If a, b, c are non-coplanar vectors and lambd | vedclass

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(D) Given the equation: $[\lambda(a + b), \lambda^2 b, \lambda c] = [a, b + c, b]$Using the property of scalar triple product $[x, y, z] = x \cdot (y

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No value of $\lambda$

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(D) Given the equation: $[\lambda(a + b), \lambda^2 b, \lambda c] = [a, b + c, b]$Using the property of scalar triple product $[x, y, z] = x \cdot (y 	imes z)$,we have:$\lambda(a + b) \cdot (\lambda^2 b 	imes \lambda c) = a \cdot ((b + c) 	imes b)$Simplify the left side:$\lambda(a + b) \cdot (\lambda^3 (b 	imes c)) = \lambda^4 (a \cdot (b 	imes c) + b \cdot (b 	imes c))$Since $b 	imes b = 0$,this becomes $\lambda^4 [a, b, c]$.Simplify the right side:$a \cdot (b 	imes b + c 	imes b) = a \cdot (0 + c 	imes b) = a \cdot (c 	imes b) = [a, c, b]$Using the property $[a, c, b] = -[a, b, c]$,we get:$\lambda^4 [a, b, c] = -[a, b, c]$Rearranging gives:$[a, b, c](\lambda^4 + 1) = 0$Since $a, b, c$ are non-coplanar,$[a, b, c] 
eq 0$. Therefore,$\lambda^4 + 1 = 0$,which implies $\lambda^4 = -1$.Since $\lambda$ is a real number,$\lambda^4$ must be non-negative. Thus,there is no real value of $\lambda$ that satisfies the equation.

If $a, b, c$ are non-coplanar vectors and $\lambda$ is a real number,then $[\lambda(a + b), \lambda^2 b, \lambda c] = [a, b + c, b]$ for

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