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If $\vec{a}=2\vec{p}+3\vec{q}-\vec{r}$ , $\vec{b}=\vec{p}-2\vec{q}+2\vec{r }$ , $\vec{c}=-2\vec{p}+\vec{q}-2\: \vec{r }$ and $\vec{R}=3\vec{p}-\vec{q}+2\: \vec{r }$ , then $\vec{R}$ equals ( $\vec{p},\vec{q},\vec{r}$ are non-coplanar)
Option: 1
$2\vec{a}+5\vec{b}+3\vec{c}$

Option: 2
$2\vec{a}-5\vec{b}-3\vec{c}$

Option: 3
$2\vec{a}-5\vec{b}+3\vec{c}$

Option: 4
$2\vec{a}+5\vec{b}-3\vec{c}$

Answers (1)

best_answer

AS we learned

Linear combination of vectors -

$\vec{r}=l\vec{a}+m\vec{b}+n\vec{c}$

- wherein

Any vectors $\vec{r}$ in space can be written as linear combination of 3 non-coplanar vectors.

Let $\vec{R}=x\vec{a}+y\vec{b}+z\vec{c}$

$\Rightarrow 3\vec{p}-\vec{q}+2\vec{r}=(2x+y-2z)\vec{p}+(3x-2y+z)\vec{q}+(-x+2y-2z)\vec{r }$

$\because \vec{p},\vec{q},\, and\, \gamma$ are non-coplanar, so

$\alpha \vec{p}+\beta \vec{q}+\gamma \vec{r}=\vec{0}$ only when $\alpha =0,\, \, \beta =0,\, \, \gamma =0$

2x+y-2z=3; 3x-2y+z=-1; -x+2y-2z=2

on solving, we get

x=2, y=5, z=3

$\therefore \vec{R}=2\vec{a}+5\vec{b}+3\vec{c}$

Posted by

Devendra Khairwa

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