Question

Calculating orthogonal projections For the given vectors \(\mathbf{u}\) and \(\mathbf{v},\) calculate proj\(_{\mathbf{v}} \mathbf{u}\) and \(\operatorname{scal}_{\mathbf{v}} \mathbf{u}\). $$\mathbf{u}=\langle 5,0,15\rangle \text { and } \mathbf{v}=\langle 0,4,-2\rangle$$

Short answer

Provide a step-by-step solution and calculate the orthogonal projection of vector \(u = \langle5, 0, 15\rangle\) onto vector \(v = \langle0, 4, -2\rangle\), and find the scalar component of vector \(u\) along vector \(v\). The solution involves the following steps: 1. Find the dot product of vectors \(u\) and \(v\): \((\mathbf{u} \cdot \mathbf{v}) = -30\) 2. Calculate the magnitude of vector \(v\): \(\|\mathbf{v}\| = \sqrt{20}\) 3. Compute the orthogonal projection of \(u\) onto \(v\): \(\text{proj}_{\mathbf{v}} \mathbf{u} = \langle 0, -6, 3\rangle\) 4. Calculate the scalar component of \(u\) along \(v\): \(\operatorname{scal}_{\mathbf{v}} \mathbf{u} = -\frac{6\sqrt{5}}{2}\)

Step-by-step solution

Compute the dot product of u and v

To find the dot product of \(\mathbf{u}\) and \(\mathbf{v}\), multiply the corresponding components and add them together: $$(\mathbf{u} \cdot \mathbf{v}) = 5 \cdot 0 + 0 \cdot 4 + 15 \cdot (-2) = -30$$

Compute the magnitude of v

Now find the magnitude of \(\mathbf{v}\) using the formula \(\|\mathbf{v}\| = \sqrt{x^2 + y^2 + z^2}\), where \(x, y, z\) are the components of \(\mathbf{v}\): $$\|\mathbf{v}\| = \sqrt{0^2 + 4^2 + (-2)^2} = \sqrt{20}$$

Calculate proj\(_{\mathbf{v}} \mathbf{u}\)

Now, calculate the orthogonal projection of \(\mathbf{u}\) onto \(\mathbf{v}\) using the formula from the analysis: $$\text{proj}_{\mathbf{v}} \mathbf{u} =\frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2}\mathbf{v} = \frac{-30}{20}\langle 0, 4, -2\rangle = \langle 0, -6, 3\rangle$$ So, \(\text{proj}_{\mathbf{v}} \mathbf{u} = \langle 0, -6, 3\rangle\).

Calculate \(\operatorname{scal}_{\mathbf{v}} \mathbf{u}\)

Finally, calculate the scalar component of \(\mathbf{u}\) along \(\mathbf{v}\) using the formula from the analysis: $$\operatorname{scal}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|} = \frac{-30}{\sqrt{20}} = -\frac{6\sqrt{5}}{2}$$ So, \(\operatorname{scal}_{\mathbf{v}} \mathbf{u} = -\frac{6\sqrt{5}}{2}\).