Porisma

Worked solution

Check a point using one consistent parameter

Determine whether (−1,3,−3)(-1,3,-3) lies on the line in Example 21: r=(5,1,−2)+λ′(6,−2,1)\mathbf r=(5,1,-2)+\lambda'(6,-2,1), λ′∈R\lambda'\in\mathbb R, equivalently (x−5)/3=1−y=2z+4(x-5)/3=1-y=2z+4.

CJC · Lecture · Example 22

Method 1: compare all three components

Assume the point is on the line. Then (−1,3,−3)=(5,1,−2)+λ′(6,−2,1)(-1,3,-3)=(5,1,-2)+\lambda'(6,-2,1). The components give −1=5+6λ′-1=5+6\lambda', 3=1−2λ′3=1-2\lambda' and −3=−2+λ′-3=-2+\lambda'. Each gives λ′=−1\lambda'=-1, so the equations are consistent and the point lies on the line.

How did this part go?

Method 2: compare the Cartesian expressions

At x=−1,y=3,z=−3x=-1,y=3,z=-3, (x−5)/3=((−1)−5)/3=−2(x-5)/3=((-1)-5)/3=-2, 1−y=1−3=−21-y=1-3=-2 and 2z+4=2(−3)+4=−22z+4=2(-3)+4=-2. All three Cartesian expressions agree, so the point lies on the line. This common Cartesian value is λ=−2, while the rescaled vector parameter is λ′=−1, as in Example 21.

How did this part go?

Common mistakes

Check every coordinate using the same parameter. Different equivalent line equations can have differently scaled parameters; compare consistency within one representation.

Read the hint