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 \((e) u \cdot(2 v)\). $$ \mathbf{u}=\langle 2,-3,4\rangle, \quad \mathbf{v}=\langle 0,6,5\rangle $$

Short answer

The calculations result in the following: (a) \( \mathbf{u} \cdot \mathbf{v} = 8 \) (b) \( \mathbf{u} \cdot \mathbf{u} = 29 \) (c) \( \|\mathbf{u}\|^2 = 29 \) (d) \( (\mathbf{u} \cdot \mathbf{v}) \mathbf{v} = \langle 0,48,40 \rangle \) (e) \( u \cdot(2 v) = 16\)

Step-by-step solution

Find the dot product \(\mathbf{u} \cdot \mathbf{v}\)

To find this, multiply together the corresponding elements in vectors \(\mathbf{u}\) and \(\mathbf{v}\), and then add up those products. This results in \( (2)(0) + (-3)(6) + (4)(5) = 8 \)

Find the dot product \(\mathbf{u} \cdot \mathbf{u}\)

This involves finding the dot product of \(\mathbf{u}\) vector with itself. Carry out the operation as before, but now with the elements of \(\mathbf{u}\) itself which results in \( (2)(2) + (-3)(-3) + (4)(4) = 29 \)

Find the squared norm \(\|\mathbf{u}\|^2\)

The squared norm of a vector \(\mathbf{u}\) is found by squaring each of its components, and then adding them up. This results in \( (2)^2 + (-3)^2 + (4)^2 = 29 \)

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

This involves multiplying the scalar \(\mathbf{u} \cdot \mathbf{v}\) found in step 1 with each component of vector \(\mathbf{v}\). Hence we get \(8 \times \langle 0,6,5 \rangle = \langle 0,48,40 \rangle \)

Find \( u \cdot(2 v)\)

To carry out this operation, multiply each component of vector \(\mathbf{v}\) by 2 then use the modified \(\mathbf{v}\) to find the dot product with \(\mathbf{u}\) as \( (2)(0) + (-3)(2 \times 6) + (4)(2 \times 5) = 16 \)