Porisma

Worked solution

Read a common Cartesian parameter and rescale it consistently

A line has equation (x−5)/3=1−y=2z+4(x-5)/3=1-y=2z+4. Find a vector equation of the line.

CJC · Lecture · Example 21

Set the common value equal to a parameter

Let (x−5)/3=1−y=2z+4=λ(x-5)/3=1-y=2z+4=\lambda, with λ∈R\lambda\in\mathbb R. Then (x−5)/3=λ(x-5)/3=\lambda gives x=5+3λ; 1−y=λ gives y=1−λ; and 2 z+4=λ gives z=−2+λ/2.

How did this part go?

Collect the constant and parameter coefficients

Therefore r=(5,1,−2)+λ(3,−1,1/2)\mathbf r=(5,1,-2)+\lambda(3,-1,1/2), λ∈R\lambda\in\mathbb R.

How did this part go?

Write an equivalent equation with an integer direction

The same line is r=(5,1,−2)+λ′(6,−2,1)\mathbf r=(5,1,-2)+\lambda'(6,-2,1), with λ′=λ/2\lambda'=\lambda/2. Both parameters range over all real values. Converting between Cartesian, parametric and vector forms preserves the line, even when the direction vector and parameter change together.

How did this part go?

Common mistakes

After doubling a direction vector, halve its parameter for the same point. A coefficient such as 2 z+4 must first be solved for z.

Read the hint