Problem

Find a vector equation and parametric equations f…

01:56

Problem 4 Easy Difficulty

Find a vector equation and parametric equations for the line.

The line through the point $ (0, 14, -10) $ and parallel to the line $ x = -1 + 2t $ , $ y = 6 - 3t $ , $ z = 3 + 9t $

Answer

$\mathbf{r}=(14 \mathbf{j}-10 \mathbf{k})+t(2 \mathbf{i}-3 \mathbf{j}+9 \mathbf{k})$
$x=2 t \quad y=14-3 t \quad z=-10+9 t$

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Video Transcript

No I got a problem for again. The problem for is the same thing. It passes to the going to Judo 14 and -10. And again it's parlor too. X equals 2 -1 plus two. T Why equal to 6- Treaty? And G equals 23 plus 90. Right now again with the same thing for the parameter equation. What we do can you do the same thing? This is my remember and this the vicar be here is two -3 and nine so He can parliamentary question. It just takes -0/2. The same as Y -14/-3. The same as G plus 10 overnight. And again to use a secret to T. So access to T. Why is -3 t plus 14 And G. is 90 -10. This is how the question looks like. And for the better equation like we said before this is a not some T. V. So this is due to wake up plus 14 Jacob Afghanistan K cap nasty times to wake up minus T. Jacob plus nine. Kick and dissect. That's the answer.

Tattwamasi A.

Topics

Vectors

Calculus: Early Transcendentals

Chapter 12

Vectors and the Geometry of Space

Section 5

Equations of Lines and Planes

Problem