Algebra Of Vectors

i. a displacement of 40 km, 30° east of north

Step 1: Draw north, south, east and west as shown below:

Step 2: Plot a line 30° east of north as shown below:

Step 3: Define scale and mark 40km on line

Let the scale be 10km = 1cm

∴ represents the displacement of 40 km, 30^o East of North

ii. a displacement of 50 km south - east

Step 1: Draw north, south, east and west as shown below:

Step 2: As the displacement should be south - east, the angle between the displacement and east (or south) will be 45°. Now, plot a line 45° east of south as shown below:

Step 3: Define scale and mark point R such that OR = 50km on line . Let the scale be 10km = 1cm

∴ represents the displacement of 50 km south – east

iii. A displacement of 70 km, 40^o north of west.

Step 1: Draw north, south, east and west as shown below:

Step 2: Plot a line 40° north of west as shown below:

Step 3: Define scale and mark point R such that OR = 70km on line .Let the scale be 10km = 1cm

∴ represents the displacement of 70 km, 40^o north of west

i. 15 kg - is a scalar quantity as this involves only mass. A scalar quantity is a one - dimensional measurement of a quantity, like temperature, or mass.

ii. 20 kg weight - is a vector quantity as it involves both magnitude and direction. Weight is a force which is a vector and has a magnitude and direction.

iii. 45^o is a scalar quantity as it involves the only magnitude. A scalar quantity is a one - dimensional measurement of a quantity, like temperature, or mass.

iv. 10 meters south - east is a vector quantity as it involves both magnitude and direction.

v. 50 m/sec² is a scalar quantity as it involves a magnitude of acceleration. A scalar quantity is a one - dimensional measurement of a quantity.

i. Time period - is a scalar quantity as it involves only magnitude. A scalar quantity is a one - dimensional measurement of a quantity. Eg: 10 seconds has only magnitude, i.e., 10 and no direction.

ii. Distance - is a scalar quantity as it involves only magnitude. A scalar quantity is a one dimensional measurement of a quantity. Eg: 5meters has only magnitude 5 and no direction.

iii. Displacement - is vector quantity as it involves both magnitude and direction. Vector quantity has both magnitude and direction.

iv. Force - is a vector quantity as it involves both magnitude and direction. Vector quantity has both magnitude and direction. Eg., 5N downward has magnitude of 5 and direction is downward.

v. Work done - is a scalar quantity as it involves only magnitude and no particular direction. A scalar quantity is a one dimensional measurement of a quantity.

vi. Velocity - is a vector quantity as it involves both magnitude as well as direction. Vector quantity has both magnitude and direction. Eg., 5m/s east has magnitude of 5m/s and also direction towards east.

vii. Acceleration is a vector quantity because it involves both magnitude as well as direction.

i. Collinear

Two or more vectors that lie on the same line or on a parallel line to this are called collinear vectors. Two collinear vectors may point in either same or opposite direction. But, they cannot be inclined at some angle from each other.

Hence FE (), AD () and BC () are collinear vectors.

And also AF () and CD () are collinear vectors.

And AB () and ED () are collinear vectors.

ii. Equal

Equal vectors are vectors that have the same magnitude and the same direction. Equal vectors may start at different positions.

Hence AF () and CD () are equal vectors.

And also FE () and BC () are equal vectors.

And AB () and ED () are equal vectors.

iii. Co - initial

Any given two vectors are called co - initial vectors if both the given vectors have the same initial point.

Hence, AB (), AF () and AD () are co - initial vectors.

iv. Collinear but not equal.

And AD () and BC () are collinear but not equal vectors.

And AD () and FE () are collinear but not equal vectors

i. and are collinear. (True)

Two or more vectors that lie on the same line or on a parallel line to this are called collinear vectors.

and are collinear.

ii. Two collinear vectors are always equal in magnitude. (False)

Two or more vectors that lie on the same line or on a parallel line to this are called collinear vectors. Two collinear vectors may point in either same or opposite direction. And they are not necessarily equal in magnitude they can be of different magnitude also.

iii. Zero vector is unique.(True)

There is only one zero - vector in a vector space. Hence zero vector is unique.

iv. Two vectors having same magnitude are collinear. (False)

It is not necessary for two vectors having the same magnitude to be parallel to the same line. Hence two vectors having same magnitude need not be collinear.

v. Two collinear vectors having the same magnitude are equal.(False)

Two vectors are said to be equal if they have the same magnitude and direction, regardless of the positions of their initial points.

Zero vector is a vector which has magnitude is 0. It is denoted by .

A unit vector is a vector whose magnitude is 1. It is denoted by capping the vector whose unit vector is required. For instance, the unit vector of will be .

A position vector is a vector which tells the relative position of any point in space with respect to origin. This vector starts from origin and its head lies on the point itself. If The x, y, z coordinates of the point is x₁, y₁, z₁, the position vector will be equal to .

Any vector , if the position vectors of point P () and Q () are known, can be written as .

Let the position vectors of points P, Q, R be .

Then =

Since and are non-colinear vectors, the only way the equality will hold is if x=y=0.

In the above figure, , , and

Using parallelogram law of vector addition, we can say that

and or

Also, are the diagonals of the parallelogram.

Hence the diagonal vectors of a parallelogram formed by vectors and will be ) and .

Let ∆ABC be the required triangle with and .

Any vector , if the position vectors of point A () and B () are known, can be written as .

Let the position vectors of points A, B, C be .

Then

Any vector , if the position vectors of point A () and B () are known, can be written as .

Since the position vectors of points A, B, C are , we get

In the figure, D is the mid-point of AB, so it divides AB in 1:1 ratio. CD is a median of ∆ABC. G is the centroid of the triangle and by the property of triangle, G divides CD in 2:1 ratio.

The position vector of point D can be calculated using the section formula for vector, which states that the position vector of a point() dividing two position vectors() in ration m:n, internally is

So,

Similarly, using section formula for G between points C and D, we get

Let the position vector points A, B, C be and . Then the position vector of G will be .

Any vector , if the position vectors of point A () and B () are known, can be written as .

Then,

Given that the position vector points A, B are .

Let us assume that C lies between A and B.

Then AB=AC+BC

Given that 3AC=2AB=2(AC+BC)

⇒ AC=2BC

Therefore AC: BC=1:2

Also, since the ratio is positive, our assumption was correct.

Using section formula,

Let the position vectors of A, B and C be and

Then the position vector of will be

Any vector , if the position vectors of point A () and B () are known, can be written as .

Then,

⇒

Substituting value of , we get

⇒ or λ=2

Let the position vectors of A, B and C be and

Then the position vector of and will be , and respectively.

Any vector , if the position vectors of point A () and B () are known, can be written as .

Then,

The modulus of is a, therefore can be written as modulus×unit-direction=

Given that m has the magnitude of 1, therefore has magnitude of 1 or ma=1.Hence

Since in an equilateral triangle, orthocenter and centroid coincide, therefore the position vector of centroid is .

Also, the position vector of centroid G () can be defined as

Therefore, hence

Let the angle made be α. We know that the sum of squares of direction cosines of a vector is 1. SO, we get

cos²α+cos²α+cos²α=1

Since, α is acute therefore

Any vector, if it’s magnitude and direction cosines are given can be written as

So the required vector is

Rationalizing, we get

The sum of squares of direction cosines of a vector is 1.

Let the angles made by vector be α, β, γ. Then, we get

cos²α+cos²β+cos²γ=1

using cos²θ=1-sin²θ, we get

(1-sin²α) +(1-sin²β) +(1-sin²γ) =1

Or, sin²α+sin²β+sin²γ=2

Any vector , if the position vectors of point A () and B () are known, can be written as .

Since the position vectors of points A, B, C are , we get

Let the angles made by vector be α, β, γ and the magnitude be m.

Given that α=45°, β=60° and m= 12. We have to figure out the vector.

Since cos²α+cos²β+cos²γ=1, we get

cos²45°+cos²60°+cos²γ=1

Since γ is obtuse, .

Any vector, if it’s magnitude and direction cosines are given can be written as

So the required vector is

Rationalizing, we get

Since L_x=12, L_y=3 and L_z=4 are given, we can find out L by

L²= L_x²+L_y²+L_z²

=12²+3²+4²

=144+9+16=169

Hence L=13 units.

If a point R() divides the vector joining point P() and Q() externally in the ratio m:n, then

Here, , , m=1 and n=4

Then

The direction cosines of a vector are

In this question, a₁=6, a₂=-2 and a₃=3, Substituting in formulas we get

Given that , we get

The unit vector of any vector can be written as

Let

Then,

Both will be parallel to , therefore the answer is

Given that , we get

The unit vector of any vector can be written as

Let

Then,

The unit vector in direction of is

. If a point R() divides the vector joining point P() and Q() externally in the ratio m:n, then

Here, , , m=2 and n=3

Then

Given that , we get

The unit vector of any vector can be written as

Let

Then,

Given that , we get

The magnitude of any vector can be written as

The sum of squares of direction cosines of a vector is 1.

Let the angles made by vector be α, β, θ. Then, we get

cos²α+cos²β+cos²θ=1

Given that , we have to calculate θ

⇒ (Since θ is acute)

Hence,

The unit vector of any vector can be written as

Given that

We get

Given that , we get

The unit vector of any vector can be written as

Let

Then,

Both will be parallel to , therefore the answer is

The unit vector of any vector can be written as

Given that

We get

If the co-ordinates of a point A≡ (x₁, y₁, z₁), then the position vector of A() is

Given that A≡(3,4,2) and B≡(1,2,4), we get position vector of A() and B(). Let the midpoint be C()

The position vector of midpoint of two vectors is defined by

Given that

The unit vector of any vector can be written as

Let the required vector be .

Any vector () with magnitude m and unit vector can be written as .

Since the magnitude of is 6 and it’s unit vector is , we get

The angle that a vector makes with y-axis is

In this question, a₁=, a₂=1 and a₃=1, Substituting in formulas we get

and both have magnitude 1 but different directions. is along x-axis and is along y-axis.

and both have the same direction but different magnitudes, 1 and 2.

Given that

The unit vector of any vector can be written as

Let the required vector be .

Any vector () with magnitude m and unit vector can be written as .

Since the magnitude of is 8 and it’s unit vector is , we get

The direction cosines of a vector are

In this question, a₁=1, a₂=2 and a₃=3, Substituting in formulas we get

Given that

The unit vector of any vector can be written as

Two vectors and are collinear , if and only if, .

Here and

Hence,

Solving this equality, we get

a=-4

The direction cosines of a vector are

In this question, a₁=-2, a₂=1 and a₃=-5, Substituting in formulas we get

Then

Given that

The unit vector of any vector can be written as

Two vectors and are equal , if and only if, a₁=b₁, a₂=b₂, a₃=b₃

Here and

Hence, we get x=3, -y=2⇒y=-2 and -z=1⇒ z=-1

and

Let

The unit vector of any vector can be written as

Hence,

Two vectors and are parallel , if and only if, .

Here and

Hence,

Solving this equality, we get

Let the angles made by vector be α, β, θ and the magnitude be m.

Given that . We have to figure out the vector.

Since cos²α+cos²β+cos²θ=1, we get

Since θ is acute, .

Any vector, if it’s magnitude and direction cosines are given can be written as

So, the required vector is

If the co-ordinates of points A≡ (x₁, y₁, z₁) and B≡ (x₂, y₂, z₂), then the vector is

Given that P≡(1,3,0) and Q≡(4,5,6), we get

The unit vector of any vector can be written as

Hence,

Given that

The unit vector of any vector can be written as

Let the required vector be .

Any vector () with magnitude m and unit vector can be written as .

Since the magnitude of is 21 and it’s unit vector is , we get

Given that and -3≤λ≤2

We have to figure out range of

In calculating the modulus of a vector multiplied by a scalar quantity, the sign of the scalar quantity does not matter, only it’s absolute value does.

Hence the minimum value of when λ=0 and maximum value of when λ=-3.

Given that the position vectors of point A and B are and . Let the position vector of point C be .

The position vector of B will be defined as

⇒

If a point R() divides the vector joining point P() and Q() externally in the ratio m:n, then

Here, , , m=2 and n=1

We get,

Slope of a line joining two points

Slope of AC

Slope of AB

Product of Slopes (AC AB) =

= -1

As the Product of Slopes (AC AB) = -1, so AC ⟘ AB, ie.., CAB = 90°.

Circumcentre (O) of Triangle ABC = Mid-Point of BC

Mid-Point of BC =

Now,

Unit Vector

Vector along , whose magnitude is

Option (A) is the answer.

--------- (i)

As, AD = 2BC {Properties of a regular hexagon, also AD || BC (Parallel)}

Putting in equation (i),

Option(C)is the answer.

Draw ON perpendicular to the line AB.

Let be the unit vector along ON,

The resultant force --------- (i)

The angles between and the forces , , are CON, AON & BON respectively,

R = 8

---------- (i)

Multiplying the equation (i) by 3,

----------(ii)

Also, ----------(iv)

Multiplying the equation (iv) by 5,

----------(v)

Adding equation (iv) & (v) respectively,

3AC = 5CB

Option(C)is the answer.

As is collinear with ,

-------- (i)

As is collinear with ,

-------- (ii)

Adding both sides of the equation (i),

-------- (iii)

Adding both sides of the equation (iii),

--------- (iv)

Equating the RHS of equation (iii)& (iv), being their LHS equal,

As, a is not collinear with c,

{From equation (iv)}

Option (D)is the answer.

A (, B () & C () are collinear,

Then

Comparing the LHS & RHS of the above mentioned equation,

-80 = a – 40

a = 40 – 80

a = -40

a = -40

Option (B) is the answer.

Let us consider the point O as origin.

G is the mid – point of AC.

------- (i)

Also, G is the mid− point of BD,

------- (ii)

Adding eq. (i) & eq. (ii),

Option (B) is the answer

=1

Hence, the given vector is a unit vector.

Option (B) is the answer

------- (i)

In the triangle ADE,

{}

Option (C) is the answer

As a plane passing through

Lines and lie on the plane.

The parametric equation of the plane can be expressed as,

As,

Option (B) is the answer.

Let the vertices of the triangle ABC be A (), B () & C (), with respect to the origin.

O (x,y) is the circumcentre & O’ (0,0) is the orthocenter.

As the centroid ‘G’ divides the orthocentre ‘C’ (x,y) and circumcentre (0,0) in the ratio 2 : 1.

By using Section Formula,

Option (B)is the answer.

And,

(Given)

Multiplying the aqbove mentioned equation by ,

So, the position vector of mid – point of BD = Position Vector of mid - point of AC.

Hence the diagonals bisect each other.

Therefore the given figure ABCD is a parallelogram.

Option (D) is the answer

Let A be the origin, then , implies that the position vectors of B and C are & respectively.

Let AD be the median and G be the centroid.

Then,

Position Vector of

Position Vector of

Option (A) is the answer.

As,

In triangle AOF,

And,

Option (D) is the answer

Hence the triangle is isosceles with two sides equal.

Option (A) is the answer

the given vectors are collinear.

After comparing the equations,

y – 3 = -6

y = -6 + 3

y = -3

4 – x = 2

x = 4 – 2

x = 2

(x, y) = (2, -3)

Option (A) is the answer.

ABCD is a parallelogram with diagonals AC and BD.

&

{}

Option (C) is the answer.

,

{}

-------(i)

Option (D) is the answer.

If & are collinear vectors, then they are parallel,

Then,

For some scalar

If , then,

If

Thus, respective components of are proportional.

However the vectors can have different directions.

Statement given in D is incorrect.

Option (D) is the answer.

Subtracting, from both the sides of the above mentioned equation,

Solving RHS,

LHS ≠ RHS

Hence, it is not true.

Option (C) is the answer.

As P, Q and R are three collinear points.

Hence, as shown in above fig

And given and .

Therefore

Given that, and are three sides of a triangle.

Hence from the above figure we get,

AB = , BC = and AC =

So

[since]

[Since CA = - AC]

Triangle law says that, if vectors are represented in magnitude and direction by the two sides of a triangle is same order, then their sum is represented by the third side took in reverse order. Thus,

or or

Given and are two non - collinear vectors having the same initial point.

Let and

Let us draw a parallelogram with AB and AD as any of the two sides of the parallelogram as shown below.

We know in parallelogram opposite sides are equal hence,

and

Now consider ΔABC, applying triangles law of vectors, we get

Similarly in ΔABD, applying triangles law of vectors, we get

Looking at the two equations (i) and (ii) we can conclude that

andare the diagonals of a parallelogram whose adjacent sides areand.

Given is a vector and m is a scalar such that

Let then according to the given question

Compare the coefficients of, we get

ma₁ = 0 ⇒ m = 0 or a₁ = 0

Similarly, mb₁ = 0 ⇒ m = 0 or b₁ = 0

And, mc₁ = 0 ⇒ m = 0 or c₁ = 0

From the above three conditions,

m = 0 or a₁ = b₁ = c₁ = 0

⇒ m = 0 or

Hence the alternatives for m and are m = 0 or

Given:

Let and

So according to the given criteria,

Compare the coefficients of, we get

a₁ = a₂, b₁ = b₂ and c₁ = c₂……….(i)

and

Substitute the values from eqn (i) in above eqn we get

But,

Hence

Therefore,

Given:

It means the magnitude of the vector is equal to the magnitude of the vector_, but we cannot conclude anything about the direction of the vector.

So it is false that

Given:

It means the magnitude of the vector is equal to the magnitude of the vector_, but we cannot conclude anything about the direction of the vector.

And we know that means magnitude and same direction. So, it is false that

Given: ABCD is a quadrilateral as shown below

Consider ΔADC and apply triangle law of vector, we get

Similarly, consider ΔABC and apply triangle law of vector, we get

Substituting the value of from eqn(i) into eqn(ii), we get

Now add on both sides, we get

Hence sum of the vectors and is

Given: ABCDE is a pentagon as shown below

Consider ΔABC and apply triangle law of vector, we get

Similarly, consider ΔACD and apply triangle law of vector, we get

And, consider ΔADE and apply triangle law of vector, we get

Adding (i), (ii) and (iii), we get

[as

Hence proved

Given: ABCDE is a pentagon as shown below

Consider ΔABC and apply triangle law of vector, we get

Similarly, consider ΔADE and apply triangle law of vector, we get

And, consider ΔADC and apply triangle law of vector, we get

Adding (i), (ii) and (iii), we get

Add on both sides we get,

Or

Hence proved.

Given: a regular octagon

To prove the sum of all vectors drawn from the centre of a regular octagon to its vertices is the zero vector

Proof:

Let O be the centre of a regular octagon, we know that the centre of a regular octagon bisects all the diagonals passing through it as shown in figure below

Thus,

The sum of all vectors drawn from the centre of a regular octagon to its vertices is

Substitute the values from eqn(i) in above eqn, we get

Hence, the sum of all vectors drawn from the centre of a regular octagon to its vertices is a zero vector.

Hence, proved.

Given a quadrilateral ABCD, P is a point outside the quadrilateral and

[given]

Or,

………..(i)[as ]

Consider ΔAPB and apply triangle law of vector, we get

And consider ΔDPC and apply triangle law of vector, we get

Substitute the values from eqn(ii) an eqn(iii) in eqn(i), we get

Therefore, AB is parallel to DC and equal is magnitude.

Hence, ABCD is a parallelogram.

Hence proved

To prove

We know that centre O of the hexagon bisects the diagonals

∴, ,

Consider ΔABO and apply triangle law of vector, we get

And consider ΔACO and apply triangle law of vector, we get

And consider ΔAEO and apply triangle law of vector, we get

And consider ΔAFO and apply triangle law of vector, we get

Now,

Substitute the corresponding values from eqn(i) to eqn(v) in above eqn, we get

[from eqn(i)]

[as and ]

Hence

Therefore the resultant of the five forces and is

Hence proved

Let the position vectors of points P, Q and R be, and respectively.

Given and

(i) R divides PQ internally in the ratio 1:2

Recall the position vector of point P which divides AB, the line joining points A and B with position vectors and respectively, internally in the ratio m : n is

Here, m = 1 and n = 2.

We have and

Thus, the position vector of point R is .

(ii) R divides PQ externally in the ratio 1:2

Recall the position vector of point P which divides AB, the line joining points A and B with position vectors and respectively, externally in the ratio m : n is

Here, m = 1 and n = 2.

We have and

Thus, the position vector of point R is .

Given the position vectors of points A and B are and.

Let the position vectors of points C and D be and.

We have AC = 3AB.

From the above figure, observe AB = AC – BC

⇒ AC = 3 (AC – BC)

⇒ AC = 3AC – 3BC

⇒ 2AC = 3BC

∴ AC : BC = 3 : 2

So, C divides AB externally in the ratio 3:2.

Recall the position vector of point P which divides AB, the line joining points A and B with position vectors and respectively, externally in the ratio m : n is

Here, m = 3 and n = 2

So, the position vector of C is

We also have BD = 2BA.

From the figure, observe BA = BD – AD

⇒ BD = 2 (BD – AD)

⇒ BD = 2BD – 2AD

⇒ BD = 2AD

∴ BD : AD = 2 : 1

So, D divides BA externally in the ratio 2:1.

We now use the same formula as earlier to find the position vector of D.

Here, m = 2 and n = 1

Thus, the position vector of point C is and the position vector of point D is .

Given the position vectors of points A, B, C and D are,, and respectively.

We have

Rearranging the terms in the above equation,

Observe that the sum of coefficients on the LHS of this equation (3 + 5 = 8) is equal to that on the RHS (2 + 6 = 8).

We now divide the equation with 8 on both sides.

Now, consider the LHS of this equation.

Let , the position vector of some point X.

Recall the position vector of point P which divides AB, the line joining points A and B with position vectors and respectively, internally in the ratio m : n is

Here, m = 3 and n = 5

So, X divides CA internally in the ratio 3:5.

Similarly, considering the RHS of this equation, we have the same point X dividing DB in the ratio 2:6.

So, the point X lies on both the line segments AC and BD making it the point of intersection of AC and BD.

As AC and BD are two straight lines having a common point, we have all the points A, B, C and D lying in the same plane.

Thus, the points A, B, C and D are coplanar and in addition, the position vector of the point of intersection of line segments AC and BD is or.

Given the position vectors of points P, Q, R and S are,, and respectively.

We have

Rearranging the terms in the above equation,

Observe that the sum of coefficients on the LHS of this equation (5 + 6 = 11) is equal to that on the RHS (2 + 9 = 11).

We now divide the equation with 11 on both sides.

Now, consider the LHS of this equation.

Let , the position vector of some point X.

Recall the position vector of point P which divides AB, the line joining points A and B with position vectors and respectively, internally in the ratio m : n is

Here, m = 5 and n = 6

So, X divides RP internally in the ratio 5:6.

Similarly, considering the RHS of this equation, we have the same point X dividing SQ in the ratio 2:9.

So, the point X lies on both the line segments PR and QS making it the point of intersection of PR and QS.

As PR and QS are two straight lines having a common point, we have all the points P, Q, R and S lying in the same plane.

Thus, the points P, Q, R and S are coplanar and in addition, the position vector of the point of intersection of line segments PR and QS is or.

Given the position vectors of vertices A, B and C of ΔABC are and respectively.

D is point on BC with position vector such that AD is the bisector of ∠A. I is the incenter of ΔABC.

Observe from the figure that D divides BC in the ratio BD:DC.

Using the angular bisector theorem, we know that the angle bisector of an angle in a triangle bisects the opposite side in the ratio equal to the ratio of the other two sides.

But, and.

Recall the vector is given by

Similarly,

So, we have .

Recall the position vector of point P which divides AB, the line joining points A and B with position vectors and respectively, internally in the ratio m : n is

Here, we have D dividing BC internally in the ratio m:n where m = BD = and n = DC =

Suppose and.

From angular bisector theorem above, we have.

Adding 1 to both sides,

In addition, as CI is the angular bisector of ∠C in ΔACD, using the angular bisector theorem, we have

So, we get

We have and

Assume

So, I divides AD in the ratio (β + γ):α.

Let the position vector of I be .

Using the aforementioned section formula, we can write

But, we already found.

Thus, and the position vector of the incenter is , where, and .

Given the position vectors of points A, B, C and D are,, and respectively.

Recall the vector is given by

Similarly, the vector is given by

But, it is given that.

Two vectors are equal only when both their magnitudes and directions are equal.

and.

This means that the opposite sides in quadrilateral ABCD are parallel and equal.

Thus, ABCD is a parallelogram.

Let position vectors of the vertices A, B and C of ΔABC with respect to O be , and respectively.

⇒

Let us also assume the position vectors of the midpoints D, E and F with respect to O are, and respectively.

⇒

Now, D is the midpoint of side BC.

This means D divides BC in the ratio 1:1.

Recall the position vector of point P which divides AB, the line joining points A and B with position vectors and respectively, internally in the ratio m : n is

Here, m = n = 1

Similarly, for midpoint E and side CA, we get and for midpoint F and side AB, we get .

Adding these three equations, we get

Thus, .

Consider a ΔABC with D, E and F being the midpoints of sides BC, CA and AB respectively.

Let the position vectors of these vertices and midpoints be as shown in the figure.

We need to prove.

As D is the midpoint of BC, using midpoint formula, we have

Similarly, and.

Recall the vector is given by

Similarly, and

Now, consider the vector.

But, and

Thus, the sum of the three vectors determined by the medians of a triangle is zero.

Let position vectors of the vertices A, B, C and D of the parallelogram ABCD with respect to O be , , and respectively.

⇒

Also, let us assume position vector of P is .

Given ABCD is a parallelogram.

We know that the two diagonals of a parallelogram bisect each other. So, P is the midpoint of AC and BD.

As P is the midpoint of AC, using midpoint formula, we have

P is also the midpoint of BC.

So,

Now we have and.

Adding these two equations, we get

Thus.

Let ABCD be a quadrilateral. E, F, G and H are the midpoints of sides AB, BC, CD and DA respectively.

We need to prove EG and HF bisect each other. It is sufficient to show EFGH is a parallelogram, as the diagonals in a parallelogram bisect each other.

Let the position vectors of these vertices and midpoints be as shown in the figure.

As E is the midpoint of AB, using midpoint formula, we have

Similarly, , and .

Recall the vector is given by

Similarly

So, we have .

Two vectors are equal only when both their magnitudes and directions are equal.

and.

This means that the opposite sides in quadrilateral EFGH are parallel and equal, making EFGH a parallelogram.

EG and HF are diagonals of parallelogram EFGH. So, EG and HF bisect each other.

Thus, the line segments joining the midpoints of opposite sides of a quadrilateral bisect each other.

Let E, F, G and H be the midpoints of sides AB, BC, CD and DA respectively of quadrilateral ABCD.

Let the position vectors of these vertices and midpoints be as shown in the figure.

As E is the midpoint of AB, using midpoint formula, we have

Similarly, , and .

We know that the line segments joining the midpoints of opposite sides of a quadrilateral bisect each other.

⇒ Q is the midpoint of EG and HF.

Once again using midpoint formula, we get

But, we found and .

Now, consider the vector .

Let the position vector of point P be .

Recall the vector is given by

Similarly, , and .

But, we found

Observe,

Thus,

Consider ΔABC with vertices A, B, C and sides BC = α, AC = β and AB = γ.

Let the position vectors of A, B and C be and respectively.

Let D and E (with position vectors and) be points on BC and AB such that AD and CE are the bisectors of ∠A and ∠C. Let, AB and CE meet at point I.

Observe from the figure that D divides BC in the ratio BD:DC.

Using the angular bisector theorem, we know that the angle bisector of an angle in a triangle bisects the opposite side in the ratio equal to the ratio of the other two sides.

(from our initial assumption)

Recall the position vector of point P which divides AB, the line joining points A and B with position vectors and respectively, internally in the ratio m : n is

Here, we have D dividing BC internally in the ratio m:n where m = γ and n = β.

From angular bisector theorem above, we had.

Adding 1 to both sides,

In addition, as CI is the angular bisector of ∠C in ΔACD, using the angular bisector theorem, we have

So, we get

So, I divides AD in the ratio (β + γ):α.

Let the position vector of I be .

Using the aforementioned section formula, we can write

But, we already found.

Now, observe E divides AB in the ratio AE:EB.

(from angular bisector theorem)

So, (using section formula)

By doing similar calculations as above for ∠C, we get

So, I divides CE in the ratio (α + β):γ.

Let the position vector of I now be .

Using the aforementioned section formula, we can write

But, we already found.

Observe that meaning the point I with position vector lies on both AB and CE.

Similarly, it can be shown that this point I also lies on the third angular bisector.

Thus, the internal bisectors of the angles of a triangle are concurrent with the point of concurrency given by the position vector where α,β and γ are sides of the ΔABC opposite to the vertices A, B and C respectively.

Given is the position vector of point (–4, –3).

We know position vector of a point (x, y) is given by, where and are unit vectors in X and Y directions.

Now, we need to find magnitude of i.e. .

Recall the magnitude of the vector is given as

Here, x = –4 and y = –3

Thus, .

Given is the position vector of point (12, n).

We know position vector of a point (x, y) is given by, where and are unit vectors in X and Y directions.

Now, we need to find n such that.

Recall the magnitude of the vector is given as

Here, x = 12 and y = n

Squaring both the sides, we have

Thus, n = 5 or –5.

Let be the required vector that is parallel to.

We know any vector parallel to a given vector is of the form, where λ is a real number.

Now, we need to find λ such that.

Recall the magnitude of the vector is given as

Here, x = and y = λ

Squaring both the sides, we have

Thus, the required vector is.

(i) Given A = (4, –1) and B = (1, 3)

We know position vector of a point (x, y) is given by, where and are unit vectors in X and Y directions.

Let position vectors of points A and B be and respectively.

We also have.

Recall the vector is given by

Recall the magnitude of the vector is given as

Here, x = –3 and y = 4

Thus, and.

(ii) Given A = (–6, 3) and B = (–2, –5)

We know position vector of a point (x, y) is given by, where and are unit vectors in X and Y directions.

Let position vectors of points A and B be and respectively.

We also have.

Recall the vector is given by

Recall the magnitude of the vector is given as

Here, x = 4 and y = –8

Thus, and.

Given A = (–1, 3) and B = (–2, 1)

We know position vector of a point (x, y) is given by, where and are unit vectors in X and Y directions.

Let position vectors of points A and B be and respectively.

We also have.

Recall the vector is given by

Now, it is given that there exists a point say (x, y) whose position vector is same as.

We know position vector of a point (x, y) is given by.

By comparing both the sides, we get x = –1 and y = –2

Thus, (–1, –2) is the tip of position vector that is same as .

Given A = (–2, –1), B = (3, 0) and C = (1, –2)

Let the other vertex D = (x, y)

We know position vector of a point (x, y) is given by, where and are unit vectors in X and Y directions.

Let position vectors of points A, B, C and D be,, and respectively.

We also have.

Similarly and.

Recall the vector is given by

Similarly, the vector is given by

But, it is given that ABCD is a parallelogram.

(as the opposite sides are parallel and equal)

By comparing both sides, we get 1 – x = 5 and 2 + y = –1

⇒ x = 1 – 5 = –4

and y = –1 – 2 = –3

So, x = –4 and y = –3

Thus, vertex D of parallelogram ABCD = (–4, –3).

Given A = (3, 4), B = (5, –6) and C = (4, –1)

We know position vector of a point (x, y) is given by, where and are unit vectors in X and Y directions.

(position vector of point A)

We also have

Similarly.

We need to compute.

Thus, .

We know position vector of a point (x, y) is given by, where and are unit vectors in X and Y directions.

So, position vector of (5, –3) is

Given A = (4, –1) and let the coordinates of B = (x, y)

Let position vectors of points A and B be and respectively.

We also have.

Recall the vector is given by

But, it is given that

By comparing both sides, we get x – 4 = 5 and y + 1 = –3

⇒ x = 5 + 4 = 9

and y = –3 – 1 = –4

So, x = 9 and y = –4

Thus, coordinates of point B are (9, –4).

Let, and be the position vectors corresponding to the vertices A, B and C of ΔABC.

Recall the vector is given by

Recall the magnitude of the vector is given as

Now, we find the magnitude of .

Similarly, the vector is given by

Now, we find the magnitude of .

Similarly, the vector is given by

Now, we find the magnitude of .

Observe that which means the sides AB and AC of ΔABC are equal in length, making it an isosceles triangle.

Thus, the triangle formed by the given points is isosceles.

Let be the required vector that is parallel to.

We know any vector parallel to a given vector is of the form, where λ is a real number.

Now, we need to find λ such that.

Recall the magnitude of the vector is given as

Here, x = λ and y =

Squaring both the sides, we have

Thus, the required vector is.

Let the position vectors of points A, B and C be, and respectively.

Given:, and

C divides AB internally in the ratio 3:1.

Recall the position vector of point P which divides AB, the line joining points A and B with position vectors and respectively, internally in the ratio m : n is

Here, m = 3 and n = 1.

We have, and

By comparing both sides, we get 36 + λ = –44

⇒ λ = –44 – 36

∴ λ = –80

We also have 3μ + 3 = –12

⇒ 3μ = –15

∴ μ = –5

Thus, λ = –80 and μ = –5

(i) Given P = (3, 2)

We know position vector of a point (x, y) is given by, where and are unit vectors in X and Y directions.

Let position vector of point P be.

So, component of along the X-axis is , that is a vector of magnitude 3 along the positive direction of the X-axis.

Also, component of along the Y-axis is , that is a vector of magnitude 2 along the positive direction of the Y-axis.

(ii) Given Q = (5, 1)

We know position vector of a point (x, y) is given by, where and are unit vectors in X and Y directions.

Let position vector of point Q be.

So, component of along the X-axis is 5, that is a vector of magnitude 5 along the positive direction of the X-axis.

Also, component of along the Y-axis is , that is a vector of magnitude 1 along the positive direction of the Y-axis.

(iii) Given R = (–11, –9)

We know position vector of a point (x, y) is given by, where and are unit vectors in X and Y directions.

Let position vector of point R be.

So, component of along the X-axis is , that is a vector of magnitude 11 along the negative direction of the X-axis.

Also, component of along the Y-axis is , that is a vector of magnitude 9 along the negative direction of the Y-axis.

(iv) Given S = (4, –3)

We know position vector of a point (x, y) is given by, where and are unit vectors in X and Y directions.

Let position vector of point S be.

So, component of along the X-axis is , that is a vector of magnitude 4 along the positive direction of the X-axis.

Also, component of along the Y-axis is , that is a vector of magnitude 3 along the negative direction of the Y-axis.

If a vector is given by then the magnitude of vector is generally denoted by which is equal to

So the magnitude

So the magnitude of the vector is 7.

Let the unit vector in the direction of

So any unit vector in the direction of

So the magnitude of the vector

So, the unit vector

To find the resultant vector we add all the vector by vector addition.

So, resultant vector is

So, the unit vector

Magnitude of

Side BC parallel to

So resultant vector c = b + a

So, vector

So unit vector along the diagonal of

Parallelogram is

We want to find the magnitude of vector

So,

If a vector is given by then the magnitude of vector is generally denoted by which is equal to

Position vector of ‘P’ is

Let the position vector of point ‘Q’ is ‘a.’

So we need to find the value of ‘a.’

Position vector of ‘Q’ - Position vector of ‘P’

So the position vector of ‘Q” is (4, 1, 1)

In a right angle triangle

Where CA is the hypotenuse

BC is the perpendicular and AB is the base

Vertices of the triangle are given below

A = (1, -1, 0), B = (4, -3, 1), C = (2, -4, 5)

So,

………..(1)

Similarly,

…………(2)

…………..(3)

35 = 14 + 21

35 = 35

LHS = RHS

So, these point form a right angle triangle

Let the position vector of the vertex ‘A’ is ,

And similarly B and C

Side AB is

………..(1)

Equation (1) vector representation of the side AB

Magnitude of side AB,

And similarly for side BC and CA

……….(2)

Length of side BC and CA

Centeroid of the triangle with Vertices (x₁, y₁, z₁), (x₂, y₂, z₂), and (x₃, y₃, z₃) is given by,

In vector algebra, ‘x’ consider as a coefficient of and ‘y’ as a coefficient of and ‘z’ as a coefficient of

So the position vector of the centroid,

So the location of the centroid is

And the vector is,

By using section formula,

(1) Internally

Position vectors of P and Q are given as

and

The position vector of point R which divides the line joining two points P and Q internally in the ratio 2 : 1 is given by,

The position vector of point R which divides the line joining two points P and Q externally in the ratio 2 : 1 is given by,

(2)Externally

If P and Q are two points with position vector and then the position vector of mid point A is given by

Let A is the mid point of PQ.

So, position vector of A

First we need to create vector PQ

Position vector of P = OP = (1, 2, 3) and position vector of Q = OQ = (4, 5, 6)

So unit vector in the direction PQ,

If A, B and C are the vertices of the right angle triangle

So,

In a right angle triangle

Where AB is the hypotenuse

BC is the perpendicular and CA is the base

Vertices of the triangle are given bellow

, and

So,

……(1)

Similarly,

……(2)

……(3)

41 = 35 + 6

41 = 41

LHS = RHS

If P and Q are two points with position vector and then the position vector of mid point A is given by

Let A is the mid point of PQ.

So, position vector of A

We need to find the value of ‘x’ for which is a unit vector

If any vector is a unit vector, then its magnitude should be one.

So, the magnitude of the vector is,

For this value of ‘x’ the above vector is a unit vector

First, we need to create a vector in the direction of

So,

So the unit vector in the direction of is,

This is the unit vector in the direction of

Vector parallel to is,

If a vector parallel to other vector, so we can write a scalar multiple of the other so,

=

So,

=

this is given in the question

So the vector parallel to the is =

Let resultant vector is ‘R’ so the resultant vector by using the vector triangle law

If a vector parallel to other vector so we can write scalar multiple of the other so,

=

= has a magnitude of 5 unit so

So the vector is =

Let D be the point at which median drawn from A touches side BC.

Let be the position vectors of the vertices A, B and C.

So position vector of D =

So we creating a vector in the direction of AD

Position vector of D – position vector of A

So length of AD

Let us understand that, two more points are said to be collinear if they all lie on a single straight line.

We have been given that,

So, in this case if we prove that and are parallel to each other, then we can easily show that A, B and C are collinear.

Therefore, is given by

And is given by

Let us note the relation between and .

We know,

Or

Or [∵, ]

This relation shows that and are parallel to each other.

But also, is the common vector in and .

⇒ and are not parallel but lies on a straight line.

Thus, A, B and C are collinear.

Let us understand that, two more points are said to be collinear if they all lie on a single straight line.

Given that, , and are non-coplanar vectors.

And we know that, vectors that do not lie on the same plane or line are called non-coplanar vectors.

To Prove: , and are collinear.

Proof: Let the points be A, B and C.

Then,

So, in this case if we prove that and are parallel to each other, then we can easily show that A, B and C are collinear.

Therefore, is given by

And is given by

Let us note the relation between and .

We know,

Or

Or

Or [∵, ]

This relation shows that and are parallel to each other.

But also, is the common vector in and .

⇒ and are not parallel but lies on a straight line.

Thus, A, B and C are collinear.

Let us understand that, two more points are said to be collinear if they all lie on a single straight line.

Given that, , and are non-coplanar vectors.

And we know that, vectors that do not lie on the same plane or line are called non-coplanar vectors.

To Prove: , and are collinear.

Proof: Let the points be A, B and C.

Then,

So, in this case if we prove that and are parallel to each other, then we can easily show that A, B and C are collinear.

Therefore, is given by

And is given by

Let us note the relation between and .

We know,

Or

Or [∵, ]

This relation shows that and are parallel to each other.

But also, is the common vector in and .

⇒ and are not parallel but lies on a straight line.

Thus, A, B and C are collinear.

Let us understand that, two more points are said to be collinear if they all lie on a single straight line.

Let the points be A, B and C having position vectors such that,

So, in this case if we prove that and are parallel to each other, then we can easily show that A, B and C are collinear.

Therefore, is given by

And is given by

Let us note the relation between and .

We know,

Or

Or [∵, ]

This relation shows that and are parallel to each other.

But also, is the common vector in and .

⇒ and are not parallel but lies on a straight line.

Thus, A, B and C are collinear.

Let us understand that, two more points are said to be collinear if they all lie on a single straight line.

Let the points be A, B and C having position vectors such that,

So, let us find and .

Therefore, is given by

…(i)

And is given by

…(ii)

Since, it has been given that points A, B and C are collinear.

So, we can write as

Where λ = a scalar quantity

Put the values of and from (i) and (ii), we get

Comparing the vectors and respectively, we get

a – 12 = 2λ …(iii)

and, 16 = –8λ

From –8λ = 16, we can find the value of λ.

–8λ = 16

⇒ λ = –2

Put λ = –2 in equation (iii), we get

a – 12 = 2λ

⇒ a – 12 = 2(–2)

⇒ a – 12 = –4

⇒ a = –4 + 12

⇒ a = 8

Thus, we have got a = 8.

Let us understand that, two more points are said to be collinear if they all lie on a single straight line.

Given that, and are two non-collinear vectors.

Let the points be A, B and C having position vectors such that,

So, in this case if we prove that and are parallel to each other, then we can easily show that A, B and C are collinear.

Therefore, is given by

…(i)

And is given by

Let us note the relation between and .

We know,

Or

Or [∵, from (i)]

Or …(ii)

If λ is any real value, then is also a real value.

Then, for any real value , we can write

From (ii) equation, we can write

This relation shows that and are parallel to each other.

But also, is the common vector in and .

⇒ and are not parallel but lies on a straight line.

Thus, A, B and C are collinear.

Let us understand that, two more points are said to be collinear if they all lie on a single straight line.

Given:

To Prove: A, B and C are collinear points.

Proof: We have been given that,

Rearrange it so that we get a relationship between and .

…(i)

Now, we know that

But actually we are doing , such that O is the point of origin so that the difference between the two vectors is a displacement.

So, …(ii)

Similarly,
…(iii)

Substituting equation (ii) & (iii) in equation (i), we get

Thus, this relation shows that and are parallel to each other.

But also, is the common vector in and .

⇒ and are not parallel but lies on a straight line.

Hence, A, B and C are collinear.

Let us understand that, two more points are said to be collinear if they all lie on a single straight line.

We have been given position vectors and .

Let

Also, let O be the initial point having position vector as

Now, let us find and .

is given by

is given by

We have as

[∵, ]

Thus, this relation shows that and are parallel to each other.

But also, is the common vector in and .

⇒ and are not parallel but lies on a straight line.

⇒ A and B are collinear.

Hence, and are collinear.

Let us understand that, two more points are said to be collinear if they all lie on a single straight line.

We have been given points:

A (m, –1), B (2, 1) and C (4, 5).

These points are collinear.

Let us define the position vectors as,

Now, we need to find the vectors and .

is given by

And is given by

Since, A, B, C and D are collinear. We can draw a relation between and .

Putting the values of and , we get

Comparing L.H.S and R.H.S, we get

2 – m = 2λ

And 2 = 4λ

We need to find the value of λ in order to find m.

We have

2 = 4λ

Putting the value of λ in equation (2 – m) = 2λ

⇒ 2 – m = 1

⇒ m = 2 – 1

⇒ m = 1

Thus, the value of m = 1.

Let us understand that, two more points are said to be collinear if they all lie on a single straight line.

Let the points be A (3, 4), B (–5, 16) and C (5, 1).

Let

So, in this case if we prove that and are parallel to each other, then we can easily show that A, B and C are collinear.

Therefore, is given by

And is given by

Let us note the relation between and .

We know,

Or …(i)

And we know,

Or

Or …(ii)

Substituting the value of in equation (i), we get

This relation shows that and are parallel to each other.

But also, is the common vector in and .

⇒ and are not parallel but lies on a straight line.

Thus, A, B and C are collinear.

Let us understand that, two more points are said to be collinear if they all lie on a single straight line.

We have the position vectors as,

Since, a and b are collinear. We can draw a relation between and .

Putting the values of and , we get

Comparing L.H.S and R.H.S, we get

2 = –6λ

And –3 = mλ

We need to find the value of λ in order to find m.

We have

2 = –6λ

Putting the value of λ in equation –3 = mλ

⇒ m = 3 × 3

⇒ m = 9

Thus, the value of m = 9.

We have been given the points A (1, –2, –8), B (5, 0, –2) and C (11, 3, 7).

We need to show that A, B and C are collinear.

Let us define the position vector.

So, in this case if we find a relation between , and , then we can easily show that A, B and C are collinear.

Therefore, is given by

And is given by

And is given by

Let us add and , we get

Thus, clearly A, B and C are collinear.

We need to find the ratio in which B divides AC.

Let the ratio at which B divides AC be λ : 1. Then, position vector of B is:

But the position vector of B is .

So, by comparing the position vectors of B, we can write

Solving these equations separately, we get

⇒ 11λ + 1 = 5(λ + 1)

⇒ 11λ + 1 = 5λ + 5

⇒ 11λ – 5λ = 5 – 1

⇒ 6λ = 4

The ratio at which B divides AC is λ : 1.

Since,

We can say

Solving it further, multiply the ratio by 3.