Consider two vectors u=3i-j and v=2i+j- k, 0.

Question

Consider two vectors
$\vec{u}=3\hat{i}-\hat{j}$ and $\vec{v}=2\hat{i}+\hat{j}-\lambda \hat{k},\lambda >0$. The angle between them is given by ${\cos }^{-1}(\frac{\sqrt{5}}{2\sqrt{7}})$.
Let $\vec{v}={\vec{v}}_{1}+\overset{	o }{{v}_{2}}$, where ${\vec{v}}_{1}$ is parallel to $\vec{u}$ and $\overset{	o }{{v}_{2}}$ is perpendicular to $\vec{u}$. Then the value ${|\overset{	o }{{v}_{1}}|}^{2}+{|\overset{	o }{{v}_{2}}|}^{2}$ is equal to

Accepted Answer

$\begin{gathered} \vec{u}\cdot \vec{v}=|u|\cdot |v|\cdot \cos 	heta \ \Rightarrow 6-1=\sqrt{10}\cdot \sqrt{5+{\lambda }^{2}}\cdot \frac{\sqrt{5}}{2\sqrt{7}} \ \Rightarrow 1=\sqrt{2}\cdot \sqrt{5+{\lambda }^{2}}\cdot \frac{1}{2\sqrt{7}} \ \Rightarrow 14=5+{\lambda }^{2} \ \Rightarrow {\lambda }^{2}=9 \ \Rightarrow \lambda =3 \ {v}_{1}=k\vec{u} \ \vec{v}={\vec{v}}_{1}+{\vec{v}}_{2} \ \Rightarrow \vec{v}=k\vec{u}+{\vec{v}}_{2} \ \vec{v}\cdot \vec{u}=k\cdot |\vec{u}{|}^{2} \end{gathered}$

$\begin{gathered} \Rightarrow 5=k\cdot 10\Rightarrow k=\frac{1}{2} \ 	herefore \, {\vec{v}}_{1}=\frac{\vec{u}}{2}=\frac{3\hat{i}}{2}-\frac{\hat{j}}{2} \ {|{\vec{v}}_{1}|}^{2}=\frac{10}{4} \ {\vec{v}}_{2}=\vec{v}-{\vec{v}}_{1} \ =\frac{1}{2}\hat{i}+\frac{3\hat{j}}{2}-3\hat{k} \ {|{\vec{v}}_{2}|}^{2}=\frac{10}{4}+9 \ {|{\vec{v}}_{1}|}^{2}+{|{\vec{v}}_{2}|}^{2}=\frac{10}{4}+\frac{10}{4}+9=14 \end{gathered}$

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