Question

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>`

Answer 1

`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`