Question

Find the component of \(u\) that is orthogonal to \(\mathbf{v},\) given \(\mathbf{w}_{\mathbf{1}}=\operatorname{proj}_{\mathbf{v}} \mathbf{u}\). $$ \mathbf{u}=\langle 8,2,0\rangle, \quad \mathbf{v}=\langle 2,1,-1\rangle, \quad \operatorname{proj}_{\mathbf{v}} \mathbf{u}=\langle 6,3,-3\rangle $$

Short answer

The component of vector \(u\) that is orthogonal to vector \(v\) is \(\langle 2,-1,3\rangle\).

Step-by-step solution

Understanding given vectors

In the problem, three vectors are given: \(\mathbf{u}=\langle 8,2,0\rangle\), \(\mathbf{v}=\langle 2,1,-1\rangle\), and the projection of \(u\) onto \(v\), denoted as \(\operatorname{proj}_{\mathbf{v}} \mathbf{u}=\langle 6,3,-3\rangle\).

Calculating the orthogonal vector

To find the orthogonal vector, a simple calculation is required: \(\mathbf{w}_{2}=\mathbf{u}-\mathbf{w}_{1}\), where \(w_2\) is the vector that is orthogonal to \(v\), and \(w_1\) is the projection of \(u\) onto \(v\). Substituting the given vectors into this expression, we get \(\mathbf{w}_{2}=\langle 8,2,0\rangle - \langle 6,3,-3\rangle\).

Deriving the vector components

We continue by subtracting the components of the vectors from one another. The result is \(\mathbf{w}_{2}=\langle 8-6,2-3,0-(-3)\rangle = \langle 2,-1,3\rangle\). This is the vector that is orthogonal to \(v\).