If the vectors vec{a} = lambda hat{i} + 2hat{ | vedclass

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(C) Given that the magnitudes of the vectors are equal,$|\vec{a}| = |\vec{b}|$. Calculating the magnitude of $\vec{a}$: $|\vec{a}| = \sqrt{\lambda^2 +

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$0$ or $1$

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(C) Given that the magnitudes of the vectors are equal,$|\vec{a}| = |\vec{b}|$. Calculating the magnitude of $\vec{a}$: $|\vec{a}| = \sqrt{\lambda^2 + 2^2 + (-3)^2} = \sqrt{\lambda^2 + 4 + 9} = \sqrt{\lambda^2 + 13}$. Calculating the magnitude of $\vec{b}$: $|\vec{b}| = \sqrt{(\sqrt{\lambda})^2 + (\sqrt{13})^2} = \sqrt{\lambda + 13}$. Equating the magnitudes: $\sqrt{\lambda^2 + 13} = \sqrt{\lambda + 13}$. Squaring both sides: $\lambda^2 + 13 = \lambda + 13$. Subtracting $13$ from both sides: $\lambda^2 - \lambda = 0$. Factoring the equation: $\lambda(\lambda - 1) = 0$. Thus,$\lambda = 0$ or $\lambda = 1$.

If the vectors $\vec{a} = \lambda \hat{i} + 2\hat{j} - 3\hat{k}$ and $\vec{b} = \sqrt{\lambda} \hat{i} + \sqrt{13} \hat{j}$ have the same magnitude,then the value of $\lambda$ is:

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