If non-zero vectors a and b are perpendicular | vedclass

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(A) Since $a$,$b$,and $a 	imes b$ form a basis for $3D$ space,we can write $r = xa + yb + z(a 	imes b)$ for some scalars $x, y, z$.Given $r 	imes a

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

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(A) Since $a$,$b$,and $a 	imes b$ form a basis for $3D$ space,we can write $r = xa + yb + z(a 	imes b)$ for some scalars $x, y, z$.Given $r 	imes a = b$,we substitute $r$:$b = (xa + yb + z(a 	imes b)) 	imes a$$b = x(a 	imes a) + y(b 	imes a) + z((a 	imes b) 	imes a)$Since $a 	imes a = 0$ and $b 	imes a = -(a 	imes b)$:$b = -y(a 	imes b) + z((a \cdot a)b - (a \cdot b)a)$Given $a \perp b$,$a \cdot b = 0$,so:$b = -y(a 	imes b) + z(a \cdot a)b$Comparing coefficients of $b$ and $(a 	imes b)$ on both sides:$-y = 0 \implies y = 0$$z(a \cdot a) = 1 \implies z = \frac{1}{a \cdot a}$Thus,$r = xa + \frac{1}{a \cdot a}(a 	imes b)$.

If non-zero vectors $a$ and $b$ are perpendicular to each other,then what is the solution for $r \times a = b$?

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