If vec{a} and vec{b} are non-zero vectors whi | vedclass

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(A) Since $\vec{a}$ and $\vec{b}$ are linearly dependent,we can write $\vec{b} = \lambda \vec{a}$ for some scalar $\lambda 
eq 0$.Substituting this i

Vedclass · Accepted Answer

$\vec{b} = 3\vec{a}$

Vedclass · Answer

(A) Since $\vec{a}$ and $\vec{b}$ are linearly dependent,we can write $\vec{b} = \lambda \vec{a}$ for some scalar $\lambda 
eq 0$.Substituting this into the given equation:$\frac{|\vec{a} + \lambda \vec{a}|}{|\vec{a} - \lambda \vec{a}|} = 2$$\frac{|1 + \lambda| |\vec{a}|}{|1 - \lambda| |\vec{a}|} = 2$Since $\vec{a} 
eq 0$,we have $\frac{|1 + \lambda|}{|1 - \lambda|} = 2$.Squaring both sides:$\frac{(1 + \lambda)^2}{(1 - \lambda)^2} = 4$$(1 + \lambda)^2 = 4(1 - \lambda)^2$$1 + 2\lambda + \lambda^2 = 4(1 - 2\lambda + \lambda^2)$$1 + 2\lambda + \lambda^2 = 4 - 8\lambda + 4\lambda^2$$3\lambda^2 - 10\lambda + 3 = 0$$(3\lambda - 1)(\lambda - 3) = 0$Thus,$\lambda = 3$ or $\lambda = \frac{1}{3}$.Given the condition $|\vec{b}| > |\vec{a}|$,we have $|\lambda \vec{a}| > |\vec{a}|$,which implies $|\lambda| > 1$.Therefore,$\lambda = 3$ is the only valid solution.Hence,$\vec{b} = 3\vec{a}$.

If $\vec{a}$ and $\vec{b}$ are non-zero vectors which are linearly dependent such that $\frac{|\vec{a} + \vec{b}|}{|\vec{a} - \vec{b}|} = 2$ and $|\vec{b}| > |\vec{a}|$,then:

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