Question

find the cross product \\( \mathbf{a} \times \mathbf{b} \\). \\( \mathbf{a} = t\mathbf{i} + \cos(t)\mathbf{j} + \sin(t)\mathbf{k} \\), \\( \mathbf{b} = \mathbf{i} - \sin(t)\mathbf{j} + \cos(t)\mathbf{k} \\). verify that it is orthogonal to both \\( \mathbf{a} \\) and \\( \mathbf{b} \\). \\( (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a} = 0 \\) (with checkmark), \\( (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} = 0 \\) (with checkmark).

Explanation:

Step1: Recall cross product formula

The cross product of two vectors \(\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}\) and \(\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}\) is given by:

$$ \mathbf{a} \times \mathbf{b} =
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= (a_2b_3 - a_3b_2)\mathbf{i} - (a_1b_3 - a_3b_1)\mathbf{j} + (a_1b_2 - a_2b_1)\mathbf{k} $$

For \(\mathbf{a} = t\mathbf{i} + \cos(t)\mathbf{j} + \sin(t)\mathbf{k}\) and \(\mathbf{b} = \mathbf{i} - \sin(t)\mathbf{j} + \cos(t)\mathbf{k}\), we have \(a_1 = t\), \(a_2 = \cos(t)\), \(a_3 = \sin(t)\), \(b_1 = 1\), \(b_2 = -\sin(t)\), \(b_3 = \cos(t)\).

Step2: Calculate the \(i\)-component

The \(i\)-component is \(a_2b_3 - a_3b_2\):

$$ \cos(t)\cdot\cos(t) - \sin(t)\cdot(-\sin(t)) = \cos^2(t) + \sin^2(t) = 1 $$

Step3: Calculate the \(j\)-component

The \(j\)-component is \(-(a_1b_3 - a_3b_1)\):

$$ -(t\cdot\cos(t) - \sin(t)\cdot1) = -t\cos(t) + \sin(t) $$

Step4: Calculate the \(k\)-component

The \(k\)-component is \(a_1b_2 - a_2b_1\):

$$ t\cdot(-\sin(t)) - \cos(t)\cdot1 = -t\sin(t) - \cos(t) $$

Step5: Combine the components

Putting it all together, the cross product \(\mathbf{a} \times \mathbf{b}\) is:

$$ \mathbf{a} \times \mathbf{b} = 1\mathbf{i} + (-\ t\cos(t) + \sin(t))\mathbf{j} + (-t\sin(t) - \cos(t))\mathbf{k} $$

Which can be written as:

$$ \mathbf{a} \times \mathbf{b} = \mathbf{i} + (\sin(t) - t\cos(t))\mathbf{j} + (-t\sin(t) - \cos(t))\mathbf{k} $$

Answer:

\(\mathbf{a} \times \mathbf{b} = \mathbf{i} + (\sin(t) - t\cos(t))\mathbf{j} + (-t\sin(t) - \cos(t))\mathbf{k}\)