If the non-zero vectors a and b are perpendic | vedclass

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(A) Since $a, b$ and $a 	imes b$ are non-coplanar,we can express $r$ as a linear combination: $r = xa + yb + z(a 	imes b)$ for some scalars $x, y, z

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$r = xa + \frac{1}{a \cdot a}(a 	imes b)$

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(A) Since $a, b$ and $a 	imes b$ are non-coplanar,we can express $r$ as a linear combination: $r = xa + yb + z(a 	imes b)$ for some scalars $x, y, z$.Given the equation $r 	imes a = b$,we substitute the expression for $r$:$b = (xa + yb + z(a 	imes b)) 	imes a$Using the distributive property of the cross product:$b = x(a 	imes a) + y(b 	imes a) + z((a 	imes b) 	imes a)$Since $a 	imes a = 0$ and $b 	imes a = -(a 	imes b)$:$b = -y(a 	imes b) + z((a \cdot a)b - (a \cdot b)a)$Given that $a$ and $b$ are perpendicular,$a \cdot b = 0$:$b = -y(a 	imes b) + z(a \cdot a)b$Comparing the coefficients of $b$ and $(a 	imes b)$ on both sides:For the coefficient of $(a 	imes b)$,we have $-y = 0$,so $y = 0$.For the coefficient of $b$,we have $z(a \cdot a) = 1$,so $z = \frac{1}{a \cdot a}$.Substituting these values back into the expression for $r$:$r = xa + 0b + \frac{1}{a \cdot a}(a 	imes b) = xa + \frac{1}{a \cdot a}(a 	imes b)$.

If the non-zero vectors $a$ and $b$ are perpendicular to each other,then the solution of the equation $r \times a = b$ is given by

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