AcademyVectors In Rn
Academy
Angles And Orthogonality
Level 1 - Linear Algebra topic page in Vectors In Rn.
Principle
Angles in \(\mathbb R^n\) are defined using the scalar product. Two non-zero vectors are orthogonal when their angle is \(\pi/2\), which is equivalent to having scalar product zero.
Orthogonality is the algebraic version of perpendicularity.
Notation
\(\theta\)
the angle between two non-zero vectors, measured in radians
\(\cos\theta\)
cosine of the angle between the vectors
\(\mathbf u\perp\mathbf v\)
\mathbf u and \mathbf v are orthogonal
\(|\mathbf u|,|\mathbf v|\)
norms of the vectors
\(\mathbf u\cdot\mathbf v\)
scalar product
The Core Test
For non-zero vectors \(\mathbf u\) and \(\mathbf v\), define the angle by
Angle formula
\[\cos\theta=\frac{\mathbf u\cdot\mathbf v}{|\mathbf u|\,|\mathbf v|}\]
The orthogonality test is
Orthogonality
\[\mathbf u\perp\mathbf v\Longleftrightarrow \mathbf u\cdot\mathbf v=0\]
The angle formula requires both vectors to be non-zero. If one vector is \(\mathbf 0\), the denominator is zero and the vector has no direction.
For non-zero vectors, the sign of the scalar product classifies the angle:
- \(\mathbf u\cdot\mathbf v>0\): acute angle.
- \(\mathbf u\cdot\mathbf v=0\): right angle.
- \(\mathbf u\cdot\mathbf v<0\): obtuse angle.
Worked Cases
Question
Find the angle between \(\mathbf u=(1,1,0)\) and \(\mathbf v=(1,0,1)\).
Answer
Compute the scalar product: \(\mathbf u\cdot\mathbf v=1\cdot1+1\cdot0+0\cdot1=1\). Compute the norms: \(|\mathbf u|=\sqrt{1^2+1^2+0^2}=\sqrt2\), and \(|\mathbf v|=\sqrt{1^2+0^2+1^2}=\sqrt2\). Use the angle formula: \(\cos\theta=1/(\sqrt2\sqrt2)=1/2\). Therefore \(\theta=\pi/3\).
Examples
Question
Classify the angle between \((1,2)\) and \((-3,1)\) as acute, right, or obtuse.
Answer
Compute the scalar product: \((1,2)\cdot(-3,1)=1(-3)+2(1)=-3+2=-1\). The scalar product is negative, so the cosine of the angle is negative. Therefore the angle is obtuse.