I cant get the correct answer and I can't figure out what I am doing wrong. If a=(2,2k) and b=(4,k) are perpendicular, find k. I have worked it out, and I got -2 but it doesn't work when I check it in the original equation. It should be a.b=0 and then plug in the numbers and rearrange to find k but it won't work. Any ideas? Thanks.

If a and b are two dimensionsal vectors with components (2,2k) and (4,k)respectively, and if they are perpendicular, then the dot product must be zero.

Therefore 8 + 2k^2 = 0; however this lerads to an imaginary number for k.

There is no solution in this case. Are you sure there was not a minus sign in front of one of the components?

I am sure, because I tried looking at the question again. Because the only thing that would make sense would be if k=2 but there would need to be a negative somewhere. Could be a typo. Thanks!

Another way to look at it:

a=(2,2k) and b=(4,k)
slope of a = 2k/2 = k
slope of b = k/4
This is not going to work :)
To be perpendicular, slope of b must be -1/slope of a but obviously the sign change is impossible.
In other words
k/4 = -1/k

To determine whether two vectors are perpendicular, you can use the dot product. Recall that the dot product of two vectors a = (a1, a2) and b = (b1, b2) is given by the formula: a · b = a1 * b1 + a2 * b2.

In this case, you have vector a = (2, 2k) and vector b = (4, k). To find k such that a and b are perpendicular, we set their dot product equal to zero and solve for k:

a · b = (2 * 4) + (2k * k) = 8 + 2k^2 = 0

Now, let's solve the quadratic equation 2k^2 + 8 = 0. Subtracting 8 from both sides, we get: 2k^2 = -8. Dividing both sides by 2, we have: k^2 = -4.

Here is where the issue arises. In the equation k^2 = -4, we encounter a problem. The square of any real number is always non-negative, so the equation k^2 = -4 has no real solutions. Therefore, there is no value of k that makes the vectors a and b perpendicular.

You mentioned that you got k = -2 initially, but upon checking, it didn't work in the original equation. This result is expected because there is no valid solution for k that would satisfy the condition of the vectors being perpendicular.

In summary, the vectors a and b are not perpendicular, and there is no real value for k that makes them perpendicular.