Question

The angle between two vectors a= (1,1,2) and b = (-1, 2, k) is 60°. Find the value(s) of k.

Answer 1

We have to use the vector dot theorem

`\begin{gathered} a\circ b=\lbrack a\rbrack\lbrack b\rbrack cos\theta \\ (1,1,2)(\text{ -1, }2,k)=√((1^2+1^2+2^2)(\sqrt{(\text{ -}1)^2+2^2+k^2}cos60 \\ \\ \text{ -1 + 2 + 2}k=√(4)(√(5+k^2))((1)/(2)) \\ 1+2k=2((1)/(2))(√(5+k^2)) \\ 1+2k=√(5+k^2) \\ (1+2k)^2=(√(5+k^2))^2 \\ 1+4k+4k^2=5+k^2 \\ 0=4\text{ - }4k\text{ - }3k^2 \\ 3k^(^2)+4k\text{ - }4=0 \\ (3k\text{ - }2)(k\text{ + }2)=0 \\ \\ 3k\text{ - }2=0 \\ k=(2)/(3) \\ \\ k+2=0 \\ k=\text{ -}2 \end{gathered}`

So the values k can have for the angle to be 60º is -2 and 2/3