Question

In Exercises 31 and 32 , find the component of \(u\) that is orthogonal to \(\mathbf{v},\) given \(\mathbf{w}_{\mathbf{1}}=\operatorname{proj}_{\mathbf{v}} \mathbf{u}\). $$ \mathbf{u}=\langle 6,7\rangle, \quad \mathbf{v}=\langle 1,4\rangle, \quad \operatorname{proj}_{\mathbf{v}} \mathbf{u}=\langle 2,8\rangle $$

Short answer

The component of the vector \(u\) that is orthogonal to the vector \(v\) is \(\langle 4, -1 \rangle\).

Step-by-step solution

Understand the Concept of Orthogonal Component

When a vector \(u\) is projected onto another vector, \(v\), there are two components formed. One is the projection, \(\mathbf{w}_{\mathbf{1}}\), and the other is the orthogonal (perpendicular) component. The orthogonal component can be found by subtracting the projection from the original vector \(u\).

Substituting the Given Vectors

Next, we need to substitute the given vectors into the formula for the orthogonal component. Here, we have \(\mathbf{u}=\langle 6,7 \rangle\), \(\mathbf{v}=\langle 1,4\rangle\), and \(\operatorname{proj}_{\mathbf{v}} \mathbf{u}=\langle 2,8\rangle\). So, the orthogonal component is found by subtracting the projection of \(u\) onto \(v\) from \(u\) itself, that is \(u - \operatorname{proj}_{v}u.\)

Compute the Orthogonal Component

Subtract the coordinates of the projection vector from those of vector \(u\) to get the orthogonal component. The computation will be as follows, \(\langle 6,7 \rangle - \langle 2,8 \rangle = \langle 4,-1 \rangle\). So the orthogonal component of \(u\) relative to \(v\) is \(\langle 4, -1 \rangle\).