If vec{a} and vec{b} are mutually perpendicul | vedclass

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(B) Since $\vec{a}$ and $\vec{b}$ are mutually perpendicular unit vectors,the set $\{\vec{a}, \vec{b}, \vec{a} 	imes \vec{b}\}$ forms an orthogonal b

Vedclass · Accepted Answer

$\vec{b} + (\vec{a} 	imes \vec{b})$

Vedclass · Answer

(B) Since $\vec{a}$ and $\vec{b}$ are mutually perpendicular unit vectors,the set $\{\vec{a}, \vec{b}, \vec{a} 	imes \vec{b}\}$ forms an orthogonal basis for the space.Let $\vec{r} = x_1 \vec{a} + x_2 \vec{b} + x_3 (\vec{a} 	imes \vec{b})$.Taking the dot product with $\vec{a}$: $\vec{r} \cdot \vec{a} = x_1 |\vec{a}|^2 = x_1(1) = 0 \implies x_1 = 0$.Taking the dot product with $\vec{b}$: $\vec{r} \cdot \vec{b} = x_2 |\vec{b}|^2 = x_2(1) = 1 \implies x_2 = 1$.Taking the dot product with $(\vec{a} 	imes \vec{b})$: $\vec{r} \cdot (\vec{a} 	imes \vec{b}) = x_3 |\vec{a} 	imes \vec{b}|^2 = x_3(1)^2 = 1 \implies x_3 = 1$.Substituting the values of $x_1, x_2, x_3$ into the expression for $\vec{r}$,we get $\vec{r} = 0\vec{a} + 1\vec{b} + 1(\vec{a} 	imes \vec{b}) = \vec{b} + (\vec{a} 	imes \vec{b})$.

If $\vec{a}$ and $\vec{b}$ are mutually perpendicular unit vectors and $\vec{r}$ is a vector such that $\vec{r} \cdot \vec{a} = 0$,$\vec{r} \cdot \vec{b} = 1$,and $[\vec{r} \, \vec{a} \, \vec{b}] = 1$,then $\vec{r} = \dots$

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