Question

If α =i+j−2k, β =−i+2j+3k andγ =5i+8k, then find c and d such that γ−cα−dβ is perpendicular to both α and β

Answer 1

If
`\hat\gamma-c\hat\alpha-d\hat\beta` is perpendicular to both
`\hat\alpha` and
`\hat\beta`, then their dot products are zero.

Recall that for any vector
`\hat x`,
`\hat x\cdot \hat x = \|\hat x\|^2`.

Now, we have

`(\hat\gamma - c\hat\alpha - d\hat\beta) \cdot \hat\alpha = (\hat\gamma\cdot\hat\alpha) - c\|\hat\alpha\|^2 - d (\hat\beta\cdot\hat\alpha) = 0`

`\hat\gamma\cdot\hat\alpha = 5 + 0 - 16 = -11`

`\|\hat\alpha\|^2 = 1^2 + 1^2 + (-2)^2 = 6`

`\hat\beta\cdot\hat\alpha = -1 + 2 - 6 = -5`

`\implies 6c - 5d = -11`

and

`(\hat\gamma - c\hat\alpha - d\hat\beta) \cdot \hat\beta = (\hat\gamma\cdot\hat\beta) - c(\hat\alpha\cdot\hat\beta) - d \|\hat\beta\|^2 = 0`

`\hat\gamma\cdot\hat\beta = -5 + 0 + 24 = 19`

`\|\hat\beta\|^2 = (-1)^2 + 2^2 + 3^2 = 14`

`\implies 5c - 14d = -19`

Solve for
`c` and
`d`. By elimination,

`5(6c - 5d) - 6(5c - 14d) = 5(-11) - 6(-19)`

`-25d + 84d = -55 + 104`

`59d = 59`

`\boxed{d=1}`

`6c-5 = -11`

`6c=-6`

`\boxed{c=-1}`