Find all numbers k?
Find all numbers k if any so that |\| kvecu -vecv |\| = sqrt(6
vec u = <2,-1,0>
vec v =<1,2,-1>
Find all numbers
1 Answer
Explanation:
Given:
vec(u) = < 2, -1, 0 >
vec(v) = < 1, 2, -1 >
Then:
k vec(u) - vec(v) = k < 2, -1, 0 > - < 1, 2, -1 >
color(white)(k vec(u) - vec(v)) = < 2k-1, -k-2, 1 >
So we want to solve:
6 = abs(abs(k vec(u) - vec(v)))^2
color(white)(6) = (color(blue)(2k-1))^2+(color(blue)(-k-2))^2+(color(blue)(1))^2
color(white)(6) = 4k^2-color(red)(cancel(color(black)(4k)))+1+k^2+color(red)(cancel(color(black)(4k)))+4+1
color(white)(6) = 5k^2+6
Hence:
k = 0
Another way...
Actually we could have noticed a couple of things and reasoned this in a different way...
abs(abs(vec(v))) = abs(abs(< 1, 2, -1 >))
color(white)(abs(abs(vec(v)))) = sqrt((color(blue)(1))^2+(color(blue)(2))^2+(color(blue)(-1))^2)
color(white)(abs(abs(vec(v)))) = sqrt(1+4+1)
color(white)(abs(abs(vec(v)))) = sqrt(6)
vec(u) * vec(v) = < 2, -1, 0 > * < 1, 2, -1 >
color(white)(vec(u) * vec(v)) = (color(blue)(2))(color(blue)(1))+(color(blue)(-1))(color(blue)(2))+(color(blue)(0))(color(blue)(-1)) = 2-2+0 = 0
So
So adding any non-zero multiple of
Hence the only solution is