Question

Given a cube $ABCDA'B'C'D'$ with lower base $ABCD$ , upper base $A'B'C'D'$ and the lateral edges $AA',BB',CC'{\text{ and }}DD'$ ; $M{\text{ and }}M'$ are the centers of the faces $ABCD$ and $A'B'C'D'$ respectively. $O$ is a point on the line $MM'$ such that $O\bar A + O\bar B + O\bar C + O\bar D = \bar M'O$ , then $O\bar M = \lambda \bar M'O$ if $\lambda$ is equal to

$$A.{\text{ }}\dfrac{1}{{16}} \\\ B.{\text{ }}\dfrac{1}{8} \\\ C.{\text{ }}\dfrac{1}{4} \\\ D.{\text{ }}\dfrac{1}{2} \\\$$

Answer 1

Hint:In order to solve the problem first draw the given cube and mark the points of the cube as represented in the problem statement. Further with the help of diagram make the substitution in the given statement in terms of center point and further simplify the equation on the basis of vector algebra concept to bring it in the form of required equation to find the value of unknown variable $\lambda$.

Complete step-by-step answer:

Let us solve the problem on the basis of the diagram shown above.
Given that:
A cube $ABCDA'B'C'D'$ with lower base $ABCD$ , upper base $A'B'C'D'$
$O\bar A + O\bar B + O\bar C + O\bar D = \bar M'O$ , and $O\bar M = \lambda \bar M'O$
As we know that for any three points X, Y and Z according to concept of vector algebra we can write the sum of vectors as:
$XY + YZ = XZ$
Similarly
As we consider point M for each of the base point and the point O we can write:

$$OM + MA = OA \\\ OM + MB = OB \\\ OM + MC = OC \\\ OM + MD = OD \\\$$

Let us substitute these equations in the given equation (1). So we get:

$$\because O\bar A + O\bar B + O\bar C + O\bar D = \bar M'O \\\ \Rightarrow \left( {OM + MA} \right) + \left( {OM + MB} \right) + \left( {OM + MC} \right) + \left( {OM + MD} \right) = \bar M'O \\\$$

Now let us separate the common term and further simplify.

$$\Rightarrow 4OM + MA + MB + MC + MD = \bar M'O \\\ \Rightarrow 4OM + \left( {MA + MC} \right) + \left( {MB + MD} \right) = \bar M'O \\\$$

As we know from the concept of vector that if two vectors $OX{\text{ and }}OY$ are equal in magnitude and opposite in direction then they can be written as:
$OX = - OY$
From the figure and the question it is clear that since point M is the centre of the face of cube and the points A and C are at opposite edges so we have:

$$MA = - MC \\\ \Rightarrow MA + MC = 0 \\\$$

Similarly for points B and D we have:

$$MB = - MD \\\ \Rightarrow MB + MD = 0 \\\$$

Let us substitute the above terms in the given equation. So we get:

$$\because 4OM + \left( {MA + MC} \right) + \left( {MB + MD} \right) = \bar M'O \\\ \Rightarrow 4OM + \left( 0 \right) + \left( 0 \right) = \bar M'O \\\ \Rightarrow 4OM = \bar M'O \\\$$

Let us further simplify the term to get the required form as mentioned in the problem:
$\Rightarrow OM = \dfrac{1}{4}\bar M'O$
So after comparing from the given problem equation we have:

$$\because OM = \dfrac{1}{4}\bar M'O{\text{ }}\& {\text{ }}O\bar M = \lambda \bar M'O \\\ \Rightarrow \lambda = \dfrac{1}{4} \\\$$

Hence the value of $\lambda = \dfrac{1}{4}$
So, the correct answer is option C.

Note-:This problem can also be solved by directly substituting the given equation with a variable term in the given problem statement and further simplifying the problem in the reverse order as we did. But that would become a troublesome method as regularly we will have to deal with unknown variables. In order to solve such problems the drawing of the figure and the basic concepts of vector algebra is a must for the students.