If vec{a}, vec{b}, vec{c} are non-coplanar ve | vedclass

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(B) The scalar triple product is defined as $[\vec{x}, \vec{y}, \vec{z}] = (\vec{x} 	imes \vec{y}) \cdot \vec{z}$.Given the equation: $[\lambda(\vec{

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For no value of $\lambda$

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(B) The scalar triple product is defined as $[\vec{x}, \vec{y}, \vec{z}] = (\vec{x} 	imes \vec{y}) \cdot \vec{z}$.Given the equation: $[\lambda(\vec{a} + \vec{b}), \lambda^2\vec{b}, \lambda\vec{c}] = [\vec{a}, \vec{b} + \vec{c}, \vec{b}]$.Using the property $[k\vec{u}, m\vec{v}, n\vec{w}] = kmn[\vec{u}, \vec{v}, \vec{w}]$,the left side becomes:$\lambda \cdot \lambda^2 \cdot \lambda [\vec{a} + \vec{b}, \vec{b}, \vec{c}] = \lambda^4 [\vec{a} + \vec{b}, \vec{b}, \vec{c}]$.Since $[\vec{a} + \vec{b}, \vec{b}, \vec{c}] = [\vec{a}, \vec{b}, \vec{c}] + [\vec{b}, \vec{b}, \vec{c}] = [\vec{a}, \vec{b}, \vec{c}] + 0 = [\vec{a}, \vec{b}, \vec{c}]$,the left side is $\lambda^4 [\vec{a}, \vec{b}, \vec{c}]$.Now,the right side is $[\vec{a}, \vec{b} + \vec{c}, \vec{b}] = [\vec{a}, \vec{b}, \vec{b}] + [\vec{a}, \vec{c}, \vec{b}] = 0 + [\vec{a}, \vec{c}, \vec{b}] = -[\vec{a}, \vec{b}, \vec{c}]$.Equating both sides: $\lambda^4 [\vec{a}, \vec{b}, \vec{c}] = -[\vec{a}, \vec{b}, \vec{c}]$.Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar,$[\vec{a}, \vec{b}, \vec{c}] 
eq 0$.Thus,$\lambda^4 = -1$.Since $\lambda$ is a real number,$\lambda^4$ must be non-negative,so $\lambda^4 = -1$ has no real solution.

If $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors and $\lambda$ is a real number,then for what value of $\lambda$ does the equation $[\lambda(\vec{a} + \vec{b}), \lambda^2\vec{b}, \lambda\vec{c}] = [\vec{a}, \vec{b} + \vec{c}, \vec{b}]$ hold?

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