Question

$P$ is the point of intersection of the diagonals of the parallelogram $ABCD$. If $S$ is any point in space and $SA+SB+SC+SD=\lambda SP$, then $\lambda$ equals

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

Answer: (B)

Taking $S$ as origin, let the position vectors of $A,B$,
$C$ and $D$ be $a,b,c$ and $d$ respectively.
In $\Delta SAC$, $P$ is the mid-point of $AC$.
$\therefore SA+SC=2SP\dots (i)$

In $\Delta SBD,P$ is the mid-point of $BC$.
$\therefore$ $SB+SD=2SP......(ii)$
On adding Eqs. (i) and (ii), we get
$SA+SB+SC+SD=4SP$
But it is given, $SA+SB+SC+SD=\lambda SP$
On comparing, we get $\lambda =4$