Is there a way to do one of these with the cross product?
Yes, you can take the magnitude of the cross product of the direction vector for the line and the vector from
any point on the line to the origin, then divide that by the magnitude of the direction vector to get your final answer. To solve your problem using this method:
Given:
x = 1 + t
y = 2 - t
z = -1 + 2tThe direction vector is <1, -1, 2>. (The numbers are the coefficient of the t's from x,y, and z.)
Now let P be any point on the line and Q(0,0,0) be the origin. Since P can be
any point on the line, we should make it easy on ourselves and let t=0, and we get the point P(1,2,-1). It follows that the vector
PQ is <-1, -2, 1>. (Just subtract P from Q)
Now, you take the cross product of vector
PQ and the direction vector. You should get <-3, 3, 3>.
To get your answer, you take the magnitude of <-3,3,3> and divide it by the magnitude for the direction vector <1,-1,2>.
The answer should sqrt(9/2) or 2.12132.Why does this method work? Geometrically, the magnitude of the cross product of two vectors will give you the area of the parallelogram. The shortest distance from any point to a line is basically the height of the parallelogram, which is what you are trying to find. The area of a parallelogram is A=bh. The length of the base is just the magnitude of the direction vector. That's why you can get the height (the shortest distance) by dividing the area of the parallelogram by the its base. Don't mean to make it sound very easy, but my university hired me as a Peer Tutor and Mentor, so I have a habit of making things as easy as I can.
Now make up a few random numbers and practice so you can ace the exam