Question

Two vectors $\stackrel{\rightharpoonup }{A}$and $\stackrel{\rightharpoonup }{B}$ are such that $\stackrel{\rightharpoonup }{A}+\stackrel{\rightharpoonup }{B}=\stackrel{\rightharpoonup }{A}-\stackrel{\rightharpoonup }{B}$. Then

• A. $\stackrel{\rightharpoonup }{A}.\stackrel{\rightharpoonup }{B}=0$
• B. $\stackrel{\rightharpoonup }{A}\times \stackrel{\rightharpoonup }{B}=0$
• C. $\stackrel{\rightharpoonup }{A}=0$
• D. $\stackrel{\rightharpoonup }{B}=0$

Answer 1

The correct option is D

$\stackrel{\rightharpoonup }{B}=0$

Explanation for the correct option

Given that $\stackrel{\rightharpoonup }{A}+\stackrel{\rightharpoonup }{B}=\stackrel{\rightharpoonup }{A}-\stackrel{\rightharpoonup }{B}$

$\Rightarrow$ $\stackrel{\rightharpoonup }{B}=-\stackrel{\rightharpoonup }{B}$

Let $\stackrel{\rightharpoonup }{B}=x\hat{i}+y\hat{j}+z\hat{k}$

$\Rightarrow$$-\stackrel{\rightharpoonup }{B}=-x\hat{i}-y\hat{j}-z\hat{k}$

$\Rightarrow$$x=-x,y=-y,z=-z$

This is only possible when the vector $\stackrel{\rightharpoonup }{B}$is a null vector.

$\Rightarrow$$\stackrel{\rightharpoonup }{B}=0$

Thus $\stackrel{\rightharpoonup }{B}=0$ when $\stackrel{\rightharpoonup }{A}+\stackrel{\rightharpoonup }{B}=\stackrel{\rightharpoonup }{A}-\stackrel{\rightharpoonup }{B}$.

Hence option(D) is the correct answer.