Question

verify that the two lines are parallel, and find the distance between the lines. (round your answer to three decimal places.) l1: x = 2 - t, y = 3 + 4t, z = 4 + 2t l2: x = 5t, y = 1 - 20t, z = 4 - 10t

Explanation:

Step1: Find direction vectors

For line $L_1:x = 2 - t,y = 3+4t,z = 4 + 2t$, the direction vector $\vec{v_1}=\langle - 1,4,2
angle$. For line $L_2:x = 5t,y = 1-20t,z = 4-10t$, the direction vector $\vec{v_2}=\langle5,-20,-10
angle$. Since $\vec{v_2}=- 5\vec{v_1}$, the two lines are parallel.

Step2: Select points on the lines

On $L_1$, let $t = 0$, we get the point $P_1(2,3,4)$. On $L_2$, let $t = 0$, we get the point $P_2(0,1,4)$. Then $\vec{P_1P_2}=\langle0 - 2,1 - 3,4 - 4
angle=\langle-2,-2,0
angle$.

Step3: Use the distance formula for parallel lines

The distance $d$ between two parallel lines with direction - vector $\vec{v}$ and a vector $\vec{a}$ connecting two points on the lines is $d=\frac{\vert\vec{a}\times\vec{v}\vert}{\vert\vec{v}\vert}$. Here $\vec{v}=\vec{v_1}=\langle - 1,4,2
angle$ and $\vec{a}=\vec{P_1P_2}=\langle-2,-2,0
angle$.
First, calculate $\vec{a}\times\vec{v}=

$$\begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\-2&-2&0\\-1&4&2\end{vmatrix}$$

=\vec{i}(- 4-0)-\vec{j}(-4 - 0)+\vec{k}(-8 - 2)=\langle-4,4,-10
angle$.
Then $\vert\vec{a}\times\vec{v}\vert=\sqrt{(-4)^2 + 4^2+(-10)^2}=\sqrt{16 + 16 + 100}=\sqrt{132}$. And $\vert\vec{v}\vert=\sqrt{(-1)^2+4^2 + 2^2}=\sqrt{1 + 16+4}=\sqrt{21}$.
So $d=\frac{\sqrt{132}}{\sqrt{21}}=\sqrt{\frac{132}{21}}\approx2.509$.

Answer:

$2.509$