Question

Vectors:
question below
only 7b. I need explanation why the answer is k is greater than or equal to zero

Answer 1

answer: \(k \ge 0\)

Answer 2

\[|a||b|~cos\alpha=a\cdot b\]

|a| = sqrt(2), |b| = k, a.b = k

Answer 3

but the real question is in the fact that (0,0) is 90 degrees to all other vectors, and cannot possibly be 45 degree with 1,1

Answer 4

why it can't be 45 degrees with 1,1?

Answer 5

0,0 dot 1,1 = 0, and when a dot product is zero, the vectors are 90 degrees

Answer 6

consider this in 3 dimensions:

1,1,0
0,k,0
and
0,0,1

when k=0, all we are left with is 1,1,0 and 0,0,1

Answer 7

sorry, i 'm still trying to understand this :/
where did you get 0,0,1?

Answer 8

its a vector that is perpendicular to all other vectors in the xy plane; the zero vector in they xy plane is actually a part of the 0,0,1 vector, or line, in xyz space

Answer 9

oh ok :)

Answer 10

k=0 seems problematic. But here is what they are thinking:

\[ a \cdot b = |a| |b| \cos \theta \]
\[ |a| = \sqrt{2} , |b|= |k| , \cos \theta= \frac{1}{\sqrt{2}}\]
\[ a \cdot b = k = \sqrt{2} \ |k| \frac{1}{\sqrt{2}} \\
k= |k|\]
k≥0 is the solution

Answer 11

to clrify the picture with @amistre64 permition:

the angle will always be 45º if k is positive.