Question

If $\overline{DA}=\overline{a};\overline{AB}=\overline{b}$ and $\overline{CB}=k\overline{a}$ where $k>0$ and $x,y$ are the midpoints of $DB$ and $AC$ respectively such that $|\overline{a}|=17$ and $|\overline{XY}|=4$, then $k=$

• A. $\frac{8}{17}$
• B. $\frac{9}{17}$
• C. $\frac{11}{17}$
• D. $\frac{4}{17}$

Answer 1

Answer: (B)

$\frac{9}{17}$

$AY=CY$ ....... $[y$ is the midpoint of $AC]$
$DX=BX$ ....... $[x$ is the midpoint of $DB]$
Let D be the origin
Position vector(P.V.) of $A=\overrightarrow{a}$

P.V. of $B=\overrightarrow{a}+\overrightarrow{b}$ ....... $[\because \overline{AB}=\overline{b}]$
P.V. of $C=-(k-1)\overrightarrow{a}+\overrightarrow{b}$ ....... $[\because \overline{CB}=k\overline{a}]$
P.V. of $x=\frac{\overrightarrow{a}+\overrightarrow{b}}{2}$
P.V. of $y=\frac{-k\overrightarrow{a}+2\overrightarrow{a}}{2}+\frac{\overrightarrow{b}}{2}$
$\overrightarrow{XY}=\frac{\overrightarrow{a}-k\overrightarrow{a}}{2}$
$\left|\overrightarrow{XY}\right|=\frac{\left|a\right|}{2}\sqrt{(1-k{)}^2}$
$⟹\frac{8}{17}=1-k$
$⟹k=\frac{9}{17}$