Vector Algebra

North (N), South (S), East (E), and West (W) are plotted on the paper as shown.

Let be the displacement vector such that = 40 km

Vector makes an angle of 30° east of North i.e. 30° with North in North-East direction.

Scalars are defined as quantities that have magnitude only.

Vectors are defined as a quantity that has magnitude as well as direction.

(i) 10 kg: It is a measure of mass. It is a scalar quantity as it has magnitude only and no direction.

(ii) 2 meters north-west: It is a measure of distance in a particular direction. ∴ It is a vector quantity as it has magnitude as well as direction.

(iii) 40°: It is a measure of angle. It is a scalar quantity as it has magnitude only and no direction.

(iv) 40 Watt: It is a measure of power. It is a scalar quantity as it has magnitude only and no direction.

(v) 10^-19 coulomb: It is a measure of electric charge. It is a scalar quantity as it has magnitude only and no direction.

(vi) 20 m/sec²: It is a measure of acceleration. It is a vector quantity as it is a measure of rate of change of velocity.

Scalars are defined as quantities that have magnitude only.

Vectors are defined as a quantity that has magnitude as well as direction.

(i) time period: It is a scalar quantity as it has magnitude only and no direction.

(ii) distance: It is a scalar quantity as it has magnitude only and no direction.

(iii) force: It is a vector quantity as it has magnitude as well as direction.

(iv) velocity: It is a vector quantity as it has magnitude as well as direction.

(v) work done: It is a scalar quantity as it has magnitude as well as direction.

(i) Coinitial vectors: Two or more vectors having same initial point are called coinitial vectors.

So, in the above figure and are coinitial vectors as they have same initial point.

∴ Coinitial vectors: and

(ii) Equal vectors: Two or more vectors having same direction and same magnitude are called equal vectors.

So, in the above figure and are equal vectors as they have same magnitude and same direction.

∴ Equal vectors: and

(iii) Collinear but not equal:

∵ and are parallel vectors, so, they are collinear. But they have opposite direction, so, they are not equal.

Hence, and are collinear but not equal.

(i) and are collinear: True

Explanation: and are parallel vectors, so, they are collinear.

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

Explanation: we know, and are unit vectors, so they have same magnitude but and are vectors along x – axis and y – axis respectively.

(iii) Two vectors having same magnitude are collinear: False

Explanation: Two or more vectors are said to be collinear if they are parallel to the same line, irrespective of their magnitudes and directions.

We know, and are unit vectors, so they have same magnitude but and are vectors along x – axis and y – axis respectively. Since, they are not parallel to each other, so, they are not collinear.

(iv) Two collinear vectors having the same magnitude are equal: False

Explanation: Two vector are said to be equal, if they have the same magnitude and direction.

and are collinear vectors as they are parallel to each other and also they have same magnitude. But they do not have same direction, so, they are unequal vectors.

Let

And

We can see clearly because the all the coefficients of and are not same. In and , the coefficients are different. Now, we check the magnitude of both,

And

So, magnitude of both vectors is same but they are different.

Let

And

So, here where m = 2 > 0

Therefore, both vectors have same direction.

Now, we check for the magnitude,

So, they have same direction but same magnitude.

For two vectors to be equal, the coefficients of both vectors should be equal.

Comparing the -coefficient, we get x=2

Comparing the -coefficient, we get y=3

Let A be the initial point (2,1) and B be the final point (-5,7).

So, and

Now we want to find the

Therefore, the scalar components of are -7 and 6. The vector components of are .

And

We want to find

To find the sum, we add coefficients of to each other.

So,

We know that unit vector means that the magnitude of the vector is 1(unit).

It is defined as

So, we find the magnitude of first.

So, and

Now we want to find the

Now, we have to find the unit direction in the direction of .We know that unit vector means that the magnitude of the vector is 1(unit).

It is defined as

So, we find the magnitude of first.

We want to find

To find the sum, we add coefficients of to each other.

So,

Now, we have to find the unit direction in the direction of.We know that unit vector means that the magnitude of the vector is 1(unit).

So, we find the magnitude of first.

Let

The vector in the direction of having unit magnitude is .

So, The vector in the direction of having magnitude 8 units =.

We know that two vectors are collinear if they have the same direction or are parallel or anti-parallel. They can be expressed in the form where a and b are vectors and 'm' is a scalar quantity.

From the question,

Let and

Here,

So, where, m=-2.

∴ The given two vectors are collinear.

The direction cosines of a vector are defined as the coefficients of in the unit vector in the direction of the vector.

So, first we find the unit vector in the direction of the vector.

Let

Therefore, The direction cosines of the given vector are .

So, and

Now we want to find the

Now, we have to find the direction cosines which are the coefficients of the unit vector in the direction of .We know that unit vector means that the magnitude of the vector is 1(unit).

It is defined as

So, we find the magnitude of first.

Therefore, The direction cosines of the given vector are .

To find the inclination of the vector with OX, OY, OZ. We find the direction cosines of the vector.

We know that the direction cosines of a vector are defined as the coefficients of in the unit vector in the direction of the vector.

So, first we find the unit vector in the direction of the vector.

Let

Therefore, The direction cosines of the given vector are .

Let be the angle between and OX.

Therefore,

Similarly, Let be the angle between and OY.

Therefore,

And, be the angle between and OZ.

Therefore,

Therefore,

Hence proved that the vector is equally inclined with the axes OX, OY, and OZ.

Given that and

(i) The lies on the segment PQ(internal division).If m:n is the ratio in which divides PQ, then

Given m:n=2:1, m=2 and n=1

(ii) The does not lie on the segment PQ(external division).If m:n is the ratio in which divides PQ,then

Given m:n=2:1, m=2 and n=1

So, and

Let be the mid point of .

O be the origin,
Let

And

Now we find the vectors

……….(1)

………….(2)

……….(3)

Adding equations (1), (2) and (3)

Therefore, ABC form a triangle.

Now, we want to prove that it is a right angled triangle.

Therefore, the triangle satisfies the Pythagoras theorem. Hence, proved the given vectors form a right angled triangle.

We know by triangle law of vectors,

…………..(1)

Therefore,

Hence (B) is true.

We know that

Putting in eq(1), we get

…….(2)

Hence (A) is also true.

Now,

Putting in eq(2)

Hence (D) is correct.

We are asked the option which is not true.

Therefore, the correct option is (C).

We know that two vectors are collinear if they have the same direction or are parallel or anti-parallel. They can be expressed in the form where a and b are vectors and ' m ' is a scalar quantity.

Therefore, (a) is true.

In (b),

So, (b) is also true.

The vectors and are proportional,
Therefore, (c) is not true.

The vectors and can have different magnitude as well as different direction.

Therefore, (d) is not true.

We are asked the options which are not true.

∴ The correct answer is (c) and (d).

Given,

We know,

Putting the value of

∴

So, the angle between the vectors is π/4

Given vectors are

Let

=

∴ We know,

⇒

⇒

⇒ cos θ = 10/14

⇒ θ = cos^-1(5/7)

Let and

Projection of is given by

Now

∴ Projection of

= 0

∴ Projection of is 0.

Let

Projection of is given by

Now

∴ Projection of

∴ Projection of .

Let

Now we need to find out the magnitude of

Since,

∴ the three given vectors are unit vectors.

To show that each of three vectors are mutually perpendicular to each other

We have to show

Since,

∴ are mutually perpendicular to each other.

Given,

Now,

∴

⇒

∴

To find

Let θ be the angle between

Given,

and θ = 60°

Scalar product of

i.e

We know

∴ [magnitude of vector is positive]

So,

Hence,

Given, vector is a unit vector

∴

∴

Given,

It is given is perpendicular to , then

∴ λ = 8

To show is perpendicular to for any two non zero vectors , we need to show dot product of is zero.

= 0

Since, dot product of is zero.

∴ is perpendicular to

Given,

⇒

⇒

∴ is a zero vector.

Hence, can be any vector.

Given are unit vectors and

Now

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

∴

If either = 0

Now let

So,

∴

∴

Since

Hence, the converse of given statement need not be true.

Given points are A (1, 2, 3), B (-1, 0, 0) and C (0, 1, 2)

∠ABC is the angle between vectors

We know

⇒ ∠ABC = cos^-1(10/√102)

Given points are A (1, 2, 7), B (2, 6, 3) and C (3, 10, -1)

So,

Since,

∴ The given points A, B and C are collinear.

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

Then,

Now vectors represent sides of ΔABC

Now

And

∴

From Pythagoras theorem we know

If then ΔABC is a right-angled triangle

∴ Δ ABC is a right-angled triangle.

Given λ is a unit vector.

And

Since

∴

λ is a unit vector if a = 1/|λ|

Given that
and

Expanding along first row,

Given that
and

Let

Let

Now, we want to find a vector which is perpendicular to both and . It is given by .

Let

Expanding along first row,

Therefore,

Now we want to find the unit vector. We know that unit vector means that the magnitude of the vector is 1(unit).

It is defined as

So, we find the magnitude of first.

Let

Given that it is a unit vector,

So, ……(1)

Let be the angle with respectively.

Then,

(given)

(given)

Putting value of x,y, z in equation (1)

(Squaring both sides)

As should be acute,

Components of are the coefficients of , which are,

We solve for the left-hand side,

L.H.S.=

We know and

Also,

Putting these in our equation, we get

which is equal to R.H.S.

Hence proved.

Let
and

Given that

Expanding along first row,

By comparing the y-coefficients,

By comparing the z-coefficients,

These values should also satisfy the equation we will get from comparing the x-coefficients,

Therefore, .

Given that

(where θ is the angle between the vectors)

Also given that,

(where θ is the angle between the vectors)

As there is no value of θ for which both sin θ and cos θ are zero.

Therefore, the condition for which and is:

.

Given that

and

Solving for left hand side,

First, we calculate

(Property of determinant)

Which is equal to R.H.S.

Hence proved.

(where θ is the angle between the vectors)

If

Similarly, If

Now, If

(where θ is the angle between the vectors)

sin θ = 0, this implies θ = 0°

This implies that the converse is not always true. The vectors may not be zero but the angle between is 0°, i.e. the vectors are parallel.

So,

Now we want to find the

Now we want to find the

We know that

So, we find first.

Expanding along first row,

Given
and

We know that

So, we find ,

Expanding along first row,

Given that:

(Magnitude of unit vector is 1)

Putting the values in the equation (1), we get

(where θ is the angle between the vectors)

Therefore, The correct option is (B).

Let

Now we want to find the

Now we want to find the

We know that

So, we find ,

Expanding along first row,

Given: A unit vector in XY- plane.

Let is a unit vector in the given XY- plane then the value of

Will be :

θ is the angle given, which is made by unit vector with positive direction of x-axis.

∴ for θ = 30°

Hence, the required unit vector is .

Given: points P(x₁, y₁, z₁) and Q(x₂, y₂, z₂) are given.

The vector obtained by joining the given points P and Q:

= position vector of Q – position vector of P

Hence the scalar component of the vector obtained by joining the points are

[(x₂-x₁), (y₂-y₁), (z₂-z₁)]

And the magnitude of the vector obtained by joining the points is

let O be the initial position and B be the final position of the girl.

Position of girl will be as shown in figure:

Hence, and ∠ BAO = 60°

Now, by using triangle law of vector addition,

Hence girls displacement from initial to final position is :

Let in given triangle

Now, by using triangle law of vector addition,

As we can see that represent the sides of triangle.

Also we know that sum of two sides of a triangle must be greater than its third side.

is not true.

Given: as a unit vector.

Now, if is a unit vector then

Given:

Let the resultant of

Then

Then

Now the vector of magnitude 5 units and parallel to is:

Given:

Let the resultant of

Then

Then

∴ unit vector along

Given: A(1, – 2, – 8), B(5, 0, –2) and C(11, 3, 7)

Then

Then

Thus the given points are collinear.

Now to find the ratio in which B divides AC. Let it be λ :1

On equating the terms, we get:

5(λ + 1) = 11λ + 1

⇒ 5λ + 5 = 11λ + 1

⇒ 4 = 6λ

⇒ λ = 4/6 = 2/3

Hence, B divides AC in the ratio 2:3.

Given: points are given.

Point R is given which divides P and Q in the ratio 1:2.

Then

∴ position vector of R is

And position vector of mid-point of RQ =

Hence, P is mid-point of the line segment RQ.

Given: Two adjacent sides of a parallelogram are

Then the diagonal of parallelogram is given by the resultant of .

Let the diagonal is .

Then

Then

∴ unit vector parallel to its diagonal is

And area of parallelogram ABCD is

∴

Hence, area of parallelogram ABCD is .

let the vector equally inclined to OX, OY and OZ at angle α.

Then the direction cosine of the vectors are cos α, cos α and cos α.

Because,

Cos²α + Cos²α + Cos²α = 1

⇒ 3 Cos²α = 1

⇒ Cos²α = 1/3

Hence, the direction cosines of the vector which are equally inclined to the axis are

.

Given:

Let

Because is perpendicular to both and .

Then

d₁ + 4d₂ + 2d₃ = 0 .......(1)

And

3d₁ - 2d₂ + 7d₃ = 0 ......(2)

And (given)

2d₁ - d₂ + 4d₃ = 15.......(3)

Hence

Hence, the required vector is

Given: given vectors are

Then sum of vector is given by the resultant of .

Let the sum is .

Then

Then magnitude of is :

Scalar product of

Square on both sides:

⇒ (λ + 6)² = (λ)² + 4λ + 44

⇒ (λ)² + 12λ + 36 =(λ)² + 4λ + 44

⇒ 8λ = 8

⇒ λ = 1

Given: vectors are mutually perpendicular to each other and are of equal magnitude.

And

Let the vector be inclined to at angles α, β and γ respectively.

Then, we have

From (1), (2) and (3)

As

Hence,

Hence, the vector are equal inclined to .

given: (

To prove: vectors are mutually perpendicular to each other.

Hence, are mutually perpendicular to each other.as is given.

let θ is the angle between two vectors .

Then are non-zero vectors so that are positive.

As we know

For

As are positive.

Hence,

The correct answer is (B).

let the two unit vectors are and is the angle between.

Then

Then this is () is unit vector if

scalar product is cumulative.

Hence, () is unit vector if .

The correct answer is (D).

given:

=1-1 + 1

=1

The correct answer is (C).

let θ is the angle between two vectors .

Then are non zero vectors so that are positive.

Thus when .

The correct answer is (B).