1
$\begingroup$

I know these two lines are parallel, but I don't know how to find the distance between them. Any suggestions?

Line 1: (3, 4, 7) + {6, 2, 4}t

Line 2: (-1, 5, -1) + {3, 1, 2}t

I tried creating a triangle composed of an orthogonal line to both lines (the shortest distance between the lines), a hypotenuse that connects the two known points (3, 4, 7) and (-1, 5, -1), and a leg that connects (-1, 5, -1) to the intersection of the orthogonal line on line 2. On line 1, the point (3, 4, 7) was the intersection of the orthogonal line and the hypotenuse.

I'm sorry if that doesn't make any sense, I did my best. I have no idea how to include pictures on this website, this is my first post. And this is not a homework assignment, it's a practice problem for a Multivariable Calculus class. I'm studying for a test we have next week. Any and all help is appreciated!

$\endgroup$

closed as off-topic by Ian MacDonald, 2012rcampion, Matt, Deusovi, f'' May 26 '16 at 17:14

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – Ian MacDonald, 2012rcampion, Matt, Deusovi, f''
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 5
    $\begingroup$ This is off-topic here. Perhaps try asking on math.SE. $\endgroup$ – Ian MacDonald May 26 '16 at 17:04
  • $\begingroup$ Hint: the component of a vector $\vec{u}$ perpendicular to $\vec{v}$ is $\vec{u}-\vec{v}\frac{\vec{u}\cdot\vec{v}}{\vec{v}\cdot\vec{v}}$. $\endgroup$ – f'' May 26 '16 at 17:17

Browse other questions tagged or ask your own question.