Question

Find values of t for which vectors a and c are orthogonal.

a=<2,-2,4> c=<2t, 0, -t^3>

Answer 1

I know we use dot product

Answer 2

I already found one of the values, supposedly there is more than one value of t.

Answer 3

I get an equation such as

Answer 4

If 'a' and 'c' are orthogonal then their dot product is 0. Let's evaluate using the algebraic form of the dot-product and determine the value of 't' accordingly:

a · c = 2(2t) + (-2)(0) + (4)(t^3) = 0

=> 4t^3 + 4t = 0

Now factor and determine the possible for 't'.

@raffle_snaffle

Answer 5

What do you get for 't'? @raffle_snaffle

Answer 6

I used my calculator and got t=0

Answer 7

I think my equation is wrong...

Answer 8

Correct. t = 0 is the right answer. And so it turns out that 'c' is just the zero vector since t = 0.

@raffle_snaffle

Answer 9

Is there another value of t? My previous professor told me there was another value of t

Answer 10

There is another value of 't' but it's complex.

Answer 11

Well for one thing I didn't get the same equation as you.

Answer 12

oh wait

Answer 13

The equation should -4t^3 + 4t = 0 instead. I forgot the negative.

Answer 14

I messed up when you mult. the magnitude of a and c

Answer 15

Now factoring out -4t we get the following:

-4t(t^2 - 1) = 0

But note that t^2 - 1 can further be factored using difference of squares so we get:

-4t(t-1)(t+1) = 0

And now we conclude that t = 0 or t = 1 or t = -1.

@raffle_snaffle