Vector Algebra

Given is the vector .

Let this vector be , such that

Let us first find the unit vector in the direction of this vector .

We know that, a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.

To find a unit vector with the same direction as a given vector, we divide by the magnitude of the vector.

Unit vector in the direction of the vector is given as,

As, we have .

Then,

[∵ if

]

Therefore,

[∵ and ]

We have found unit vector in the direction of the vector , but we need to find the unit vector in the direction of but also with the magnitude 9.

We have the formula:

Vector in the direction ofwith a magnitude of 9

And as just found.

So,

⇒Vector in the direction ofwith a magnitude of 9=

Thus, vector in the direction of vector and has magnitude 9 is .

We have,

Since, unit vector is needed to be found in the direction of the sum of vectors and .

So, add vectors and .

Let,

Substituting the values of vectors and .

We know that, a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.

To find a unit vector with the same direction as a given vector, we divide by the magnitude of the vector.

For finding unit vector, we have the formula:

Substitute the value of .

Here, .

Thus, unit vector in the direction of sum of vectors and is .

We have,

(i). We need to find the unit vector in the direction of .

First, let us calculate .

As we have,

Multiply it by 6 on both sides.

We can easily multiply vector by a scalar by multiplying similar components, that is, vector’s magnitude by the scalar’s magnitude.

We know that, a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.

To find a unit vector with the same direction as a given vector, we divide by the magnitude of the vector.

For finding unit vector, we have the formula:

Now we know the value of , so just substitute the value in the above equation.

Here, .

Let us simplify.

Thus, unit vector in the direction of is .

We need to find the unit vector in the direction of .

First, let us calculate .

As we have,

…(a)

…(b)

Then multiply equation (a) by 2 on both sides,

We can easily multiply vector by a scalar by multiplying similar components, that is, vector’s magnitude by the scalar’s magnitude.

…(c)

Subtract (b) from (c). We get,

We know that, a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.

To find a unit vector with the same direction as a given vector, we divide by the magnitude of the vector.

For finding unit vector, we have the formula:

Now we know the value of , so we just need to substitute in the above equation.

Here, .

Thus, unit vector in the direction of is .

We have,

Coordinates of P is (5, 0, 8).

Coordinates of Q is (3, 3, 2).

So,

Position vector of P is given by,

Position vector of Q is given by,

To find unit vector in the direction of PQ, we need to find position vector of PQ.

Position vector of PQ is given by,

We know that, a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.

To find a unit vector with the same direction as a given vector, we divide by the magnitude of the vector.

For finding unit vector, we have the formula:

Here, .

Thus, unit vector in the direction of PQ is .

We have been given that,

Position vector of A

Position vector of B

Here, O is the origin.

We need to find the position vector of C, that is, .

Also, we have

…(i)

Here,

Position vector of C-Position vector of B

…(ii)

And,

Position vector of A-Position vector of B

…(iii)

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

[∵ it is given that ]

Thus, position vector of point C is .

Let the points be A (k, -10, 3), B (1, -1, 3) and C (3, 5, 3).

Let us find the position vectors of these points.

Assume that O is the origin.

Position vector of A is given by,

Position vector of B is given by,

Position vector of C is given by,

Know that, two vectors are said to be collinear, if they lie on the same line or parallel lines.

Since, A (k, -10, 3), B (1, -1, 3) and C (3, 5, 3) are collinear, we can say that:

Sum of modulus of any two vectors will be equal to the modulus of third vector.

This means, we need to find .

To find :

Position vector of B-Position vector of A

Now,

…(i)

To find :

Position vector of C-Position vector of B

Now,

…(ii)

To find :

Position vector of C-Position vector of A

Now,

…(iii)

Take,

Substitute values of from (i), (ii) and (iii) respectively. We get,

Or

[∵ by algebraic identity, (a – b)² = a² + b² – 2ab]

Squaring on both sides,

[∵ by algebraic identity, (a – b)² = a² + b² – 2ab]

Again, squaring on both sides, we get

⇒ (48)² + k² – 2(48)(k) = (k² – 6k + 234)(10) [∵ by algebraic identity, (a – b)² = a² + b² – 2ab]

⇒ 2304 + k² – 96k = 10k² – 60k + 2340

⇒ 10k² – k² – 60k + 96k + 2340 – 2304 = 0

⇒ 9k² + 36k + 36 = 0

⇒ 9 (k² + 4k + 4) = 0

⇒ k² + 4k + 4 = 0

⇒ k² + 2k + 2k + 4 = 0

⇒ k (k + 2) + 2 (k + 2) = 0

⇒ (k + 2)(k + 2) = 0

⇒ k = -2 or k = -2

Thus, value of k is -2.

Given that,

Magnitude of = 2√3

Also, given that

Vector is equally inclined to the three axes.

This means, direction cosines of the unit vector will be same. The direction cosines are (l, m, n).

⇒ l = m = n

The direction cosines of a vector are simply the cosines of the angles between the vector and the three coordinate axes.

We know the relationship between direction cosines is,

l² + m² + n² = 1

⇒ l² + l² + l² = 1 [∵ l = m = n]

⇒ 3.l² = 1

Also, we know that is represented in terms of direction cosines as,

We are familiar with the formula,

To find ,

Substituting values of and .

Thus, the value of is .

Given that,

Magnitude of vector = 14

Also, direction ratios = 2 : 3 : -6

Also can be defined as,

Know that, the direction cosines of a vector are the cosines of the angles between the vector and the three coordinate axes.

∴, the direction cosines l, m and n are

[∵ ]

[∵ ]

[∵ ]

And we know that,

l² + m² + n² = 1

⇒ 49k² = 196

⇒ k² = 4

⇒ k = ±√4

⇒ k = ±2

Since, makes an acute angle with x-axis, then k will be positive.

⇒ k = 2

The direction cosines are

The components of can be found out by,

Thus, the direction cosines (l, m, n) are ; and the components of are .

Let the vectors be and , such that

We need to find a vector perpendicular to both the vectors and .

Any vector perpendicular to both and can be given as,

Let

As we know, a unit vector in a normed vector space is a vector (often a spatial vector) of length 1.

So,

A vector of magnitude 6 in the direction of is given by,

Here, .

Thus, required vector is .

Let the vectors be and , such that

We know that angle between two vectors, here, and is given by,

Here,

Since, and .

So,

Thus, required angle is .

Given that,

Find the value of .

Take .

[∵ ]

…(i)

[∵ by anti-commutative law, ]

Now, take .

[∵ ]

…(ii)

[∵ by anti-commutative law, ]

From equations (i) and (ii), we have

Now, let us interpret the result graphically.

Let there be a parallelogram ABCD.

Here, and .

And AB and AD sides are making angle θ between them.

Area of parallelogram is given by,

Area of parallelogram = Base × Height

So from the diagram, area of parallelogram ABCD can be written as,

Or,

Since, parallelogram on the same base and between the same parallels are equal in area, so we have

This also implies that,

Thus, it is represented graphically.

Given are points, A, B, C and D.

Let O be the origin.

We have,

Position vector of A

Position vector of B

Position vector of C

Position vector of D

Now, let us find out and .

Position vector of B-Position vector of A

And,

Position vector of D-Position vector of C

The projection of along is given by,

[∵ we know that, and .

So,

Also, we know that, ]

Multiply numerator and denominator by √21.

⇒ Projection = √21

Thus, projection of along is √21 units.

We have,

The coordinates of points A, B and C are (1, 2, 3), (2, -1, 4) and (4, 5, -1) respectively.

We need to find the area of this triangle ABC.

We have the formula given as,

…(i)

Let us find out and first.

We can say,

Position vector of A

Position vector of B

Position vector of C

For :

Position vector of B-Position vector of A

For :

Position vector of C-Position vector of A

Now, substitute values of and in . We can find out as,

And,

From equation (i), we get

Thus, area of ∆ABC is .

We have,

Given:

There are more than 1 parallelogram, and their bases can be taken as common and they are between same parallels.

To Prove:

These parallelograms whose bases are same and are between the same parallel sides have equal area.

Proof:

Let ABCD and ABFE be two parallelograms on the same base AB and between same parallel lines AB and DF.

Here,

AB ∥ DC and AE ∥ BF

We can represent area of parallelogram ABCD as,

…(i)

Now, area of parallelogram ABFE can be represented as,

Area of parallelogram ABFE

[∵ in right-angled ∆ADE, ]

⇒Area of parallelogram ABFE

[∵ , where k is scalar; is parallel to and hence ]

[∵ a scalar term can be taken out of a vector product]

[∵ ]

⇒Area of parallelogram ABFE …(ii)

From equation (i) and (ii), we can conclude that

Area of parallelogram ABCD = Area of parallelogram ABFE

Thus, parallelogram on same base and between same parallels are equal in area.

Hence, proved.

Given:

a, b, c are magnitudes of the sides opposite to the vertices A, B, C respectively.

⇒ AB = c, BC = a and CA = b

To Prove:

In triangle ABC,

Construction: We have constructed a triangle ABC and named the vertices according to the question.

Note the height of the triangle, BD.

If ∠BAD = A

Then, BD = c sin A

[∵ in ∆BAD

⇒ BD = c sin A]

And, AD = c cos A

[∵ in ∆BAD

⇒ AD = c cos A]

Proof:

Here, components of c which are:

c sin A

c cos A

are drawn on the diagram.

Using Pythagoras theorem which says that,

(hypotenuse)² =(perpendicular)² + (base)²

Take ∆BDC, which is a right-angled triangle.

Here,

Hypotenuse = BC

Base = CD

Perpendicular = BD

We get,

(BC)² = (BD)² + (CD)²

⇒ a² = (c sin A)² + (CD)² [∵ from the diagram, BD = c sin A]

⇒ a² = c² sin² A + (b – c cos A)²

[∵ from the diagram, AC = CD + AD

⇒ CD = AC – AD

⇒ CD = b – c cos A]

⇒ a² = c² sin² A + (b² + (-c cos A)² – 2bc cos A) [∵ from algebraic identity, (a – b)² = a² + b² – 2ab]

⇒ a² = c² sin² A + b² + c² cos² A – 2bc cos A

⇒ a² = c² sin² A + c² cos² A + b² – 2bc cos A

⇒ a² = c² (sin² A + cos² A) + b² – 2bc cos A

⇒ a² = c² + b² – 2bc cos A [∵ from trigonometric identity, sin² θ + cos² θ = 1]

⇒ 2bc cos A = c² + b² – a²

⇒ 2bc cos A = b² + c² – a²

Hence, proved.

Let are vertices of a triangle ABC.

Also, we get

Position vector of A

Position vector of B

Position vector of C

We need to show that,

gives the vector are of the triangle.

We know that,

Vector area of ∆ABC is given as,

Here,

…(i)

Thus, shown.

We know that, two vectors are collinear if they lie on the same line or parallel lines.

For to be collinear, area of the ∆ABC should be equal to 0.

⇒ Area of ∆ABC = 0

…(ii)

Thus, this is the required condition for to be collinear.

Now, we need to find the unit vector normal to the plane of the triangle.

Let be the unit vector normal to the plane of the triangle.

Note that, from equation (i).

And, from equation (i).

So,

Thus, unit vector normal to the plane of the triangle is .

We have,

Let ABCD be a parallelogram.

In ABCD,

And since, AD ∥ BC

So,

We need to show that,

Where, and are diagonals of the parallelogram ABCD.

Now, by triangle law of addition, we get

…(i)

Similarly,

…(ii)

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

…(iii)

And,

…(iv)

Now, can be written as,

[∵ and ]

We know that,

Vector area of parallelogram ABCD is given by,

Area of parallelogram ABCD

Hence, shown.

Now, we need to find the area of parallelogram whose diagonals are and .

We have already derived the relationship between area of parallelogram and diagonals of parallelogram, which is

Here,

And,

⇒Area of parallelogram

Thus, area of required parallelogram is .

Given that,

We need to find vector .

Let , where x, y, z be any scalars.

Now, for :

Comparing Left Hand Side and Right Hand Side, we get

From coefficient of ⇒ z – y = 0 …(i)

From coefficient of ⇒ -(z – x) = 1

⇒ x – z = 1 …(ii)

From coefficient of ⇒ y – x = -1

⇒ x – y = 1 …(iii)

Also, for :

Since, , as and other dot multiplication is zero. We get,

x + y + z = 3 …(iv)

Now, add equations (ii) and (iii), we get

(x – z) + (x – y) = 1 + 1

⇒ x + x – y – z = 2

⇒ 2x – y – z = 2 …(v)

Add equations (iv) and (v), we get

(x + y + z) + (2x – y – z) = 3 + 2

⇒ x + 2x + y – y + z – z = 5

⇒ 3x = 5

Put value of x in equation (iii), we get

Equation (iii) ⇒ x – y = 1

Put this value of y in equation (i), we get

Equation (i) ⇒ z – y = 0

Since,

By putting the values of x, y and z, we get

Thus, we have found the vector .

We are given points and .

Let these points be

and .

Also, given in the question that,

A point divides AB in the ratio of 3 : 1.

Let this point be C.

⇒ C divides AB in the ratio = 3 : 1

We need to find the position vector of C.

We know the position vector of a point C dividing the line segment joining the points P and Q, whose position vectors are p and q in the ratio m : n internally is given by,

According to the question, here

m : n = 3 : 1

⇒ m = 3 and n = 1

Also,

And

Substituting these values in the formula above, we get

Thus, position vector of the point is .

Let initial point be A(2,5,0) and terminal point be B(-3,7,4).So, the required vector joining A and B is the vector .

⇒

=

Given that,

Let θ be the angle between vector a and b.

Then,

⇒

⇒

⇒

Given that, are orthogonal.

⇒

⇒

⇒ 2 + 2λ + 3 = 0

⇒ 2λ = -5

⇒

Given that, are parallel

⇒

⇒

Given that, vector from origin to the point A, and vector from origin to the point B, .

Area of Δ OAB =

=

=

=

=

=

=

=

Let , then

⇒

⇒

⇒

Similarly, we get

⇒

⇒

∴

=

Given that,

Let θ be the angle between vector a and b.

Then,

⇒ 12 = 10×2 cosθ

⇒

⇒

⇒

Now,

⇒

Given that,

Let

Now, are coplanar

if,

⇒ λ (λ²-1) -1(λ+2) +2(-1-2 λ)=0

⇒ λ³- λ –λ-2-2-4 λ=0

⇒ λ³- 6λ-4 =0

⇒ (λ+2)( λ²-2 λ-2)=0

⇒ λ = -2 and

⇒ λ = -2 and λ = 1±√3

Given that, are unit vectors ⇒ and

⇒

⇒

⇒

⇒

⇒

Let θ be the angle between .

From figure we can see that , length OL is the projection of .

In Δ OLA ,we have

⇒ OL = OA cosθ

⇒

⇒

⇒

Now,

⇒

⇒

Given that, and

⇒

⇒

⇒

⇒

⇒

Given that, and –3≤λ≤2

We know that,

⇒ at λ = -3

⇒ at λ = 0

⇒ at λ = 2

Hence, the range of is [0,12]

Given that ,

Now, a vector which is perpendicular to both is given by

Now,

∴ the required unit vector

=

There are two perpendicular directions to any plane. Thus, another unit vector perpendicular to

⇒

Hence, there are two unit length perpendicular to the .

Let are two non-collinear vectors.

Let bisects the angle between .

⇒ θ₁ = θ₂

∴

Since, θ₁ = θ₂ ⇒ cosθ₁ = cosθ₂

⇒

⇒

Thus, the vector bisects the angle between the non-collinear vectors if they are equal.

Given that, for some non-zero vector .

⇒ is perpendicular to

⇒ are coplanar.

⇒

Given that ,

Let are two diagonals of parallelogram.

⇒

⇒

⇒

⇒

Let θ be the angle between diagonals .

Then,

⇒

⇒

⇒

⇒

Given that,

⇒

⇒ |k| < 1

⇒ -1 <k<1

Also, is parallel to

⇒ k cannot be equal to , otherwise it will become null vector and then it will not be parallel to .

Since, k is along the direction of and not in its opposite direction.

∴ k ∈ (-1,1) –

We have,

=

=

=

=

Thus ,

Given that,

⇒

⇒

⇒

⇒

⇒

⇒

⇒

Let

Now, taking dot product of with , we get

⇒

⇒

Similarly, taking dot product of with ,we get

⇒

False

Explanation:

Let

⇒

and

Now, we observe that

True

Explanation:

Consider a point P in space, having coordinates (x, y, z) with

respect to the origin O(0, 0, 0). Then, the vector having O and P as its initial and terminal points, respectively, is called the position vector of the point P with respect to O.

True

Explanation:

Given that,

On squaring both the sides, we get

⇒

⇒

⇒

⇒

⇒

⇒

Hence, are orthogonal.

False

Explanation:

⇒

⇒

⇒

False

Explanation:

Given that, ⇒ are perpendicular to each other.

But, adjacent sides of rhombus are not perpendicular.