Question

Let $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}$ be three non zero vectors which are pairwise non-collinear. If $\overrightarrow{a}+3\overrightarrow{b}$ is collinear and $\overrightarrow{b}+2\overrightarrow{c}$ is collinear with $\overrightarrow{a}$, then $\overrightarrow{a}+3\overrightarrow{b}+6\overrightarrow{c}$

• A. $\overrightarrow{a}$
• B. $\overrightarrow{c}$
• C. $\overrightarrow{0}$
• D. $\overrightarrow{a}+\overrightarrow{c}$

Answer 1

The correct option is A $\overrightarrow{a}$

$a,b,c$ are pairwise non collinear means

$a\times b\ne 0$ $|a|,|b|,|c|\ne 0$

$b\times c\ne 0$

$c\times a\ne 0$

$(a+3b)\times a=0$----- $\left(1\right)$

$(b+2c)\times a=0$------ $\left(2\right)$

$b\times a+2c\times a=0$

$b\times a=-\frac{2c\times a}{b=-2c}=2a\times c$----- $\left(3\right)$

from $\left(1\right)$

$a\times +3b\times a=0$

$b\times a=0\Rightarrow b\perp a$

From $\left(3\right)$

$2a\times c=0$

$a\perp c$

$|\overrightarrow{a}+3\overrightarrow{b}+3\overrightarrow{c}{|}^2=|\overrightarrow{a}{|}^2+9|\overrightarrow{b}{|}^2+36|\overrightarrow{c}{|}^2+2(2.3b+3b.6c+6c.a)$

$|a{|}^2+9|b{|}^2+36|\overrightarrow{c}{|}^2+2(1B.b.c)$

$=a^2+(3b+6c{)}^2$

$=a^2+9(b+2c{)}^2$ Putting $b=-2c$

$=a^2$

$|\overrightarrow{a}+3\overrightarrow{b}+6\overrightarrow{c}|=\sqrt{a^2}=\overrightarrow{a}$