Question

Find \((\mathbf{a}) \mathbf{u} \cdot \mathbf{v},(\mathbf{b}) \mathbf{u} \cdot \mathbf{u},(\mathbf{c})\|\mathbf{u}\|^{2},(\mathbf{d})(\mathbf{u} \cdot \mathbf{v}) \mathbf{v}\) and \((e) u \cdot(2 v)\). $$ \begin{array}{l} \mathbf{u}=2 \mathbf{i}+\mathbf{j}-2 \mathbf{k} \\ \mathbf{v}=\mathbf{i}-3 \mathbf{j}+2 \mathbf{k} \end{array} $$

Short answer

The answers are: (a) \(\mathbf{u} \cdot \mathbf{v} = -5\), (b) \(\mathbf{u} \cdot \mathbf{u} = 9\), (c) \(\|\mathbf{u}\|^{2} = 9\), (d) (\(\mathbf{u} \cdot \mathbf{v}) \mathbf{v} = -5 \mathbf{i} + 15 \mathbf{j} - 10 \mathbf{k}\), (e) \(\mathbf{u} \cdot (2 \mathbf{v}) = -10\).

Step-by-step solution

Calculate \(\mathbf{u} \cdot \mathbf{v}\)

Dot product is calculated as the sum of the product of corresponding components of the two vectors. So \(\mathbf{u} \cdot \mathbf{v} = (2*1) + (1*-3) + (-2*2) = 2 - 3 - 4 = -5\).

Calculate \(\mathbf{u} \cdot \mathbf{u}\)

The dot product of a vector with itself corresponds to the square of its length. So \(\mathbf{u} \cdot \mathbf{u} = (2*2) + (1*1) + (-2*-2) = 4 + 1 + 4 = 9\).

Calculate \(\|\mathbf{u}\|^{2}\)

The square of the length of a vector is calculated as the square of each of its components summed together. This is equivalent to the result obtained in step 2. So \(\|\mathbf{u}\|^{2} = 2^{2} + 1^{2} + (-2)^{2} = 9\).

Calculate \((\mathbf{u} \cdot \mathbf{v}) \mathbf{v}\)

This is the dot product of \(\mathbf{u}\) and \(\mathbf{v}\) (obtained in step 1) times the vector \(\mathbf{v}\). So \((\mathbf{u} \cdot \mathbf{v}) \mathbf{v} = -5 * (1 \mathbf{i}-3 \mathbf{j}+2 \mathbf{k}) = -5 \mathbf{i} + 15 \mathbf{j} - 10 \mathbf{k}\).

Calculate \(\mathbf{u} \cdot (2 \mathbf{v})\)

For scalar multiplication, the dot product has the distributive property. So \(\mathbf{u} \cdot (2 \mathbf{v}) = 2*(\mathbf{u} \cdot \mathbf{v}) = 2*-5 = -10\).