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>

1 Answer
Aug 2, 2018

k=0

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 vec(v) and hence -vec(v) is already of the required length sqrt(6) and vec(u) is orthogonal to it.

So adding any non-zero multiple of vec(u) to -vec(v) will result in a longer vector.

Hence the only solution is k=0