Question

In Exercises 33-36, (a) find the projection of \(\mathbf{u}\) onto \(\mathbf{v}\), and (b) find the vector component of u orthogonal to v. $$ \mathbf{u}=\langle 2,3\rangle, \quad \mathbf{v}=\langle 5,1\rangle $$

Short answer

The projection of \( \mathbf{u} \) onto \( \mathbf{v} \) is \( \langle 2.5, 0.5 \rangle \) and the component of \( \mathbf{u} \) orthogonal to \( \mathbf{v} \) is \( \langle -0.5, 2.5 \rangle \)

Step-by-step solution

Determine the projection

The projection of vector \( \mathbf{u} \) onto vector \( \mathbf{v} \) is given by the formula \( proj_{\mathbf{v}}\mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} \mathbf{v} \). Our vectors are \( \mathbf{u} = \langle 2,3 \rangle \) and \( \mathbf{v} = \langle 5,1 \rangle \). First, perform the dot product of the vectors \( \mathbf{u} \cdot \mathbf{v} = 2*5 + 3*1 = 13 \), \( \mathbf{v} \cdot \mathbf{v} = 5*5 + 1*1 = 26 \), and hence \( \frac{\mathbf{u} \cdot \mathbf{v}}{\mathbf{v} \cdot \mathbf{v}} = \frac{13}{26} = 0.5 \). Multiply this result by \( \mathbf{v} \) to get \( proj_{\mathbf{v}}\mathbf{u} = 0.5*\langle 5,1 \rangle = \langle 2.5, 0.5 \rangle \)

Determine the orthogonal component

The component of vector \( \mathbf{u} \) orthogonal to vector \( \mathbf{v} \) is found by subtracting the projection from vector \( \mathbf{u} \). As per our results, \( \mathbf{u} - proj_{\mathbf{v}}\mathbf{u} = \langle 2,3 \rangle - \langle 2.5, 0.5 \rangle = \langle -0.5, 2.5 \rangle \)