If a and b are two non-zero perpendicular vec | vedclass

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(C) Given $a \cdot y = c$ and $a 	imes y = b$. Since $a$ and $b$ are perpendicular, $a \cdot b = 0$. Taking the cross product of $a$ with the second

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$\frac{1}{|a|^2}[c a - (a 	imes b)]$

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(C) Given $a \cdot y = c$ and $a 	imes y = b$. Since $a$ and $b$ are perpendicular, $a \cdot b = 0$. Taking the cross product of $a$ with the second equation: $a 	imes (a 	imes y) = a 	imes b$. Using the vector triple product identity $a 	imes (b 	imes c) = (a \cdot c)b - (a \cdot b)c$: $(a \cdot y)a - (a \cdot a)y = a 	imes b$. Substitute $a \cdot y = c$ and $a \cdot a = |a|^2$: $c a - |a|^2 y = a 	imes b$. Rearranging for $y$: $|a|^2 y = c a - (a 	imes b)$. $y = \frac{1}{|a|^2}[c a - (a 	imes b)]$.

If $a$ and $b$ are two non-zero perpendicular vectors, then a vector $y$ satisfying the equations $a \cdot y = c$ (where $c$ is a scalar) and $a \times y = b$ is

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