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Varsity · Mathematics · First Year

Linear Algebra

Drag vectors, build matrices, and see transformations live. The abstraction becomes visual.

Vectors

A vector has both magnitude and direction. In R2\mathbb{R}^2 we write v=(xy)\mathbf{v} = \begin{pmatrix} x \\ y \end{pmatrix} or (x,y)(x,y). The zero vector 0\mathbf{0} has magnitude 0 and no direction.
Theorem 1📌 Learn for exam
Vector Operations

For vectors a=(a1,a2)\mathbf{a} = (a_1, a_2) and b=(b1,b2)\mathbf{b} = (b_1, b_2), scalar kk:

a+b=(a1+b1,  a2+b2)ka=(ka1,  ka2)a=a12+a22\begin{aligned} \mathbf{a} + \mathbf{b} &= (a_1+b_1,\; a_2+b_2) \\ k\mathbf{a} &= (ka_1,\; ka_2) \\ \|\mathbf{a}\| &= \sqrt{a_1^2 + a_2^2} \end{aligned}
▸ Vector Addition — drag the tips of a and bboth tips are draggable
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Theorem 2📌 Learn for exam
Dot Product
ab=a1b1+a2b2=abcosθ\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 = \|\mathbf{a}\|\|\mathbf{b}\|\cos\theta

Key facts: ab=0    ab\mathbf{a} \cdot \mathbf{b} = 0 \iff \mathbf{a} \perp \mathbf{b}. The dot product is a scalar, not a vector.

▸ Dot Product — drag vectors to change the anglewatch the projection (teal dot)
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Tip
When ab>0\mathbf{a} \cdot \mathbf{b} > 0 the angle between them is acute. When <0< 0 it is obtuse. When =0= 0 they are perpendicular.
Worked Example

Dot product and angle

Given a=(3,1)\mathbf{a} = (3, -1) and b=(2,4)\mathbf{b} = (2, 4), find ab\mathbf{a}\cdot\mathbf{b} and the angle between them.