If lambda 0,let theta be the angle between | vedclass

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Question

(A) Given that $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ are perpendicular,their dot product is zero: $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b})

Vedclass · Accepted Answer

$25$

Vedclass · Answer

(A) Given that $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ are perpendicular,their dot product is zero: $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 0$. This simplifies to $|\vec{a}|^2 - |\vec{b}|^2 = 0$,which means $|\vec{a}|^2 = |\vec{b}|^2$. Calculating the magnitudes: $|\vec{a}|^2 = 1^2 + \lambda^2 + (-3)^2 = 10 + \lambda^2$ and $|\vec{b}|^2 = 3^2 + (-1)^2 + 2^2 = 9 + 1 + 4 = 14$. Equating them: $10 + \lambda^2 = 14 \implies \lambda^2 = 4$. Since $\lambda > 0$,we have $\lambda = 2$. Now,$\vec{a} = \hat{i} + 2 \hat{j} - 3 \hat{k}$ and $\vec{b} = 3 \hat{i} - \hat{j} + 2 \hat{k}$. The dot product $\vec{a} \cdot \vec{b} = (1)(3) + (2)(-1) + (-3)(2) = 3 - 2 - 6 = -5$. The angle $	heta$ is given by $\cos 	heta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$. Since $|\vec{a}|^2 = 14$ and $|\vec{b}|^2 = 14$,we have $|\vec{a}| = \sqrt{14}$ and $|\vec{b}| = \sqrt{14}$. Thus,$\cos 	heta = \frac{-5}{\sqrt{14} \cdot \sqrt{14}} = \frac{-5}{14}$. Therefore,$14 \cos 	heta = -5$,and $(14 \cos 	heta)^2 = (-5)^2 = 25$.

If $\lambda > 0$,let $\theta$ be the angle between the vectors $\vec{a} = \hat{i} + \lambda \hat{j} - 3 \hat{k}$ and $\vec{b} = 3 \hat{i} - \hat{j} + 2 \hat{k}$. If the vectors $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ are mutually perpendicular,then the value of $(14 \cos \theta)^2$ is equal to

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