Consider two vectors overrightarrow{u} = 3hat | vedclass

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(B) Given $\overrightarrow{u} = 3\hat{i} - \hat{j}$ and $\overrightarrow{v} = 2\hat{i} + \hat{j} - \lambda\hat{k}$.The angle $	heta$ between $\overri

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(B) Given $\overrightarrow{u} = 3\hat{i} - \hat{j}$ and $\overrightarrow{v} = 2\hat{i} + \hat{j} - \lambda\hat{k}$.The angle $	heta$ between $\overrightarrow{u}$ and $\overrightarrow{v}$ is $\cos 	heta = \frac{\overrightarrow{u} \cdot \overrightarrow{v}}{|\overrightarrow{u}| |\overrightarrow{v}|}$.$\overrightarrow{u} \cdot \overrightarrow{v} = (3)(2) + (-1)(1) + (0)(-\lambda) = 6 - 1 = 5$.$|\overrightarrow{u}| = \sqrt{3^2 + (-1)^2} = \sqrt{10}$.$|\overrightarrow{v}| = \sqrt{2^2 + 1^2 + (-\lambda)^2} = \sqrt{5 + \lambda^2}$.Given $\cos 	heta = \frac{\sqrt{5}}{2\sqrt{7}}$,so $\frac{5}{\sqrt{10} \sqrt{5 + \lambda^2}} = \frac{\sqrt{5}}{2\sqrt{7}}$.Squaring both sides: $\frac{25}{10(5 + \lambda^2)} = \frac{5}{4 	imes 7} = \frac{5}{28}$.$\frac{5}{2(5 + \lambda^2)} = \frac{5}{28} \Rightarrow 5 + \lambda^2 = 14 \Rightarrow \lambda^2 = 9 \Rightarrow \lambda = 3$ (since $\lambda > 0$).Now,$\vec{v} = \vec{v}_1 + \vec{v}_2$,where $\vec{v}_1 \parallel \overrightarrow{u}$ and $\vec{v}_2 \perp \overrightarrow{u}$.Since $\vec{v}_1$ and $\vec{v}_2$ are orthogonal,$|\vec{v}|^2 = |\vec{v}_1|^2 + |\vec{v}_2|^2$.$|\vec{v}|^2 = 2^2 + 1^2 + (-3)^2 = 4 + 1 + 9 = 14$.Thus,$|\vec{v}_1|^2 + |\vec{v}_2|^2 = 14$.

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