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Let $\vec{a}$ and $\vec{b}$ be two non-collinear vectors of unit modulus. If $\vec{u}=\vec{a}-(\vec{a} \cdot…
Let $\vec{a}$ and $\vec{b}$ be two non-collinear vectors of unit modulus.
If $\vec{u}=\vec{a}-(\vec{a} \cdot \vec{b}) \vec{b}$ and $\vec{v}=\vec{a} \times \vec{b}$, then $|\vec{v}|=$
$|\vec{u}|+|\vec{u} \cdot \vec{v}|$ $\frac{|\vec{u}|}{2}$ $|\vec{u}|+\frac{|\vec{u} \cdot \vec{b}|}{2}$ $\frac{|\vec{u}|}{5}$ Solution
Given, $|\vec{a}|=|\vec{b}|=1$
$\begin{aligned}
& \text { Since, } \vec{u}=\vec{a}-(\vec{a} \cdot \vec{b}) \vec{b} \Rightarrow \vec{a}-\cos \theta \vec{b}=\vec{u} \\
& \Rightarrow|\vec{a}|^2+\cos ^2 \theta|\vec{b}|^2-2 \vec{a} \cdot \cos \theta \vec{b}=|\vec{u}|^2 \\
& \Rightarrow|\vec{u}|^2=\sin ^2 \theta \text { and } \vec{v}=\vec{a} \times \vec{b} \Rightarrow|\vec{v}|=\sin \theta=|\vec{u}|
\end{aligned}$
Now, $\vec{u} \cdot \vec{v}=(\vec{a}-\cos \theta \vec{b}) \cdot(\vec{a} \times \vec{b})$
$=a \cdot(\vec{a} \times \vec{b})-\cos \theta \vec{b} .(\vec{a} \times \vec{b})=0-0=0$
Now, $|\vec{v}|=\sin \theta=|\vec{u}|+0=|\vec{u}|+|\vec{u} \cdot \vec{v}|$.
Asked in: AP EAMCET 2024 (19 May Shift 2)
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