Question

(a) find the projection of \(\mathbf{u}\) onto \(\mathbf{v}\), and (b) find the vector component of u orthogonal to v. $$ \mathbf{u}=\langle 2,1,2\rangle, \quad \mathbf{v}=\langle 0,3,4\rangle $$

Short answer

The projection of \(\mathbf{u}\) onto \(\mathbf{v}\) is \(\langle 0,\frac{33}{25},\frac{44}{25}\rangle\) and the vector component of \(\mathbf{u}\) orthogonal to \(\mathbf{v}\) is \(\langle 2,\frac{-8}{25},\frac{-4}{25}\rangle\).

Step-by-step solution

Compute the Dot Product of vectors \(\mathbf{u}\) and \(\mathbf{v}\)

The dot product of two vectors \(\mathbf{u}\) and \(\mathbf{v}\) is computed by multiplying corresponding entries and then adding those products. Dot product,\(\mathbf{u} \cdot \mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3 = (2*0) + (1*3) + (2*4) = 11.

Compute the Square of Magnitude of \(\mathbf{v}\)

The square of magnitude is simply the dot product of the vector with itself. The square of magnitude of vector \(\mathbf{v}\) i.e. \(\mathbf{v}\cdot\mathbf{v} = 0^2 + 3^2 + 4^2 = 25.

Compute the Projection of \(\mathbf{u}\) onto \(\mathbf{v}\)

The projection of \(\mathbf{u}\) onto \(\mathbf{v}\) is given by the formula: \(\text{proj}_\mathbf{v}\mathbf{u} = \left(\frac{\mathbf{u}\cdot\mathbf{v}}{\mathbf{v}\cdot\mathbf{v}}\right)\mathbf{v}\). So, \(\text{proj}_\mathbf{v}\mathbf{u} = \left(\frac{11}{25}\right)\langle 0,3,4\rangle = \langle 0,\frac{33}{25},\frac{44}{25}\rangle\).

Compute the vector component of \(\mathbf{u}\) orthogonal to \(\mathbf{v}\)

The vector component of \(\mathbf{u}\) orthogonal to \(\mathbf{v}\) is obtained by subtracting the projection of \(\mathbf{u}\) onto \(\mathbf{v}\) from \(\mathbf{u}\). So \(\mathbf{u}_\bot = \mathbf{u} - \text{proj}_\mathbf{v}\mathbf{u} = \langle 2,1,2\rangle - \langle 0,\frac{33}{25},\frac{44}{25}\rangle = \langle 2,\frac{-8}{25},\frac{-4}{25}\rangle\).