Porisma

Worked solution

Recover a point and direction from Cartesian relations

Convert the equations of following lines to their vector equation form:

(a) 3y=2x−6=z−123y=2x-6=\dfrac{z-1}{2}

(b) x+5=2−4yx+5=2-4y, z=3z=3

CJC · Discussion · Question 2

Answer

(a) r=(3,0,1)+λ(3,2,12)\mathbf r=(3,0,1)+\lambda(3,2,12). (b) r=(−5,12,3)+λ(4,−1,0)\mathbf r=(-5,\frac 12,3)+\lambda(4,-1,0), with real λ\lambda.

(a) Set the common value to t

Then x=3+t/2x=3+t/2, y=t/3y=t/3, z=1+2tz=1+2 t. Hence

r=(3,0,1)+t(12,13,2),t∈R.\mathbf r=(3,0,1)+t\left(\frac 12,\frac 13,2\right),\quad t\in\mathbb R.

Rescaling t gives the equivalent printed form

r=(3,0,1)+λ(3,2,12),λ∈R.\mathbf r=(3,0,1)+\lambda(3,2,12),\quad\lambda\in\mathbb R.

How did this part go?

(b) Retain the constant coordinate

Set x+5=2−4y=tx+5=2-4 y=t. Then x=−5+tx=-5+t, y=12−t/4y=\frac 12-t/4, z=3z=3. Thus

r=(−5,12,3)+t(1,−14,0),t∈R,\mathbf r=\left(-5,\frac 12,3\right)+t\left(1,-\frac 14,0\right),\quad t\in\mathbb R,

or, rescaling the parameter,

r=(−5,12,3)+λ(4,−1,0),λ∈R.\mathbf r=\left(-5,\frac 12,3\right)+\lambda(4,-1,0),\quad\lambda\in\mathbb R.

Both printed parameterizations describe the same line.

How did this part go?

Common mistakes

The coefficients multiplying coordinates in the original equation are not directly its direction components.

Solution notes

Both fractional and integer direction forms retained; parameter names change when directions are rescaled.

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