Suppose overrightarrow{a}=lambda hat{i}-7 hat | vedclass

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(A) The angle $	heta$ between two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by $\cos 	heta = \frac{\overrightarrow{a} \cdot \ov

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$-7 < \lambda < 1$

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(A) The angle $	heta$ between two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is given by $\cos 	heta = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{a}| |\overrightarrow{b}|}$.Given that the angle $	heta > 90^{\circ}$,we know that $\cos 	heta Since the magnitudes $|\overrightarrow{a}|$ and $|\overrightarrow{b}|$ are always positive,the condition $\cos 	heta Calculating the dot product: $\overrightarrow{a} \cdot \overrightarrow{b} = (\lambda)(\lambda) + (-7)(1) + (3)(2\lambda) = \lambda^2 - 7 + 6\lambda$.Setting the dot product to be less than zero: $\lambda^2 + 6\lambda - 7 Factoring the quadratic expression: $(\lambda + 7)(\lambda - 1) The inequality holds when $\lambda$ lies between the roots of the quadratic equation,which are $-7$ and $1$.Therefore,the solution is $-7 < \lambda < 1$.

$1$. Burning candle	$i$. Line spectrum
$2$. Sodium vapour	$ii$. Continuous spectrum
$3$. Bunsen flame	$iii$. Band spectrum
$4$. Dark lines in solar spectrum	$iv$. Absorption spectrum

Suppose $\overrightarrow{a}=\lambda \hat{i}-7 \hat{j}+3 \hat{k}$ and $\overrightarrow{b}=\lambda \hat{i}+\hat{j}+2 \lambda \hat{k}$. If the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$ is greater than $90^{\circ}$,then $\lambda$ satisfies the inequality

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