**Theory Problems and Solutions in Vector :**

Here we are going to see how to solve theory problems and solutions in vector.

**Question 1 :**

If D and E are the midpoints of the sides AB and AC of a triangle ABC, prove that BE vector + DC vector = (3/2) BC vector.

**Solution :**

OA vector = a vector,

OB vector = b vector,

OC vector = c vector

Since D is the midpoint of the side AB,

OD vector = (a vector + b vector)/2

Since E is the midpoint of the side AC,

OE vector = (a vector + c vector)/2

BE vector = OE vector - OB vector

= [(a vector + c vector)/2] - b vector

= (a vector + c vector - 2b vector)/2 ----(1)

DC vector = OC vector - OD vector

= c vector - [(a vector + b vector)/2]

= (2c vector - a vector - b vector)/2 -----(2)

(1) + (2)

= (a vector+c vector-2b vector)+(2c vector-a vector-b vector) / 2

= (-3b vector + 3c vector) / 2

= (3/2) (c vector - b vector)

= (3/2) (OC vector - OB vector)

= (3/2) BC vector

Hence it is proved.

**Question 2 :**

Prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third side whose length is half of the length of the third side.

**Solution :**

In triangle ADE,

AD vector + DE vector = AE vector

DE vector = AE vector - AD vector ---(1)

E is the midpoint of the side AC

So, AE = EC

AE = (1/2) AC vector

D is the midpoint of the side AB

So, AD = DC

AD = (1/2) AB vector

By applying the value of AE and AD, we get

DE vector = [(1/2) AC vector] - [(1/2) AB vector]

DE vector = (1/2) [AC vector - AB vector] ---(2)

In triangle ABC,

AB vector + BC vector = AC vector

BC vector = AC vector - AB vector

By applying the value of BC vector in (2), we get

DE vector = (1/2) BC vector

Hence the sides DE and BC are parallel.

**Question 3 :**

Prove that the line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram.

**Solution :**

Given : ABCD is a quadrilateral

E, F, G and H are midpoints of the sides AB, BD, DC and CA respectively.

In triangle ABC :

AB vector + BD vector = AC vector ---(1)

In triangle AEH :

AE vector + EH vector = AH vector

EH vector = AH vector - AE vector

Instead of AH vector, we may write (1/2) AC vector. Instead of AE vector, we may write (1/2) AB vector.

EH vector = (1/2) AC vector - (1/2) AB vector

= (1/2) [AC vector - AB vector]

From (1), we get

BD vector = AC vector - AB vector

EH vector = (1/2) BD vector ----(A)

Hence we conclude that the sides BD and EH are parallel.

Now, let us consider the triangle BCD and triangle FCG

BC vector + BD vector = CD vector ---(2)

In triangle FCG

FC vector + FG vector = CG vector

FG vector = CG vector - FC vector

= (1/2)CD vector - (1/2)BC vector

= (1/2) [CD vector - BC vector]

From (2),

CD vector - BC vector = BD vector

= (1/2) BD vector-----(B)

From (A) and (B), EFGH is a parallelogram.

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