Question

PQRS is a quadrilateral and \(\overrightarrow{PQ}=\vec a\), \(\overrightarrow{QR}=\vec b\), \(\overrightarrow{SP}=\vec a-\vec b\). M is the midpoint of \(QR\) and X is a point on \(SM\) such that \[ \overrightarrow{SX}=\frac{4}{5}\overrightarrow{SM} \] If \[ \overrightarrow{SM}=m(4\vec a-\vec b) \] and \[ \overrightarrow{SX}=n(4\vec a-\vec b), \] then \(m+n=\)

• A. \(\dfrac{9}{10}\)
• B. \(\dfrac{10}{9}\)
• C. \(\dfrac{11}{9}\)
• D. \(\dfrac{4}{3}\)

Hint:

In vector geometry, choosing one vertex as the origin often simplifies midpoint and position-vector calculations significantly.

Answer 1

The Correct Option is A

Step 1: Express the position of \(R\) from \(P\).
Given, \[ \overrightarrow{PQ}=\vec a \] and \[ \overrightarrow{QR}=\vec b \] Hence, \[ \overrightarrow{PR} = \overrightarrow{PQ} + \overrightarrow{QR} = \vec a+\vec b \]

Step 2: Find \(\overrightarrow{PS}\).
Given, \[ \overrightarrow{SP} = \vec a-\vec b \] Therefore, \[ \overrightarrow{PS} = \vec b-\vec a \]

Step 3: Determine the midpoint \(M\) of \(QR\).
Taking \(P\) as origin, \[ Q=\vec a, \qquad R=\vec a+\vec b, \qquad S=\vec b-\vec a \] Since \(M\) is the midpoint of \(QR\), \[ M= \frac{Q+R}{2} = \frac{\vec a+(\vec a+\vec b)}{2} = \vec a+\frac{\vec b}{2} \] Thus, \[ \overrightarrow{SM} = \left(\vec a+\frac{\vec b}{2}\right) -(\vec b-\vec a) \] \[ = 2\vec a-\frac{\vec b}{2} \] \[ = \frac{1}{2}(4\vec a-\vec b) \] Hence, \[ m=\frac12 \]

Step 4: Find \(\overrightarrow{SX}\).
Given, \[ \overrightarrow{SX} = \frac45\overrightarrow{SM} \] Therefore, \[ \overrightarrow{SX} = \frac45\cdot \frac12(4\vec a-\vec b) \] \[ = \frac25(4\vec a-\vec b) \] Hence, \[ n=\frac25 \]

Step 5: Calculate \(m+n\).
\[ m+n = \frac12+\frac25 \] \[ = \frac{5+4}{10} \] \[ = \frac9{10} \]

Step 6: Final conclusion.
Therefore, \[ \boxed{\frac9{10}} \]