Samacheer Kalvi Class 11 Maths Solutions Chapter 8 Vector Algebra I Exercise 8.1

 

Question 1. Represent graphically the displacement of
(i) 45cm, 30° north of east
(ii) 80 km, 60° south of west
Answer:
(i) 45cm, 30° north of east

(ii) 80 km, 60° south of west

In simple words: To show displacement, we draw an arrow starting from the origin. The length of the arrow shows the distance, and its direction (angle from east, north, south, west) shows the angle of movement.

🎯 Exam Tip: Always start the displacement vector from the origin (the center of the compass) and measure the angle correctly from the specified cardinal direction (e.g., "north of east" means starting from east and rotating towards north).

 

Question 2. Prove that the relation R defined on the set V of all vectors by '\( \vec{a} \) R \( \vec{b} \) if \( \vec{a} = \vec{b} \)' is an equivalence relation on V.
Answer:
Given that \( V \) is the set of all vectors. The relation \( R \) is defined by \( \vec{a} R \vec{b} \) if \( \vec{a} = \vec{b} \).

**Reflexive:**
For any vector \( \vec{a} \in V \), we know that \( \vec{a} = \vec{a} \).
\( \implies \)
So, \( \vec{a} R \vec{a} \).
Therefore, \( R \) is reflexive. This means every vector is related to itself, which is a key property of an equivalence relation.

**Symmetric:**
Let \( \vec{a} R \vec{b} \).
Then, by definition, \( \vec{a} = \vec{b} \).
\( \implies \)
This also means \( \vec{b} = \vec{a} \).
\( \implies \)
So, \( \vec{b} R \vec{a} \).
Therefore, \( R \) is symmetric. If \( \vec{a} \) is related to \( \vec{b} \), then \( \vec{b} \) is also related to \( \vec{a} \).

**Transitive:**
Let \( \vec{a} R \vec{b} \) and \( \vec{b} R \vec{c} \).
Then, by definition:
\( \vec{a} = \vec{b} \)
and
\( \vec{b} = \vec{c} \)
\( \implies \)
From these two equalities, we can say that \( \vec{a} = \vec{c} \).
\( \implies \)
So, \( \vec{a} R \vec{c} \).
Therefore, \( R \) is transitive. If \( \vec{a} \) is related to \( \vec{b} \) and \( \vec{b} \) is related to \( \vec{c} \), then \( \vec{a} \) is related to \( \vec{c} \).

Since the relation \( R \) is reflexive, symmetric, and transitive, it is an equivalence relation.
In simple words: An equivalence relation has three rules: a thing is related to itself (reflexive), if A is related to B then B is related to A (symmetric), and if A is related to B and B to C, then A is related to C (transitive). Because "being equal" follows all these rules for vectors, the given relation is an equivalence relation.

🎯 Exam Tip: When proving an equivalence relation, make sure to clearly state and prove each of the three properties: reflexive, symmetric, and transitive. Use the given definition of the relation for each step.

 

Question 3. Let \( \vec{a} \) and \( \vec{b} \) be the position vectors of the points A and B. Prove that the position vectors of the points which trisect the line segment AB are \( \frac{\overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}}{3} \) and \( \frac{2 \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}}{3} \).
Answer:
Let \( O \) be the origin. The position vectors of points \( A \) and \( B \) are \( \overrightarrow{OA} = \vec{a} \) and \( \overrightarrow{OB} = \vec{b} \) respectively.
Let \( C \) and \( D \) be the points that trisect the line segment \( AB \). This means \( AC = CD = DB \).

**For point C:**
\( C \) divides \( AB \) internally in the ratio \( 1:2 \).
So, the position vector of \( C \) is \( \overrightarrow{OC} = \frac{1 \cdot \overrightarrow{OB} + 2 \cdot \overrightarrow{OA}}{1+2} \)
\( \overrightarrow{OC} = \frac{1 \cdot \vec{b} + 2 \cdot \vec{a}}{3} \)
\( \overrightarrow{OC} = \frac{\vec{b} + 2\vec{a}}{3} \)

**For point D:**
\( D \) divides \( AB \) internally in the ratio \( 2:1 \).
So, the position vector of \( D \) is \( \overrightarrow{OD} = \frac{2 \cdot \overrightarrow{OB} + 1 \cdot \overrightarrow{OA}}{2+1} \)
\( \overrightarrow{OD} = \frac{2 \cdot \vec{b} + 1 \cdot \vec{a}}{3} \)
\( \overrightarrow{OD} = \frac{2\vec{b} + \vec{a}}{3} \)
Thus, the position vectors of the points that trisect the line segment \( AB \) are \( \frac{\vec{b} + 2\vec{a}}{3} \) and \( \frac{2\vec{b} + \vec{a}}{3} \).
In simple words: When a line segment is divided into three equal parts (trisected), we use a special formula called the section formula. For the first point, it divides the line in a 1:2 ratio. For the second point, it divides the line in a 2:1 ratio. Using these ratios with the starting and ending vectors gives us the position vectors for the trisection points.

🎯 Exam Tip: Remember the section formula for internal division: if a point divides a line segment with position vectors \( \vec{a} \) and \( \vec{b} \) in the ratio \( m:n \), its position vector is \( \frac{n\vec{a} + m\vec{b}}{m+n} \).

 

Question 4. If D and E are the midpoints of the sides AB and AC of a triangle ABC, prove that \( \overrightarrow{\mathbf{B E}} + \overrightarrow{\mathbf{D C}} = \frac{3}{2} \overrightarrow{\mathrm{BC}} \).
Answer:
Let \( O \) be the origin. Let \( \vec{a}, \vec{b}, \vec{c} \) be the position vectors of the points \( A, B, C \) respectively.
So, \( \overrightarrow{OA} = \vec{a} \), \( \overrightarrow{OB} = \vec{b} \), \( \overrightarrow{OC} = \vec{c} \).

**Position vector of D (midpoint of AB):**
\( \overrightarrow{OD} = \frac{\overrightarrow{OA} + \overrightarrow{OB}}{2} \)
\( \overrightarrow{OD} = \frac{\vec{a} + \vec{b}}{2} \)

**Position vector of E (midpoint of AC):**
\( \overrightarrow{OE} = \frac{\overrightarrow{OA} + \overrightarrow{OC}}{2} \)
\( \overrightarrow{OE} = \frac{\vec{a} + \vec{c}}{2} \)

**Now, let's find \( \overrightarrow{BE} \):**
\( \overrightarrow{BE} = \overrightarrow{OE} - \overrightarrow{OB} \)
\( \overrightarrow{BE} = \frac{\vec{a} + \vec{c}}{2} - \vec{b} \)
\( \overrightarrow{BE} = \frac{\vec{a} + \vec{c} - 2\vec{b}}{2} \)

**Next, let's find \( \overrightarrow{DC} \):**
\( \overrightarrow{DC} = \overrightarrow{OC} - \overrightarrow{OD} \)
\( \overrightarrow{DC} = \vec{c} - \frac{\vec{a} + \vec{b}}{2} \)
\( \overrightarrow{DC} = \frac{2\vec{c} - \vec{a} - \vec{b}}{2} \)

**Now, we add \( \overrightarrow{BE} \) and \( \overrightarrow{DC} \):**
\( \overrightarrow{BE} + \overrightarrow{DC} = \frac{\vec{a} + \vec{c} - 2\vec{b}}{2} + \frac{2\vec{c} - \vec{a} - \vec{b}}{2} \)
\( \overrightarrow{BE} + \overrightarrow{DC} = \frac{\vec{a} + \vec{c} - 2\vec{b} + 2\vec{c} - \vec{a} - \vec{b}}{2} \)
\( \overrightarrow{BE} + \overrightarrow{DC} = \frac{(\vec{a} - \vec{a}) + (\vec{c} + 2\vec{c}) + (-2\vec{b} - \vec{b})}{2} \)
\( \overrightarrow{BE} + \overrightarrow{DC} = \frac{3\vec{c} - 3\vec{b}}{2} \)
\( \overrightarrow{BE} + \overrightarrow{DC} = \frac{3}{2}(\vec{c} - \vec{b}) \)

We know that \( \overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \vec{c} - \vec{b} \).
Therefore, \( \overrightarrow{BE} + \overrightarrow{DC} = \frac{3}{2} \overrightarrow{BC} \). This proof shows a specific vector relationship between medians and a side of a triangle.
In simple words: We used position vectors to represent the points in the triangle. First, we found the vectors for \( \overrightarrow{BE} \) and \( \overrightarrow{DC} \). Then, we added them together. After simplifying the sum, we found it was equal to \( \frac{3}{2} \) times the vector \( \overrightarrow{BC} \).

🎯 Exam Tip: When dealing with midpoints, always use the midpoint formula for position vectors. Express all required vectors (like \( \overrightarrow{BE} \) or \( \overrightarrow{DC} \)) in terms of the position vectors of the vertices relative to an origin.

 

Question 5. Prove that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.
Answer:
Let \( ABC \) be a triangle. Let \( D, E, F \) be the midpoints of the sides \( AB, BC, \) and \( AC \) respectively.
Let \( O \) be the origin, and let \( \vec{a}, \vec{b}, \vec{c} \) be the position vectors of the points \( A, B, C \) respectively.

**Position vector of D (midpoint of AB):**
\( \overrightarrow{OD} = \frac{\overrightarrow{OA} + \overrightarrow{OB}}{2} = \frac{\vec{a} + \vec{b}}{2} \)

**Position vector of E (midpoint of AC):**
\( \overrightarrow{OE} = \frac{\overrightarrow{OA} + \overrightarrow{OC}}{2} = \frac{\vec{a} + \vec{c}}{2} \)

**Position vector of F (midpoint of BC):**
\( \overrightarrow{OF} = \frac{\overrightarrow{OB} + \overrightarrow{OC}}{2} = \frac{\vec{b} + \vec{c}}{2} \)

**Let's consider the line segment \( DE \):**
\( \overrightarrow{DE} = \overrightarrow{OE} - \overrightarrow{OD} \)
\( \overrightarrow{DE} = \frac{\vec{a} + \vec{c}}{2} - \frac{\vec{a} + \vec{b}}{2} \)
\( \overrightarrow{DE} = \frac{\vec{a} + \vec{c} - \vec{a} - \vec{b}}{2} \)
\( \overrightarrow{DE} = \frac{\vec{c} - \vec{b}}{2} \)

We also know that \( \overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \vec{c} - \vec{b} \).
Therefore, \( \overrightarrow{DE} = \frac{1}{2} \overrightarrow{BC} \).
This equation shows two things:
1. Since \( \overrightarrow{DE} \) is a scalar multiple of \( \overrightarrow{BC} \), \( DE \) is parallel to \( BC \).
2. The magnitude \( |\overrightarrow{DE}| = |\frac{1}{2} \overrightarrow{BC}| = \frac{1}{2} |\overrightarrow{BC}| \), which means the length of \( DE \) is half the length of \( BC \).

Similarly, we can prove for other midpoints:
For \( EF \):
\( \overrightarrow{EF} = \overrightarrow{OF} - \overrightarrow{OE} \)
\( \overrightarrow{EF} = \frac{\vec{b} + \vec{c}}{2} - \frac{\vec{a} + \vec{c}}{2} \)
\( \overrightarrow{EF} = \frac{\vec{b} - \vec{a}}{2} = \frac{1}{2} (\overrightarrow{OB} - \overrightarrow{OA}) = \frac{1}{2} \overrightarrow{AB} \)
So, \( \overrightarrow{EF} \) is parallel to \( \overrightarrow{AB} \) and \( |\overrightarrow{EF}| = \frac{1}{2} |\overrightarrow{AB}| \).

For \( DF \):
\( \overrightarrow{DF} = \overrightarrow{OF} - \overrightarrow{OD} \)
\( \overrightarrow{DF} = \frac{\vec{b} + \vec{c}}{2} - \frac{\vec{a} + \vec{b}}{2} \)
\( \overrightarrow{DF} = \frac{\vec{c} - \vec{a}}{2} = \frac{1}{2} (\overrightarrow{OC} - \overrightarrow{OA}) = \frac{1}{2} \overrightarrow{AC} \)
So, \( \overrightarrow{DF} \) is parallel to \( \overrightarrow{AC} \) and \( |\overrightarrow{DF}| = \frac{1}{2} |\overrightarrow{AC}| \).
Thus, the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length. This is a fundamental result in geometry, known as the Midpoint Theorem, which is often proven using vector methods.
In simple words: When you connect the middle points of two sides of a triangle, the line you draw will always run straight next to the third side (parallel) and be exactly half as long as that third side. This is shown by comparing the vectors of these lines.

🎯 Exam Tip: The Midpoint Theorem is crucial. When proving it using vectors, remember that if \( \vec{u} = k\vec{v} \), then \( \vec{u} \) is parallel to \( \vec{v} \), and \( | \vec{u} | = |k| | \vec{v} | \). Clearly define your origin and position vectors for each point.

 

Question 6. Prove that the line segments joining the midpoints of the adjacent sides of a quadrilateral form a parallelogram.
Answer:
Let \( ABCD \) be any quadrilateral. Let \( D, E, F, G \) be the midpoints of the sides \( AB, BC, CD, AD \) respectively.

To prove \( DEFG \) is a parallelogram, we need to show that its opposite sides are parallel and equal in length. For instance, we can prove \( \overrightarrow{DE} = \overrightarrow{GF} \).
Let \( O \) be the origin. Let \( \vec{a}, \vec{b}, \vec{c}, \vec{d} \) be the position vectors of the points \( A, B, C, D \) respectively with respect to the origin \( O \).
So, \( \overrightarrow{OA} = \vec{a}, \overrightarrow{OB} = \vec{b}, \overrightarrow{OC} = \vec{c}, \overrightarrow{OD} = \vec{d} \).

**Position vector of D (midpoint of AB):**
\( \overrightarrow{OD} = \frac{\overrightarrow{OA} + \overrightarrow{OB}}{2} = \frac{\vec{a} + \vec{b}}{2} \)

**Position vector of E (midpoint of BC):**
\( \overrightarrow{OE} = \frac{\overrightarrow{OB} + \overrightarrow{OC}}{2} = \frac{\vec{b} + \vec{c}}{2} \)

**Position vector of F (midpoint of CD):**
\( \overrightarrow{OF} = \frac{\overrightarrow{OC} + \overrightarrow{OD}}{2} = \frac{\vec{c} + \vec{d}}{2} \)

**Position vector of G (midpoint of DA):**
\( \overrightarrow{OG} = \frac{\overrightarrow{OD} + \overrightarrow{OA}}{2} = \frac{\vec{d} + \vec{a}}{2} \)

**Let's find the vector \( \overrightarrow{DE} \):**
\( \overrightarrow{DE} = \overrightarrow{OE} - \overrightarrow{OD} \)
\( \overrightarrow{DE} = \frac{\vec{b} + \vec{c}}{2} - \frac{\vec{a} + \vec{b}}{2} \)
\( \overrightarrow{DE} = \frac{\vec{b} + \vec{c} - \vec{a} - \vec{b}}{2} \)
\( \overrightarrow{DE} = \frac{\vec{c} - \vec{a}}{2} \) -----(1)

**Now, let's find the vector \( \overrightarrow{GF} \):**
\( \overrightarrow{GF} = \overrightarrow{OF} - \overrightarrow{OG} \)
\( \overrightarrow{GF} = \frac{\vec{c} + \vec{d}}{2} - \frac{\vec{d} + \vec{a}}{2} \)
\( \overrightarrow{GF} = \frac{\vec{c} + \vec{d} - \vec{d} - \vec{a}}{2} \)
\( \overrightarrow{GF} = \frac{\vec{c} - \vec{a}}{2} \) -----(2)

From equations (1) and (2), we see that \( \overrightarrow{DE} = \overrightarrow{GF} \).
This means that \( DE \) and \( GF \) are parallel and equal in length.

**Next, let's find the vector \( \overrightarrow{EF} \):**
\( \overrightarrow{EF} = \overrightarrow{OF} - \overrightarrow{OE} \)
\( \overrightarrow{EF} = \frac{\vec{c} + \vec{d}}{2} - \frac{\vec{b} + \vec{c}}{2} \)
\( \overrightarrow{EF} = \frac{\vec{c} + \vec{d} - \vec{b} - \vec{c}}{2} \)
\( \overrightarrow{EF} = \frac{\vec{d} - \vec{b}}{2} \) -----(3)

**Now, let's find the vector \( \overrightarrow{DG} \):**
\( \overrightarrow{DG} = \overrightarrow{OG} - \overrightarrow{OD} \)
\( \overrightarrow{DG} = \frac{\vec{d} + \vec{a}}{2} - \frac{\vec{a} + \vec{b}}{2} \)
\( \overrightarrow{DG} = \frac{\vec{d} + \vec{a} - \vec{a} - \vec{b}}{2} \)
\( \overrightarrow{DG} = \frac{\vec{d} - \vec{b}}{2} \) -----(4)

From equations (3) and (4), we see that \( \overrightarrow{EF} = \overrightarrow{DG} \).
This means that \( EF \) and \( DG \) are parallel and equal in length.

Since both pairs of opposite sides are parallel and equal, the quadrilateral \( DEFG \) is a parallelogram. This principle, known as Varignon's theorem, demonstrates a beautiful property of quadrilaterals, regardless of their shape.
In simple words: We take any four-sided shape and find the middle points of each side. When we connect these middle points in order, they always form a new shape that is a parallelogram. We prove this by showing that its opposite sides are equal in length and run in the same direction using vectors.

🎯 Exam Tip: To prove a quadrilateral is a parallelogram using vectors, you need to show that one pair of opposite sides are equal and parallel (i.e., their vectors are equal) OR that both pairs of opposite sides are parallel. Using position vectors for midpoints is key here.

 

Question 7. If \( \vec{a} \) and \( \vec{b} \) represent a side and a diagonal of a parallelogram, find the other sides and the other diagonal.
Answer:
Let \( ABCD \) be a parallelogram.

Let \( \overrightarrow{AB} = \vec{a} \) be one side and \( \overrightarrow{AC} = \vec{b} \) be a diagonal.
Since \( ABCD \) is a parallelogram, we have:

**Finding the other sides:**
1. \( \overrightarrow{DC} = \overrightarrow{AB} = \vec{a} \) (Opposite sides are equal and parallel).

2. Using the triangle law of vector addition for \( \triangle ABC \):
\( \overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC} \)
\( \vec{a} + \overrightarrow{BC} = \vec{b} \)
\( \overrightarrow{BC} = \vec{b} - \vec{a} \)

3. Since \( \overrightarrow{AD} = \overrightarrow{BC} \) (Opposite sides are equal and parallel),
\( \overrightarrow{AD} = \vec{b} - \vec{a} \)

So, the other sides of the parallelogram are \( \overrightarrow{DC} = \vec{a} \) and \( \overrightarrow{BC} = \overrightarrow{AD} = \vec{b} - \vec{a} \).

**Finding the other diagonal:**
The other diagonal is \( \overrightarrow{DB} \).
Using the triangle law of vector addition for \( \triangle DAB \):
\( \overrightarrow{DA} + \overrightarrow{AB} = \overrightarrow{DB} \)
We know \( \overrightarrow{DA} = - \overrightarrow{AD} = -(\vec{b} - \vec{a}) = \vec{a} - \vec{b} \).
So, \( \overrightarrow{DB} = (\vec{a} - \vec{b}) + \vec{a} \)
\( \overrightarrow{DB} = 2\vec{a} - \vec{b} \)

Alternatively, using \( \triangle DCB \):
\( \overrightarrow{DC} + \overrightarrow{CB} = \overrightarrow{DB} \)
We know \( \overrightarrow{CB} = - \overrightarrow{BC} = -(\vec{b} - \vec{a}) = \vec{a} - \vec{b} \).
So, \( \overrightarrow{DB} = \vec{a} + (\vec{a} - \vec{b}) \)
\( \overrightarrow{DB} = 2\vec{a} - \vec{b} \)

So, the other sides are \( \vec{a} \) and \( \vec{b} - \vec{a} \), and the other diagonal is \( 2\vec{a} - \vec{b} \). This problem highlights how vectors simplify geometric proofs in quadrilaterals.
In simple words: If you know one side and one diagonal of a parallelogram as vectors, you can find the other sides and the other diagonal using vector addition and subtraction. Opposite sides are equal, and the sum of two adjacent sides gives the diagonal between them.

🎯 Exam Tip: Always draw a diagram when solving vector problems in geometry. Remember that in a parallelogram, opposite sides are represented by the same vector, and the diagonal is the vector sum of adjacent sides originating from the same point.

 

Question 8. If \( \overrightarrow{\mathbf{P O}}+\overrightarrow{\mathbf{O Q}} = \overrightarrow{\mathbf{Q O}}+\overrightarrow{\mathbf{O R}} \), prove that the points P, Q, R are collinear.
Answer:
Given the equation: \( \overrightarrow{PO} + \overrightarrow{OQ} = \overrightarrow{QO} + \overrightarrow{OR} \) -----(1)
From the figure (or triangle law of vector addition):
\( \overrightarrow{PO} + \overrightarrow{OQ} = \overrightarrow{PQ} \) -----(2)
And \( \overrightarrow{QO} + \overrightarrow{OR} = \overrightarrow{QR} \) -----(3)

Using equations (1), (2), and (3):
\( \overrightarrow{PQ} = \overrightarrow{QR} \)

Since \( \overrightarrow{PQ} = \overrightarrow{QR} \), this means that \( \overrightarrow{PQ} \) and \( \overrightarrow{QR} \) are parallel vectors and are in the same direction. Also, point \( Q \) is a common point between them.
Therefore, \( P, Q, R \) lie on a straight line.
Hence, \( P, Q, R \) are collinear. The equality of these two vectors proves that the points must lie on a single line, making the vector from P to Q identical in direction and magnitude to the vector from Q to R.
In simple words: We are given an equation that uses vectors. By changing the parts of the equation into simpler vectors, we find that the vector from P to Q is exactly the same as the vector from Q to R. Since they are the same and share point Q, all three points must be on one straight line.

🎯 Exam Tip: To prove collinearity of three points \( P, Q, R \) using vectors, you need to show that \( \overrightarrow{PQ} = k \overrightarrow{QR} \) (or \( \overrightarrow{PQ} = k \overrightarrow{PR} \)) for some scalar \( k \), and that they share a common point. In this case, \( k=1 \) and \( Q \) is common.

 

Question 9. If D is the midpoint of the side BC of a triangle ABC, prove that \( \overrightarrow{\mathbf{A B}} + \overrightarrow{\mathbf{A C}} = 2 \overrightarrow{\mathbf{A D}} \)
Answer: Let's consider a triangle ABC, where D is the midpoint of the side BC. We need to prove the relationship between the vectors \( \overrightarrow{\mathbf{A B}} \), \( \overrightarrow{\mathbf{A C}} \), and \( \overrightarrow{\mathbf{A D}} \).
Since D is the midpoint of BC, the vector \( \overrightarrow{\mathbf{B D}} \) is equal to the vector \( \overrightarrow{\mathbf{D C}} \).
Using the triangle law of vector addition:
\( \overrightarrow{\mathbf{B D}} = \overrightarrow{\mathbf{B A}} + \overrightarrow{\mathbf{A D}} \)
We also know that \( \overrightarrow{\mathbf{D C}} = \overrightarrow{\mathbf{D A}} + \overrightarrow{\mathbf{A C}} \)
Since \( \overrightarrow{\mathbf{B D}} = \overrightarrow{\mathbf{D C}} \), we can write:
\( \overrightarrow{\mathbf{B A}} + \overrightarrow{\mathbf{A D}} = \overrightarrow{\mathbf{D A}} + \overrightarrow{\mathbf{A C}} \)
We know that \( \overrightarrow{\mathbf{B A}} = - \overrightarrow{\mathbf{A B}} \) and \( \overrightarrow{\mathbf{D A}} = - \overrightarrow{\mathbf{A D}} \). Substitute these into the equation:
\( - \overrightarrow{\mathbf{A B}} + \overrightarrow{\mathbf{A D}} = - \overrightarrow{\mathbf{A D}} + \overrightarrow{\mathbf{A C}} \)
Now, bring all \( \overrightarrow{\mathbf{A D}} \) terms to one side and others to the other side:
\( \overrightarrow{\mathbf{A D}} + \overrightarrow{\mathbf{A D}} = \overrightarrow{\mathbf{A C}} + \overrightarrow{\mathbf{A B}} \)
\( 2 \overrightarrow{\mathbf{A D}} = \overrightarrow{\mathbf{A B}} + \overrightarrow{\mathbf{A C}} \)
This proves the given statement. The median of a triangle connects a vertex to the midpoint of the opposite side.
In simple words: If you add the vectors from one corner of a triangle to the other two corners, their sum is twice the vector from that same corner to the middle point of the opposite side.

🎯 Exam Tip: When proving vector relations involving midpoints, always use the section formula for position vectors or the triangle law of vector addition effectively.

 

Question 10. If G is the centroid of a triangle ABC, prove that \( \overrightarrow{\mathbf{G A}}+\overrightarrow{\mathbf{G B}}+\overrightarrow{\mathbf{G C}}=\overrightarrow{\mathbf{0}} \)
Answer: Let's consider a triangle ABC with G as its centroid. Let O be the origin. The position vectors of the vertices A, B, and C are \( \overrightarrow{\mathbf{O A}} = \vec{a} \), \( \overrightarrow{\mathbf{O B}} = \vec{b} \), and \( \overrightarrow{\mathbf{O C}} = \vec{c} \) respectively.
The position vector of the centroid G is given by the formula:
\( \overrightarrow{\mathbf{O G}} = \frac{\vec{a} + \vec{b} + \vec{c}}{3} \)
Now, we want to find the sum of vectors \( \overrightarrow{\mathbf{G A}} + \overrightarrow{\mathbf{G B}} + \overrightarrow{\mathbf{G C}} \).
We can express each vector from G to a vertex using the origin O:
\( \overrightarrow{\mathbf{G A}} = \overrightarrow{\mathbf{O A}} - \overrightarrow{\mathbf{O G}} = \vec{a} - \overrightarrow{\mathbf{O G}} \)
\( \overrightarrow{\mathbf{G B}} = \overrightarrow{\mathbf{O B}} - \overrightarrow{\mathbf{O G}} = \vec{b} - \overrightarrow{\mathbf{O G}} \)
\( \overrightarrow{\mathbf{G C}} = \overrightarrow{\mathbf{O C}} - \overrightarrow{\mathbf{O G}} = \vec{c} - \overrightarrow{\mathbf{O G}} \)
Adding these three vectors together:
\( \overrightarrow{\mathbf{G A}} + \overrightarrow{\mathbf{G B}} + \overrightarrow{\mathbf{G C}} = (\vec{a} - \overrightarrow{\mathbf{O G}}) + (\vec{b} - \overrightarrow{\mathbf{O G}}) + (\vec{c} - \overrightarrow{\mathbf{O G}}) \)
\( = \vec{a} + \vec{b} + \vec{c} - 3 \overrightarrow{\mathbf{O G}} \)
Substitute the expression for \( \overrightarrow{\mathbf{O G}} \):
\( = \vec{a} + \vec{b} + \vec{c} - 3 \left( \frac{\vec{a} + \vec{b} + \vec{c}}{3} \right) \)
\( = \vec{a} + \vec{b} + \vec{c} - (\vec{a} + \vec{b} + \vec{c}) \)
\( = \overrightarrow{\mathbf{0}} \)
Thus, the sum of the vectors from the centroid to the vertices of a triangle is the zero vector. This means these vectors perfectly balance each other out.
In simple words: The centroid is like the balance point of a triangle. If you draw arrows from this balance point to each corner, and then add those arrows up, you will always get a zero arrow, meaning they cancel each other out.

🎯 Exam Tip: Remember the formula for the position vector of a centroid. This proof highlights how it relates to vectors from the centroid to the vertices.

 

Question 11. 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 \( \overrightarrow{\mathbf{A D}}+\overrightarrow{\mathbf{B E}}+\overrightarrow{\mathbf{C F}}=\overrightarrow{\mathbf{0}} \)
Answer: Let's consider a triangle ABC, with D, E, and F as the midpoints of sides BC, CA, and AB respectively. Let O be the origin, and the position vectors of the vertices A, B, and C be \( \overrightarrow{\mathbf{O A}} = \vec{a} \), \( \overrightarrow{\mathbf{O B}} = \vec{b} \), and \( \overrightarrow{\mathbf{O C}} = \vec{c} \).
First, we find the position vectors of the midpoints D, E, and F:
Position vector of D (midpoint of BC):
\( \overrightarrow{\mathbf{O D}} = \frac{\overrightarrow{\mathbf{O B}} + \overrightarrow{\mathbf{O C}}}{2} = \frac{\vec{b} + \vec{c}}{2} \)
Position vector of E (midpoint of CA):
\( \overrightarrow{\mathbf{O E}} = \frac{\overrightarrow{\mathbf{O C}} + \overrightarrow{\mathbf{O A}}}{2} = \frac{\vec{c} + \vec{a}}{2} \)
Position vector of F (midpoint of AB):
\( \overrightarrow{\mathbf{O F}} = \frac{\overrightarrow{\mathbf{O A}} + \overrightarrow{\mathbf{O B}}}{2} = \frac{\vec{a} + \vec{b}}{2} \)
Now, we express the vectors \( \overrightarrow{\mathbf{A D}} \), \( \overrightarrow{\mathbf{B E}} \), and \( \overrightarrow{\mathbf{C F}} \) in terms of \( \vec{a}, \vec{b}, \vec{c} \):
\( \overrightarrow{\mathbf{A D}} = \overrightarrow{\mathbf{O D}} - \overrightarrow{\mathbf{O A}} = \frac{\vec{b} + \vec{c}}{2} - \vec{a} = \frac{\vec{b} + \vec{c} - 2\vec{a}}{2} \)
\( \overrightarrow{\mathbf{B E}} = \overrightarrow{\mathbf{O E}} - \overrightarrow{\mathbf{O B}} = \frac{\vec{c} + \vec{a}}{2} - \vec{b} = \frac{\vec{c} + \vec{a} - 2\vec{b}}{2} \)
\( \overrightarrow{\mathbf{C F}} = \overrightarrow{\mathbf{O F}} - \overrightarrow{\mathbf{O C}} = \frac{\vec{a} + \vec{b}}{2} - \vec{c} = \frac{\vec{a} + \vec{b} - 2\vec{c}}{2} \)
Finally, we add these three vectors together:
\( \overrightarrow{\mathbf{A D}} + \overrightarrow{\mathbf{B E}} + \overrightarrow{\mathbf{C F}} = \frac{\vec{b} + \vec{c} - 2\vec{a}}{2} + \frac{\vec{c} + \vec{a} - 2\vec{b}}{2} + \frac{\vec{a} + \vec{b} - 2\vec{c}}{2} \)
\( = \frac{(\vec{b} + \vec{c} - 2\vec{a}) + (\vec{c} + \vec{a} - 2\vec{b}) + (\vec{a} + \vec{b} - 2\vec{c})}{2} \)
\( = \frac{\vec{b} + \vec{c} - 2\vec{a} + \vec{c} + \vec{a} - 2\vec{b} + \vec{a} + \vec{b} - 2\vec{c}}{2} \)
Group the terms:
\( = \frac{(-2\vec{a} + \vec{a} + \vec{a}) + (\vec{b} - 2\vec{b} + \vec{b}) + (\vec{c} + \vec{c} - 2\vec{c})}{2} \)
\( = \frac{0\vec{a} + 0\vec{b} + 0\vec{c}}{2} \)
\( = \frac{\overrightarrow{\mathbf{0}}}{2} = \overrightarrow{\mathbf{0}} \)
This shows that the sum of the medians as vectors from the vertices to the midpoints of the opposite sides is a zero vector. This property is fundamental in vector geometry.
In simple words: When you draw lines from each corner of a triangle to the middle of the opposite side (these are called medians), and you think of them as arrows, if you add all three arrows together, they cancel each other out completely, resulting in no net movement.

🎯 Exam Tip: Break down complex vector proofs by defining position vectors for all points and then expressing the required vectors in terms of these position vectors before summing them up.

 

Question 12. If ABCD is a quadrilateral and E and F are the midpoints of AC and BD respectively, then Prove that \( \overrightarrow{\mathbf{A B}}+\overrightarrow{\mathbf{A D}}+\overrightarrow{\mathbf{C B}}+\overrightarrow{\mathbf{C D}}=4 \overrightarrow{\mathbf{E F}} \)
Answer: Let ABCD be a quadrilateral. E is the midpoint of diagonal AC, and F is the midpoint of diagonal BD. Let O be the origin, and the position vectors of the vertices A, B, C, and D be \( \overrightarrow{\mathbf{O A}} = \vec{a} \), \( \overrightarrow{\mathbf{O B}} = \vec{b} \), \( \overrightarrow{\mathbf{O C}} = \vec{c} \), and \( \overrightarrow{\mathbf{O D}} = \vec{d} \) respectively.
First, find the position vectors of the midpoints E and F:
Since E is the midpoint of AC:
\( \overrightarrow{\mathbf{O E}} = \frac{\overrightarrow{\mathbf{O A}} + \overrightarrow{\mathbf{O C}}}{2} = \frac{\vec{a} + \vec{c}}{2} \)
Since F is the midpoint of BD:
\( \overrightarrow{\mathbf{O F}} = \frac{\overrightarrow{\mathbf{O B}} + \overrightarrow{\mathbf{O D}}}{2} = \frac{\vec{b} + \vec{d}}{2} \)
Now, let's express the vector \( \overrightarrow{\mathbf{E F}} \):
\( \overrightarrow{\mathbf{E F}} = \overrightarrow{\mathbf{O F}} - \overrightarrow{\mathbf{O E}} \)
\( = \frac{\vec{b} + \vec{d}}{2} - \frac{\vec{a} + \vec{c}}{2} \)
\( = \frac{\vec{b} + \vec{d} - \vec{a} - \vec{c}}{2} \)
So, \( 4 \overrightarrow{\mathbf{E F}} = 4 \left( \frac{\vec{b} + \vec{d} - \vec{a} - \vec{c}}{2} \right) = 2(\vec{b} + \vec{d} - \vec{a} - \vec{c}) \) (Equation 1)
Next, let's calculate the left-hand side of the equation:
\( \overrightarrow{\mathbf{A B}}+\overrightarrow{\mathbf{A D}}+\overrightarrow{\mathbf{C B}}+\overrightarrow{\mathbf{C D}} \)
Express each vector using position vectors relative to the origin O:
\( \overrightarrow{\mathbf{A B}} = \overrightarrow{\mathbf{O B}} - \overrightarrow{\mathbf{O A}} = \vec{b} - \vec{a} \)
\( \overrightarrow{\mathbf{A D}} = \overrightarrow{\mathbf{O D}} - \overrightarrow{\mathbf{O A}} = \vec{d} - \vec{a} \)
\( \overrightarrow{\mathbf{C B}} = \overrightarrow{\mathbf{O B}} - \overrightarrow{\mathbf{O C}} = \vec{b} - \vec{c} \)
\( \overrightarrow{\mathbf{C D}} = \overrightarrow{\mathbf{O D}} - \overrightarrow{\mathbf{O C}} = \vec{d} - \vec{c} \)
Now, sum these four vectors:
\( \overrightarrow{\mathbf{A B}}+\overrightarrow{\mathbf{A D}}+\overrightarrow{\mathbf{C B}}+\overrightarrow{\mathbf{C D}} = (\vec{b} - \vec{a}) + (\vec{d} - \vec{a}) + (\vec{b} - \vec{c}) + (\vec{d} - \vec{c}) \)
\( = \vec{b} - \vec{a} + \vec{d} - \vec{a} + \vec{b} - \vec{c} + \vec{d} - \vec{c} \)
Group like terms:
\( = (\vec{b} + \vec{b}) + (\vec{d} + \vec{d}) + (-\vec{a} - \vec{a}) + (-\vec{c} - \vec{c}) \)
\( = 2\vec{b} + 2\vec{d} - 2\vec{a} - 2\vec{c} \)
\( = 2(\vec{b} + \vec{d} - \vec{a} - \vec{c}) \) (Equation 2)
Comparing Equation 1 and Equation 2, we see that the left-hand side equals the right-hand side. This demonstrates an interesting property relating the vectors of a quadrilateral and the segment connecting the midpoints of its diagonals.
In simple words: For any four-sided shape, if you add up the arrows along all its sides in a specific way, that sum will be four times the arrow that connects the middle points of its two diagonal lines.

🎯 Exam Tip: When dealing with quadrilateral vector proofs, consistently define an origin and position vectors. Express all given vectors and required vectors in terms of these position vectors for clarity and accuracy.