Question

If $\overrightarrow{DA}=\overrightarrow{a},A\overrightarrow{B}=\overrightarrow{b}$ and $\overrightarrow{CB}=k\overrightarrow{a}$, where $k>0$ and $X,Y$ are the mid points of $\overrightarrow{DB}$ and $\overrightarrow{AC}$ respectively such that $\left|\overrightarrow{a}\right|=17$ and $\overrightarrow{|XY}|=4,$ then $k$ is equal to?

• A. $\frac{8}{17}$
• B. $\frac{9}{17}$
• C. $\frac{25}{17}$
• D. $\frac{4}{17}$

Answer 1

The correct option is B

$\frac{9}{17}$

Explanation for the correct option:

Step 1. Find the value of $k$:

Let $D$ is the origin

$\begin{array}{rcl}\overrightarrow{DA}& =& \overrightarrow{A}–\overrightarrow{D}\\ & =& \overrightarrow{a}\end{array}$

$\Rightarrow$$\overrightarrow{A}=\overrightarrow{a}$

$\begin{array}{rcl}AB& =& \overrightarrow{B}–\overrightarrow{A}\\ & =& \overrightarrow{b}\end{array}$

$\Rightarrow$$\overrightarrow{B}=\overrightarrow{a}+\overrightarrow{b}$

$\begin{array}{rcl}CB& =& \overrightarrow{B}–\overrightarrow{C}\\ & =& k\overrightarrow{a}\end{array}$

$\Rightarrow$$\overrightarrow{C}=\overrightarrow{a}+\overrightarrow{b}–k\overrightarrow{a}$

Step 2. Find the mid points of $DB$ and $AC$

$\overrightarrow{X}=\frac{(\overrightarrow{a}+b\overrightarrow{})}{2},\overrightarrow{Y}=\frac{(2\overrightarrow{a}+\overrightarrow{b}-k\overrightarrow{a})}{2}$

$\begin{array}{rcl}\therefore \left|XY\right|& =& \left|\frac{(\overrightarrow{a}–k\overrightarrow{a})}{2}\right|\\ & =& \frac{(1-k)}{2}\left|\overrightarrow{a}\right|\end{array}$

Step 3. Put the value of $\left|XY\right|$ and $\left|a\right|$, we get

$4=\frac{(1-k)}{2}\times 17$

$\Rightarrow$$1-k=\frac{8}{17}$

$\mathbf{\therefore }\mathit{k}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{9}}{\mathbf{17}}$

Hence, Option ‘B’ is correct.