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 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.
3 answers
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
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
1 answer