Question

# ABCD is a parallelogram. The position vectors of A and C are respectively, $3\hat i + 3\hat j + 5\hat k$ and $\hat i - 5\hat j - 5\hat k$. If M is the mid-point of the diagonal DB, then the magnitude of the projection of OM on OC, where O is the origin is?

Hint: First, we need to find the midpoint of the diagonal DB, by considering the fact that both diagonals will have the same midpoint. OM and OC vectors will be with reference to the origin. Then we will obtain the projection of the vector by using the dot product of the vectors, further divided by the magnitude of OC vector.

$3\hat i + 3\hat j + 5\hat k$ and $\hat i - 5\hat j - 5\hat k$.
$O\vec M = \dfrac{{O\vec A + O\vec C}}{2} \\ = \dfrac{{(3\hat i + 3\hat j + 5\hat k) + \hat i - 5\hat j - 5\hat k}}{2} \\ = 2\hat i - \hat j \\$
So, the position vector of M is $2\hat i - \hat j$.
$\dfrac{{\left| {O\vec M.O\vec C} \right|}}{{\left| {O\vec C} \right|}} \\ = \dfrac{{\left| {2 + 5} \right|}}{{\left| {\sqrt {1 + 25 + 25} } \right|}} \\ = \dfrac{7}{{\sqrt {51} }} \\$
In the above expression we found the magnitude of OC as $\sqrt {51}$(=$\sqrt {1 + 25 + 25}$). Also, for the projection, the angle between the vectors will be zero.
The magnitude of the projection will be $\dfrac{7}{{\sqrt {51} }}$.