Find k if (4,1,k) & (5,1,-3) are perpendicular?

  • Thread starter discy
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  • #1
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Homework Statement


Vectors
find k if (4,1,k)τ & (5,1,-3)τ are perpendicular?

From the answer sheet I know the answer is k = 7

Homework Equations


I believe I need these two but I'm not certain:

Dot-product: v*w = v1w1 + v2w2 + ... vdwd
cos θ = v*w / ||v|| * ||w||


The Attempt at a Solution



[STRIKE]Because θ = arccos(v*w / ||v|| * ||w||) = the angle between two vectors. θ should be 90 = perpendicular.

90 = arccos ((4 * 5 + 1 * 1 + k * -3) / √(4²+1²+k²) * √(5²+1²+(-3)²)) =
90 = arccos (21-3k / √(17+k²) * √35)

90 = arccos(21-3k / √(17+k²) * √35)[/STRIKE]

cos(90) = 0 = perpendicular

[itex]\frac{21-3k}{\sqrt{17+k²} * \sqrt{35}} = 0[/itex]

what would be the easiest way to get to k = 7 ?
 
Last edited:

Answers and Replies

  • #2
HallsofIvy
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What you need is that cos(90)= 0! So that two vectors are perpendicular if and only if their dot product is 0.
 
  • #3
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ok. so let me reform:

(21-3k) / √(17+k²) * √35 = 0

[itex]\frac{21-3k}{\sqrt{17+k²} * \sqrt{35}} = 0[/itex]

what would be the easiest way to get to k = 7? (when I put 7 as k I get zero, so the equation is correct)
 
Last edited:
  • #4
HallsofIvy
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If you already knew that the dot product is 21- 3k, why even use that denominator?

The two vectors are perpendicular if and only if their dot product, 21- 3k= 0. Solve that for k?
 
  • #5
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[STRIKE]Ow yeah. I'm being stupid.[/STRIKE]

And if two vectors are parallel? [STRIKE]then the angle is, I suppose 180 degrees? What equation would I need to solve in that case?

(i'm sorry, my teacher is very unclear about these things, he puts questions on his sheets but doesn't always give the answers during class)[/STRIKE]

edit: cross-product = 0
 
Last edited:

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