**Vector Word Problems and Answers :**

Here we are going to see how to solve vector word problems and answers.

**Question 1 :**

If G is the centroid of a triangle ABC, prove that GA vector + GB vector + GC vector = 0.

**Solution :**

From the given information, let us draw a rough diagram.

Centroid of the triangle divides in the ratio 2 : 1.

OA vector = a vector

OB vector = b vector

OC vector = c vector

OG vector = [m (OD vector) + n (OA vector)] / (m + n)

OG vector = [2(OD vector) + 1(OA vector)] / (2+1)

OG vector = [2(OD vector) + 1(OA vector)] / 3 ---(1)

D is the midpoint of the side BC.

OD vector = (OB vector + OC vector) /2

By applying the value of OD in (1), we get

OG vector = (OB vector + OC vector + OA vector)/3

3OG vector = OA vector + OB vector + OC vector

(OA-OG) vector + (OB-OG) Vector + (OC-OG) vector = 0

GA vector + GB vector + GC vector = 0

**Question 2 :**

Let A, B, and C be the vertices of a triangle. Let D,E, and F be the midpoints of the sides BC, CA, and AB respectively. Show that AD vector + BE vector + CF vector = 0.

**Solution :**

Let us draw a picture using the given information.

OA vector = a vector

OB vector = b vector

OC vector = c vector

OD = (OB + OC)/2 = (b + c) / 2

OE = (OA + OC)/2 = (a + c) / 2

OF = (OA + OB)/2 = (a + b) / 2

**AD vector :**

AD vector = (OD - OA) vector ----(1)

By applying OD in (1), we get

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

= (b + c - 2a)/2

**BE vector :**

BE vector = (OE - OB) vector ----(2)

By applying OE in (2), we get

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

= (a + c - 2b)/2

**CF vector :**

CF vector = (OF - OC) vector ----(3)

By applying OF in (3), we get

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

= (a + b - 2c)/2

AD vector + BE vector + CF vector = 0

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

= 0

Hence it is proved

**Question 3 :**

If ABCD is a quadrilateral and E and F are the midpoints of AC and BD respectively, then prove that AB vector + AD vector + CB vector + CD vector = 4EF vector.

**Solution :**

Since F is the midpoint of the side BD,

AF vector = (AB vector + AD vector) /2

2AF vector = AB vector + AD vector ----(1)

CF vector = (CB vector + CD vector) /2

2CF vector = CB vector + CD vector ----(2)

(1) + (2)

2AF vector + 2CF vector

= (AB vector + AD vector) + (CB vector + CD vector)

2(AF vector + CF vector)

= (AB vector + AD vector) + (CB vector + CD vector)

Instead of AF + CF, we may write 2EF (using midpoint formula vice versa)

AB vector + AD vector + CB vector + CD vector = 4 EF

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