Simplify. (-6x^2 + 2y - 1) + (2x^2-5y + 3) A) -4x^2 + 3y -2 B) -4x^2 - 3y - 2 C) -4x^2 + 3y +2 D) -4x^2 -3y +2
1 answer:
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9514 1404 393
Answer:
1.3363
Step-by-step explanation:
The basic idea here is to find an expression for the direction vector between a point on L1 and a point on L2. Then, solve for the points on L1 and L2 that make that vector perpendicular to both lines L1 and L2. (The dot product of direction vectors is zero.) The distance between the points found is the shortest distance between the lines.
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Let P be a point on L1. Then the parametric equation for P is ...
P = (6t, 0, -t) . . . . . . origin + t × direction vector
Let Q be a point on L2. The direction vector for L2 is given by the difference between the given points. It is (4-1, 1-(-1), 6-1) = (3, 2, 5). Then the parametric equation for Q is ...
Q = (3s+1, 2s-1, 5s+1) . . . . (1, -1, 1) + s × direction vector
The direction vector for PQ is ...
Q -P = (3s+1-6t, 2s-1, 5s+1+t)
The dot product of this and the two lines' direction vectors will be zero:
(3s+1-6t, 2s-1, 5s+1+t)·(6, 0, -1) = 0 = 13s -37t +5 . . . perpendicular to L1
(3s+1-6t, 2s-1, 5s+1+t)·(3, 2, 5) = 0 = 38s -13t +6 . . . perpendicular to L2
The solution to these equations is ...
s = -157/1237
t = 112/1237
Then (Q-P) becomes (94, -1551, 564)/1237, and its length is ...
|PQ| = √(94² +1551² +564²)/1237 ≈ 1.3363
The distance between the two lines is about 1.3363 units.
