Cross product

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Find a unit vector perpendicular to the plane containing $\mathbf{a}=\mathbf{i}+2\mathbf{j}+3\mathbf{k}$ and $\mathbf{b}=4\mathbf{i}-\mathbf{k}$ .

\dfrac{1}{\sqrt{237}}(-2\mathbf{i}+13\mathbf{j}-8\mathbf{k})

\dfrac{1}{\sqrt{237}}(2\mathbf{i}-13\mathbf{j}+8\mathbf{k})

-2\mathbf{i}+13\mathbf{j}-8\mathbf{k}

\dfrac{1}{\sqrt{14}}(\mathbf{i}+2\mathbf{j}+3\mathbf{k})

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If $\mathbf{a}=3\mathbf{i}-\mathbf{j}+2\mathbf{k}$ and $\mathbf{b}=\mathbf{i}+3\mathbf{j}-2\mathbf{k}$ , find the magnitude and direction cosines of $\mathbf{a}\times\mathbf{b}$ .

|\mathbf{a}\times\mathbf{b}|=\sqrt{180}

, direction cosines

\left(\dfrac{-4}{\sqrt{180}},\dfrac{8}{\sqrt{180}},\dfrac{10}{\sqrt{180}}\right)

|\mathbf{a}\times\mathbf{b}|=\sqrt{180}

, direction cosines

\left(\dfrac{4}{\sqrt{180}},\dfrac{-8}{\sqrt{180}},\dfrac{-10}{\sqrt{180}}\right)

|\mathbf{a}\times\mathbf{b}|=-2

, direction cosines

(-4,8,10)

|\mathbf{a}\times\mathbf{b}|=\sqrt{180}

, direction cosines

\left(\dfrac{-4}{180},\dfrac{8}{180},\dfrac{10}{180}\right)

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What type of quantity is $\mathbf{a}\times\mathbf{b}$ , and how is its direction determined?

A.A vector perpendicular to both

\mathbf{a}

and

\mathbf{b}

, with direction given by the right-hand rule

B.A scalar equal to

|\mathbf{a}||\mathbf{b}|\cos\theta

C.A vector parallel to

\mathbf{a}

D.A vector lying in the same plane as

\mathbf{a}

and

\mathbf{b}

, bisecting the angle between them

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