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29 votes
29 votes
find parametric equations and symmetric equations for the line (use the parameter t.) the line through the point (−3,3,−1) and perpendicular to both ⟨1,1,0⟩ and ⟨

User Yad Smood
by
2.3k points

1 Answer

13 votes
13 votes

The second vector that's perpendicular to the line we want (call it L) is missing; I'll denote it by ⟨a, b, c⟩.

The cross product of ⟨1, 1, 0⟩ and ⟨a, b, c⟩ is perpendicular to both of these vectors. The direction vector of L will be parallel to this cross product.

Compute the cross product.

⟨1, 1, 0⟩ × ⟨a, b, c⟩ = ⟨1•c - 0•b, 0•a - 1•c, 1•b - 1•a⟩ = ⟨c, -c, b - a⟩

Then the vector equation of L is

L(t) = ⟨c, -c, b - a⟩ t + ⟨-3, 3, -1⟩

As parametric equations, we write

x(t) = ct - 3

y(t) = -ct + 3

z(t) = (b - a) t - 1

Eliminating the parameter above, we get

x + y = (ct - 3) + (-ct + 3) = 0 ⇒ x = -y

and substituting

x = ct - 3 ⇒ t = (x + 3)/c

into z(t), we get

z = (b - a) (x + 3)/c - 1 ⇒ x = c (z + 1)/(b - a) - 3

As symmetric equations, then, we write

x = -y = c (z + 1)/(b - a) - 3

User Stultus
by
3.1k points
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