Question

Let = (-8, 6, -5) and = (3, -2, 1) be vectors in ℝ³. Using the contents and techniques from this course, find a vector of magnitude 8 that is perpendicular to both and simultaneously. Show your step-by-step work and provide the resulting vector .

Answer 1

`\(\mathbf{v} = (1, -7, -8)\)`because The vector (1, -7, -8) is perpendicular to both (-8, 6, -5) and (3, -2, 1) because it is the cross product of the two vectors, representing their orthogonal relationship.

Step-by-step explanation:

To find a vector perpendicular to both
`\(\mathbf{u}\) and \(\mathbf{v}\)`, we can take their cross product. The cross product
`\(\mathbf{u} * \mathbf{v}\) gives a vector that is perpendicular to both \(\mathbf{u}\) and \(\mathbf{v}\)`. In this case:

`\[ \mathbf{u} * \mathbf{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -8 & 6 & -5 \\ 3 & -2 & 1 \end{vmatrix} \]`

Expanding this determinant gives the cross product vector
`\(\mathbf{u} * \mathbf{v} = (1, -7, -8)\)`. The magnitude of this vector can be found using the formula
`\(\|\mathbf{u} * \mathbf{v}\| = √(1^2 + (-7)^2 + (-8)^2) = √(1 + 49 + 64) = √(114)\)`. To obtain a vector of magnitude 8, we scale the cross product vector by \(\frac{8}
`{√(114)}\), resulting in the final answer \(\mathbf{v} = \left((8)/(√(114)), -(56)/(√(114)), -(64)/(√(114))\right)\).`