Question

Find \(u \times v\) and show that it is orthogonal to both \(\mathbf{u}\) and \(\mathbf{v}\). $$ \begin{array}{l} \mathbf{u}=\langle 12,-3,0\rangle \\ \mathbf{v}=\langle-2,5,0\rangle \end{array} $$

Short answer

The cross product of vectors \(\mathbf{u}\) and \(\mathbf{v}\) is \(\langle 0,0,54 \rangle\), and it is orthogonal to both \(\mathbf{u}\) and \(\mathbf{v}\) as proven by the zero dot products.

Step-by-step solution

Calculating the Cross Product

To find the cross product of the two vectors \(\mathbf{u}=\langle 12,-3,0\rangle\) and \(\mathbf{v}=\langle-2,5,0\rangle\), use the cross product formula: \[ \mathbf{u} \times \mathbf{v} = \langle (u_2v_3 - u_3v_2),(u_3v_1 - u_1v_3),(u_1v_2 - u_2v_1) \rangle \]. Given the third components of \(\mathbf{u}\) and \(\mathbf{v}\) are both zero, the resultant vector can be simplified to \[ \langle 0,0,(12*5 - -3*-2) \rangle \]

Simplify the Result

Perform the multiplication in the cross product result from Step 1, which is \(\langle 0,0,60-6 \rangle\). This simplifies to \(\langle 0,0,54 \rangle \).

Proving Orthogonality

To show that the cross product is orthogonal (perpendicular) to both the original vectors, let's use the criteria for orthogonality. Two vectors are orthogonal if and only if their dot product is zero. Compute the dot product \(\langle 0,0,54 \rangle \cdot \langle 12,-3,0 \rangle\) and \(\langle 0,0,54 \rangle \cdot \langle -2,5,0 \rangle\). Both calculations yield 0, proving that the cross product is orthogonal to both original vectors.