Question

The linear relation between the vectors $\overline{a}+3\overline{b}+4\overline{c}$, $\overline{a}-2\overline{b}+3\overline{c},$ $\overline{a}+5\overline{b}-2\overline{c},$ $6\overline{a}+14\overline{b}+4\overline{c}$ is

• A. $(\overline{a}+3\overline{b}+4\overline{c})+2(\overline{a}-2\overline{b}+3\overline{c})+$
  2 $(\overline{a}+5\overline{b}-2\overline{c})-1(6\overline{a}+14\overline{b}+4\overline{c})=\overline{0}$
• B. $1(\overline{a}+3\overline{b}+4\overline{c})+2(\overline{a}-2\overline{b}-3\overline{c})+$
  3 $(\overline{a}+5\overline{b}-2\overline{c})-2(7\overline{a}+14\overline{b}+4\overline{c})=\overline{0}$
• C. $1(\overline{a}+3\overline{b}+4\overline{c})+2(\overline{a}-2\overline{b}+3\overline{c})+$
  3 $(\overline{a}+5\overline{b}-2\overline{c})-1(6\overline{a}+14\overline{b}+4\overline{c})=\overline{0}$
• D. $(\overline{a}+3\overline{b}+4\overline{c})+(\overline{a}-2\overline{b}+3\overline{c})+$
  $(\overline{a}+5\overline{b}-2\overline{c})-(6\overline{a}+14\overline{b}+4\overline{c})=\overline{0}$

Answer 1

Answer: (C)

$1(\overline{a}+3\overline{b}+4\overline{c})+2(\overline{a}-2\overline{b}+3\overline{c})+$

3 $(\overline{a}+5\overline{b}-2\overline{c})-1(6\overline{a}+14\overline{b}+4\overline{c})=\overline{0}$
Since Vectors are in Linear relation we can write $6\overrightarrow{a}+14\overrightarrow{b}+4\overrightarrow{c}=v(\overrightarrow{a}+3\overrightarrow{b}+4\overrightarrow{c})+u(\overrightarrow{a}-\alpha \overrightarrow{b}+3\overrightarrow{c})+w(\overrightarrow{a}+5\overrightarrow{b}-\alpha \overrightarrow{c})$
as $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ are non linear
Solving above equation we get

$6=v+u+w$
$14=3v-2u+5w$
$4=4v+3u-2w$

Solve for $v,u,w$
$v=1,u=2,w=3$