Question

Given a cube $ABCDA _{1} B _{1} C _{1} D _{1}$ with lower base $ABCD$, upper base $A_{1} B_{1} C_{1} D_{1}$ and the lateral edges $AA _{1}, BB _{1}, CC _{1}$ and $DD _{1} ; M$ and $M _{1}$ are the centres of the faces $A B C D$ and $A_{1} B_{1} C_{1} D_{1}$ respectively. $O$ is a point on the line $MM _{1}$ such that $\overrightarrow{ OA }+\overrightarrow{ OB }+\overrightarrow{ OC }+\overrightarrow{ OD }=\overrightarrow{ OM _{1}}, \overrightarrow{ OM }=\lambda \overrightarrow{ OM _{1}}$ if $\lambda$ is equal to

• A. 1/16
• B. 1/8
• C. 1/4
• D. 1/2

Answer 1

Answer: (C)

$\overrightarrow{ OM _{1}}=\overrightarrow{ OA }+\overrightarrow{ OB }+\overrightarrow{ OC }+\overrightarrow{ OD }$
$=\overrightarrow{ OM }+\overrightarrow{ MA }+\overrightarrow{ OM }+\overrightarrow{ MB }+\overrightarrow{ OM }+\overrightarrow{ MC }+\overrightarrow{ OM }+\overrightarrow{ MD }$
$=4 \overrightarrow{ OM }$
$\overrightarrow{ OM }=(1 / 4) \overrightarrow{ OM }_{1} $
$\lambda=\frac{1}{4}$