Question

The vector $\overrightarrow{a}+\overrightarrow{b},\overrightarrow{a}-\overrightarrow{b}$ and $\overrightarrow{a}-k\overrightarrow{b}$(k scalar) are collinear for

• A. $k=0$
• B. $k=1$
• C. $k=-1$
• D. $k=2$

Answer 1

The correct option is A $k=0$

Let $\overrightarrow{OA}=\overrightarrow{a}+\overrightarrow{b},\overrightarrow{OB}=\overrightarrow{a}-\overrightarrow{b},\overrightarrow{OC}=\overrightarrow{a}-k\overrightarrow{b}$

so,

$\overrightarrow{AB}=-2\overrightarrow{b}$

$\overrightarrow{BC}=\left(-k+1\right)\overrightarrow{b}$

$\overrightarrow{CA}=\left(k+1\right)\overrightarrow{b}$

Given, $\overrightarrow{OA},\overrightarrow{OB},\overrightarrow{OC}$ are collinear

So, angle between $\overrightarrow{AB}and\overrightarrow{BC}is{180}^0$

$\overrightarrow{AB}.\overrightarrow{BC}=\left|AB\right|\left|BC\right|\cos {180}^0$

$\left(-2\overrightarrow{b}\right).\left(-k+1\right)\overrightarrow{b}=\left|-2\overrightarrow{b}\right|\left|\left(1-k\right)\overrightarrow{b}\right|\left(-1\right)$

$\left(-2\right)\left(1-k\right){\left|\overrightarrow{b}\right|}^2=-2\left(1-k\right){\left|\overrightarrow{b}\right|}^2$

So, these are collinear for any value of K=0

So Correct Answer is A.