Question

ABCD is a trapezium in which AB and CD are parallel sides. If $l\left(AB\right)=3l\left(CD\right)$ and $\overline{DC}=2\hat{i}-5\hat{k}$. Then vector $\overline{AB}$ is

• A. $\frac{3}{\sqrt{29}}(2\hat{i}-5\hat{k})$
• B. $\frac{\sqrt{29}}{3}(5\hat{i}-2\hat{k})$
• C. $-6\hat{i}+15\hat{k}$
• D. A or B

Answer 1

The correct option is C $-6\hat{i}+15\hat{k}$

Given

$\overrightarrow{DC}=2\hat{i}-5\hat{k}$

$\overrightarrow{CD}=-2\hat{i}+5\hat{k}$

AB is parallel to CD

$\overrightarrow{AB}=\lambda \overrightarrow{CD}$

$\overrightarrow{AB}=\lambda (-2\hat{i}+5\hat{k})$

$\overrightarrow{AB}=-2\lambda \hat{i}+5\lambda \hat{k}$

$l\left(AB\right)=3l\left(CD\right)$

$\sqrt{(-2\lambda {)}^2+(5\lambda {)}^2}=3\sqrt{(-2{)}^2+(5{)}^2}$

$\sqrt{4{\lambda }^2+25{\lambda }^2}=3\sqrt{4+25}$

$\sqrt{29{\lambda }^2}=3\sqrt{29}$

$\sqrt{29}\lambda =3\sqrt{29}$

on comparing both sides

$\lambda =3$

putting value of $\lambda$ in AB

$\overrightarrow{AB}=-2\times 3\hat{i}+5\times 3\hat{k}$

$\overrightarrow{AB}=-6\hat{i}+15\hat{k}$

Hence option C is correct