- $\big(\vec{\text{a}}+\vec{\text{b}}\big)$
- $\big(\vec{\text{a}}-\vec{\text{b}}\big)$
- $\frac{1}2\big(\vec{\text{b}}-\vec{\text{a}}\big)$
- $\frac{1}2\big(\vec{\text{a}}-\vec{\text{b}}\big)$
Solution:
Given a parallelogram OABC such that $\overrightarrow{\text{OC}}=\vec{\text{a}}$ and $\overrightarrow{\text{AB}}=\vec{\text{b}}$. Then,
$\overrightarrow{\text{OB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{OC}}$
$\Rightarrow\ \overrightarrow{\text{OB}}=\overrightarrow{\text{OC}}-\overrightarrow{\text{BC}}$
$\Rightarrow\ \overrightarrow{\text{OB}}=\overrightarrow{\text{OC}}-\overrightarrow{\text{OA}}$ $\Big[\because\overrightarrow{\text{BC}}=\overrightarrow{\text{OA}}\Big]$
$\Rightarrow\ \overrightarrow{\text{OB}}=\vec{\text{a}}-\overrightarrow{\text{OA}}\ \dots(1)$
Therefore,
$\overrightarrow{\text{OA}}+\overrightarrow{\text{AB}}=\overrightarrow{\text{OB}}$
$\Rightarrow\ \overrightarrow{\text{OA}}+\vec{\text{b}}=\vec{\text{a}}-\overrightarrow{\text{OA}}$ [Using (1)]
$\Rightarrow\ 2\overrightarrow{\text{OA}}=\vec{\text{a}}-\vec{\text{b}}$
$\Rightarrow\ \overrightarrow{\text{OA}}=\frac{1}2\big(\vec{\text{a}}-\vec{\text{b}}\big)$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Let f : A → B and g : B → C be the bijective functions. Then (gof)-1 is:
Statement $-1 :$ $gof $ is differentiable at $x=0$ and its derivative is continuous at that point
Statement $-2 :$ $gof $ is twice differentiable at $x=0 $