Let u and v be unit vectors inclined at an acute angle such that |u v|=√32.

Question

Let $\hat{u}$ and $\hat{v}$ be unit vectors inclined at an acute angle such that $|\hat{u}	imes \hat{v}|=\frac{\sqrt{3}}{2}$. If $\overset{	o }{A}=\lambda \hat{u}+\hat{v}+(\hat{u}	imes \hat{v})$, then $\lambda $ is equal to:

Accepted Answer

Since $\hat{u}$ and $\hat{v}$ are unit vectors,

$|\hat{u}	imes \hat{v}|=|\hat{u}||\hat{v}|\sin 	heta =\sin 	heta $

Given

$|\hat{u}	imes \hat{v}|=\frac{\sqrt{3}}{2}$

So,

$\sin 	heta =\frac{\sqrt{3}}{2}$

The angle is acute, hence

$	heta ={60}^{\circ }\, \Rightarrow \, \hat{u}\cdot \hat{v}=\cos {60}^{\circ }=\frac{1}{2}$

Now,

$\vec{A}=\lambda \hat{u}+\hat{v}+(\hat{u}	imes \hat{v})$

We take dot product with $\hat{u}$:

$\vec{A}\cdot \hat{u}=\lambda (\hat{u}\cdot \hat{u})+(\hat{v}\cdot \hat{u})+(\hat{u}	imes \hat{v})\cdot \hat{u}$

Now,

$\hat{u}\cdot \hat{u}=1,\, \hat{v}\cdot \hat{u}=\frac{1}{2},\, (\hat{u}	imes \hat{v})\cdot \hat{u}=0$

Therefore,

$\vec{A}\cdot \hat{u}=\lambda +\frac{1}{2}$

So,

$\lambda =\vec{A}\cdot \hat{u}-\frac{1}{2}\,$

Now take dot product with $\hat{v}$:

$\vec{A}\cdot \hat{v}=\lambda (\hat{u}\cdot \hat{v})+(\hat{v}\cdot \hat{v})+(\hat{u}	imes \hat{v})\cdot \hat{v}$

Using

$\hat{u}\cdot \hat{v}=\frac{1}{2},\, \hat{v}\cdot \hat{v}=1,\, (\hat{u}	imes \hat{v})\cdot \hat{v}=0$

we get

$\vec{A}\cdot \hat{v}=\frac{\lambda }{2}+1$

Hence,

$\frac{\lambda }{2}=\vec{A}\cdot \hat{v}-1\, \Rightarrow \, \lambda =2(\vec{A}\cdot \hat{v})-2\,$

Now let

$x=\vec{A}\cdot \hat{u},\, y=\vec{A}\cdot \hat{v}$

Then from above,

$x=\lambda +\frac{1}{2}\, \, y=\frac{\lambda }{2}+1$

From the first,

$\lambda =x-\frac{1}{2}$

Also,

$x-2y=(\lambda +\frac{1}{2})-2(\frac{\lambda }{2}+1)=\lambda +\frac{1}{2}-\lambda -2=-\frac{3}{2}$

So we form $\lambda $ as a linear combination of $x$ and $y$.

Check Option A:

$\frac{4}{3}x-\frac{2}{3}y=\frac{4}{3}(\lambda +\frac{1}{2})-\frac{2}{3}(\frac{\lambda }{2}+1)$

$=\frac{4\lambda }{3}+\frac{2}{3}-\frac{\lambda }{3}-\frac{2}{3}=\lambda $

Hence,

$\lambda =\frac{4}{3}(\vec{A}\cdot \hat{u})-\frac{2}{3}(\vec{A}\cdot \hat{v})$

So the correct option is:

Solutions · 1