
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?
Answer
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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.
Complete step-by-step answer:
In the question, position vectors of A and C are given as:
$3\hat i + 3\hat j + 5\hat k$ and $\hat i - 5\hat j - 5\hat k$.
As we know that, in any parallelogram the midpoints of both the diagonals are the same. Thus, given M as the midpoint of DB will imply that M will also be the midpoint of AC. Midpoint is the mean value of two vectors.
So,
\[
O\vec M = \dfrac{{O\vec A + O\vec C}}{2} \\
= \dfrac{{(3\hat i + 3\hat j + 5\hat k) + \hat i - 5\hat j - 5\hat k}}{2} \\
= 2\hat i - \hat j \\
\]
So, the position vector of M is $2\hat i - \hat j$.
Now, we will obtain the magnitude of projection of vector OM on vector C, by dividing the magnitude of the dot product of vector Om and vector OC by magnitude of vector OC.
Thus, Magnitude of the projection is,
\[
\dfrac{{\left| {O\vec M.O\vec C} \right|}}{{\left| {O\vec C} \right|}} \\
= \dfrac{{\left| {2 + 5} \right|}}{{\left| {\sqrt {1 + 25 + 25} } \right|}} \\
= \dfrac{7}{{\sqrt {51} }} \\
\]
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 $\dfrac{7}{{\sqrt {51} }}$.
Note: Dot product of the vectors is also termed as the inner product or scalar product. The vector projection of some vector b onto another vector a is in the same direction or in the opposite direction if the scalar projection is negative as of a. In another way, it is also termed as the component of b in the direction of a.
Complete step-by-step answer:
In the question, position vectors of A and C are given as:
$3\hat i + 3\hat j + 5\hat k$ and $\hat i - 5\hat j - 5\hat k$.
As we know that, in any parallelogram the midpoints of both the diagonals are the same. Thus, given M as the midpoint of DB will imply that M will also be the midpoint of AC. Midpoint is the mean value of two vectors.
So,
\[
O\vec M = \dfrac{{O\vec A + O\vec C}}{2} \\
= \dfrac{{(3\hat i + 3\hat j + 5\hat k) + \hat i - 5\hat j - 5\hat k}}{2} \\
= 2\hat i - \hat j \\
\]
So, the position vector of M is $2\hat i - \hat j$.
Now, we will obtain the magnitude of projection of vector OM on vector C, by dividing the magnitude of the dot product of vector Om and vector OC by magnitude of vector OC.
Thus, Magnitude of the projection is,
\[
\dfrac{{\left| {O\vec M.O\vec C} \right|}}{{\left| {O\vec C} \right|}} \\
= \dfrac{{\left| {2 + 5} \right|}}{{\left| {\sqrt {1 + 25 + 25} } \right|}} \\
= \dfrac{7}{{\sqrt {51} }} \\
\]
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 $\dfrac{7}{{\sqrt {51} }}$.
Note: Dot product of the vectors is also termed as the inner product or scalar product. The vector projection of some vector b onto another vector a is in the same direction or in the opposite direction if the scalar projection is negative as of a. In another way, it is also termed as the component of b in the direction of a.
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