) are perpendicular.
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\(2\cdot3 + k\cdot(-6) = 0 \Rightarrow 6 - 6k = 0 \Rightarrow k = 1\). - Write the equation of the plane through \((0,0,0)\) with normal \((1,2,-2)\).
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\(x + 2y - 2z = 0\). - Are the planes \(x + y + z = 1\) and \(2x + 2y + 2z = 5\) parallel?
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Normals \((1,1,1)\) and \((2,2,2)=2(1,1,1)\) are parallel, so yes — the planes are parallel (and distinct).
Summary
- Slope \(m = \tan\alpha\); perpendicular lines satisfy \(m_1 m_2 = -1\).
- Dot product: geometric \(\|\vec u\|\|\vec v\|\cos\theta\) = algebraic \(u_1v_1 + u_2v_2 (+ u_3v_3)\).
- \(\vec u \perp \vec v \iff \vec u \cdot \vec v = 0\).
- A plane is fixed by a point and a normal vector: \(ax+by+cz+d=0\).
- Compute \((3,-2,1)\cdot(1,4,5)\).
- Give a normal vector of the plane \(4x - y + 2z = 9\).