Showing that these two lines are parallel.

Question

$$ \dfrac{x - 1}{2} = 2 - y = 5 - z \quad \text{and} \quad \dfrac{4 - x}{4} = \dfrac{3 + y}{2} = \dfrac{5 + z}{2}. $$

I was given this math problem as homework, and I have spent the past hour spinning my wheels on how to do it. Help would be much appreciated. Thank you.

Answer 1

Hint: show that the direction vectors are parallel by finding both, and showing one is a multiple of the other.

Recall that the equation of a line is given by $$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$$ where $(x_0,y_0,z_0)$ is a point on the line, and $\langle a,b,c\rangle$ is the direction vector of the line.

Answer 2

Hint: $(2,1,1)=\frac12 (4,2,2)$

Answer 3

Why don't you write the line in (normal, vectorial) parametrized form?

$$\frac{x-1}2=2-y=5-z\iff x-1=4-2y=10-2z\implies\begin{cases}x=\;\;5-2y\\y=-3+z\end{cases}\implies$$

$$(11-2z\,,\,-3+z\,,\,z)=(11,-3,0)+t(-2,1,1)$$

Do the same with the other line:

$$\frac{4-x}4=\frac{3+y}2=\frac{5+z}2\implies4-x=6+2y=10+2z\implies\begin{cases}x=-2-2y\\y=\;\;\;2+z\end{cases}\implies$$

$$(-6-2z\,,\,2+z\,,\,z)=(-6\,,\,2\,,\,0)+t(-2,1,1)$$

And now it is obvious both lines are parallel. There are formulas to do the above faster and more directly, yet it is important, imo, to learn how to get rid of some problems by doing some basic algebra when this is possible and we don't know/remember the formulas, as in this case.

Answer 4

Remember that two lines are parallel if and only if there direction vectors are multiples of each other. Ex:

$$ \vec r_1 = t \vec r_2, \ t \in \mathbb{R}$$

For this example we have $\vec r_1 = <2, 1, 1>$, and $\vec r_2 = <4, 2, 2>$

Since $\vec r_1 = 2 \vec r_2$ we can conclude that these lines are parallel.