If the angle between the vectors bar{a}=2 lam | vedclass

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(B) Given vectors are $\bar{a}=2 \lambda^2 \hat{i}+4 \lambda \hat{j}+\hat{k}$ and $\bar{b}=7 \hat{i}-2 \hat{j}+\lambda \hat{k}$.Since the angle $	het

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$\left(0, \frac{1}{2}\right)$

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(B) Given vectors are $\bar{a}=2 \lambda^2 \hat{i}+4 \lambda \hat{j}+\hat{k}$ and $\bar{b}=7 \hat{i}-2 \hat{j}+\lambda \hat{k}$.Since the angle $	heta$ between $\bar{a}$ and $\bar{b}$ is obtuse,we have $\cos 	heta We know that $\cos 	heta = \frac{\bar{a} \cdot \bar{b}}{|\bar{a}| |\bar{b}|}$.Since $|\bar{a}| > 0$ and $|\bar{b}| > 0$,the condition $\cos 	heta Calculating the dot product: $\bar{a} \cdot \bar{b} = (2 \lambda^2)(7) + (4 \lambda)(-2) + (1)(\lambda) = 14 \lambda^2 - 8 \lambda + \lambda = 14 \lambda^2 - 7 \lambda$.Setting the dot product to be less than zero: $14 \lambda^2 - 7 \lambda $7 \lambda (2 \lambda - 1) This inequality holds when $\lambda$ lies between the roots $0$ and $\frac{1}{2}$.Therefore,$\lambda \in \left(0, \frac{1}{2}ight)$.

If the angle between the vectors $\bar{a}=2 \lambda^2 \hat{i}+4 \lambda \hat{j}+\hat{k}$ and $\bar{b}=7 \hat{i}-2 \hat{j}+\lambda \hat{k}$ is obtuse,then $\lambda \in$

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