Let a,b,c be three vectors such that a b=2(a c).

Question

Let $\vec{a},\vec{b},\vec{c}$ be three vectors such that $\vec{a}	imes \vec{b}=2(\vec{a}	imes \vec{c})$. If $|\vec{a}|=1,|\vec{b}|=4,|\vec{c}|=2$, and the angle between $\vec{b}$ and $\vec{c}$ is ${60}^{\circ }$, then $|\vec{a}\cdot \vec{c}|$ is equal to

Accepted Answer

Given that

$\begin{gathered} \vec{a}	imes \vec{b}=2(\vec{a}	imes \vec{c}) \ \vec{a}	imes \vec{b}-2(\vec{a}	imes \vec{c})=0 \ \vec{a}	imes (\vec{b}-2\vec{c})=0 \end{gathered}$

This means $\vec{a}$ and $\vec{b}-2\vec{c}$ are parallel vectors

$\begin{gathered} \vec{b}-2\vec{c}=\lambda \vec{a} \ \Rightarrow \vec{b}=\lambda \vec{a}+2\vec{c}-(1) \end{gathered}$

It is given that $|\vec{a}|=1,|b|=4$ and $|\vec{c}|=2$, angle between $\vec{b}$ and $\vec{c},	heta ={60}^{\circ }$.

$|b{|}^{2}=|\lambda \vec{a}+2\vec{c}{|}^{2}$

$\Rightarrow$ $|b{|}^{2}={\lambda }^{2}|a{|}^{2}+4|c{|}^{2}+4\lambda (\vec{a}\cdot \vec{c})$

$\Rightarrow$ $16={\lambda }^{2}(1{)}^{2}+4(2{)}^{2}+4\lambda (\vec{a}\cdot \vec{c})$

$\Rightarrow$ ${\lambda }^{2}+4\lambda (\vec{a}\cdot \vec{c})=0$ .......(2)

Taking dot product with $\overset{	o }{c}$ in equation (1)

$\vec{b}\cdot \vec{c}=(\lambda \vec{a}+2\vec{c})\cdot \vec{c}$

$\Rightarrow$ $|\vec{b}||\vec{c}|\cos {60}^{\circ }=\lambda (\vec{a}\cdot \vec{c})+2|\vec{c}{|}^{2}$

$\Rightarrow$ $4	imes 2	imes 1/2=\lambda (\vec{a}\cdot \vec{c})+2(2{)}^{2}$

$\Rightarrow$ $4=\lambda (\vec{a}\cdot \vec{c})+8$

$\Rightarrow$ $\lambda (\vec{a}\cdot \vec{c})=-4$

From (2) and (3)

${\lambda }^{2}+4(-4)=0$

$\Rightarrow$ ${\lambda }^{2}-16=0$

$\Rightarrow$ $\lambda =\pm 4$

from (3)

$\vec{a}\cdot \vec{c}=-4/\lambda $

if $\lambda =4$, then $\vec{a}\cdot \vec{c}=-1$

if $\lambda =-4$, then $\vec{a}\cdot \vec{c}=1$

$|\vec{a}\cdot c|=1$

Solutions · 1