Vector Or Cross Product

Given and

We need to find the magnitude of the vector.

Recall the cross product of two vectors and is

Here, we have (a₁, a₂, a₃) = (1, 3, –2) and (b₁, b₂, b₃) = (–1, 0, 3)

Recall the magnitude of the vector is

Now, we find.

Thus,

Given and

We need to find the magnitude of the vector.

Recall the cross product of two vectors and is

Here, we have (a₁, a₂, a₃) = (3, 4, 0) and (b₁, b₂, b₃) = (1, 1, 1)

Recall the magnitude of the vector is

Now, we find.

Thus,

Given and

We need to find the magnitude of the vector.

Recall the cross product of two vectors and is

Here, we have (a₁, a₂, a₃) = (2, 1, 0) and (b₁, b₂, b₃) = (1, 1, 1)

Recall the magnitude of the vector is

Now, we find.

Thus, the magnitude of the vector =

Given two vectors and

Let and

We need to find a unit vector perpendicular to and.

Recall a vector that is perpendicular to two vectors and is

Here, we have (a₁, a₂, a₃) = (4, –1, 3) and (b₁, b₂, b₃) = (–2, 1, –2)

Let the unit vector in the direction of be.

We know unit vector in the direction of a vector is given by .

Recall the magnitude of the vector is

Now, we find.

So, we have

Thus, the required unit vector that is perpendicular to both and is.

Given two vectors and

We need to find a unit vector perpendicular to and.

Recall a vector that is perpendicular to two vectors and is

Here, we have (a₁, a₂, a₃) = (2, 1, 1) and (b₁, b₂, b₃) = (1, 2, 1)

Let the unit vector in the direction of be.

We know unit vector in the direction of a vector is given by .

Recall the magnitude of the vector is

Now, we find.

So, we have

Thus, the required unit vector that is perpendicular to both and is.

Given

We need to find the magnitude of the vector.

Recall the cross product of two vectors and is

Here, we have (a₁, a₂, a₃) = (0, 4, 3) and (b₁, b₂, b₃) = (1, 1, –1)

Recall the magnitude of the vector is

Now, we find.

Thus, magnitude of vector =

Given and

We need to find the magnitude of vector.

We know unit vector in the direction of a vector is given by .

Recall the cross product of two vectors and is

Here, we have (a₁, a₂, a₃) = (, 0, ) and (b₁, b₂, b₃) = (4, 3, 1)

Recall the magnitude of the vector is

Now, we find.

Thus,

Given and

We need to find the vector.

Recall the cross product of two vectors and is

Here, we have (a₁, a₂, a₃) = (7, 5, 0) and (b₁, b₂, b₃) = (4, –5, –5)

Thus,

Given two vectors and

Let and

We need to find a vector of magnitude 49 that is perpendicular to and.

Recall a vector that is perpendicular to two vectors and is

Here, we have (a₁, a₂, a₃) = (2, 3, 6) and (b₁, b₂, b₃) = (3, –6, 2)

Recall the magnitude of the vector is

Now, we find.

Thus, the vector of magnitude 49 that is perpendicular to both and is.

Given two vectors and

We need to find vector of magnitude 3 that is perpendicular to and.

Recall a vector that is perpendicular to two vectors and is

Here, we have (a₁, a₂, a₃) = (3, 1, –4) and (b₁, b₂, b₃) = (6, 5, –2)

Recall the magnitude of the vector is

Now, we find.

Let the unit vector in the direction of be.

We know unit vector in the direction of a vector is given by .

So, a vector of magnitude 3 in the direction of is

Thus, the vector of magnitude 3 that is perpendicular to both and is.

Given two vectors and are sides of a parallelogram

Let and

Recall the area of the parallelogram whose adjacent sides are given by the two vectors and is where

Here, we have (a₁, a₂, a₃) = (2, 0, 0) and (b₁, b₂, b₃) = (0, 3, 0)

Recall the magnitude of the vector is

Now, we find.

Thus, area of the parallelogram is 6 square units.

Given two vectors and are sides of a parallelogram

Let and

Recall the area of the parallelogram whose adjacent sides are given by the two vectors and is where

Here, we have (a₁, a₂, a₃) = (2, 1, 3) and (b₁, b₂, b₃) = (1, –1, 0)

Recall the magnitude of the vector is

Now, we find.

Thus, area of the parallelogram is square units.

Given two vectors and are sides of a parallelogram

Let and

Recall the area of the parallelogram whose adjacent sides are given by the two vectors and is where

Here, we have (a₁, a₂, a₃) = (3, 1, –2) and (b₁, b₂, b₃) = (1, –3, 4)

Recall the magnitude of the vector is

Now, we find.

Thus, area of the parallelogram is square units.

Given two vectors and are sides of a parallelogram

Let and

Recall the area of the parallelogram whose adjacent sides are given by the two vectors and is where

Here, we have (a₁, a₂, a₃) = (1, –3, 1) and (b₁, b₂, b₃) = (1, 1, 1)

Recall the magnitude of the vector is

Now, we find.

Thus, the area of the parallelogram is square units.

Given two diagonals of a parallelogram are and

Let and

Recall the area of the parallelogram whose diagonals are given by the two vectors and is where

Here, we have (a₁, a₂, a₃) = (4, –1, –3) and (b₁, b₂, b₃) = (–2, 1, –2)

Recall the magnitude of the vector is

Now, we find.

Thus, the area of the parallelogram is 7.5 square units.

Given two diagonals of a parallelogram are and

Let and

Recall the area of the parallelogram whose diagonals are given by the two vectors and is where

Here, we have (a₁, a₂, a₃) = (2, 0, 1) and (b₁, b₂, b₃) = (1, 1, 1)

Recall the magnitude of the vector is

Now, we find.

Thus, the area of the parallelogram is square units.

Given two diagonals of a parallelogram are and

Let and

Recall the area of the parallelogram whose diagonals are given by the two vectors and is where

Here, we have (a₁, a₂, a₃) = (3, 4, 0) and (b₁, b₂, b₃) = (1, 1, 1)

Recall the magnitude of the vector is

Now, we find.

Thus, the area of the parallelogram is square units.

Given two diagonals of a parallelogram are and

Let and

Recall the area of the parallelogram whose diagonals are given by the two vectors and is where

Here, we have (a₁, a₂, a₃) = (2, 3, 6) and (b₁, b₂, b₃) = (3, –6, 2)

Recall the magnitude of the vector is

Now, we find.

Thus, area of the parallelogram is 24.5 square units.

Given, and

We need to find.

First, we will find.

Recall the cross product of two vectors and is

Here, we have (a₁, a₂, a₃) = (2, 5, –7) and (b₁, b₂, b₃) = (–3, 4, 1)

Now, we will find.

Using the formula for cross product as above, we have

Now, we need to find.

First, we will find.

Using the formula for cross product, we have

Now, we will find.

Using the formula for the cross product as above, we have

So, we found and

Therefore, we have.

Given, and

We know the cross product of two vectors and forming an angle θ is

where is a unit vector perpendicular to and

is a unit vector

⇒ 8 = 2 × 5 × sin θ × 1

⇒ 10 sin θ = 8

We also have the dot product of two vectors and forming an angle θ is

But, we have sin²θ + cos²θ = 1

Thus,

To show that,, is a right handed orthogonal system of unit vectors, we need to prove the following –

(a)

(b)

(c)

(d)

Let us consider each of these one at a time.

(a) Recall the magnitude of the vector is

First, we will find.

Now, we will find.

Finally, we will find.

Hence, we have

(b) Now, we will evaluate the vector

Recall the cross product of two vectors and is

Taking the scalar common, here, we have (a₁, a₂, a₃) = (2, 3, 6) and (b₁, b₂, b₃) = (3, –6, 2)

Hence, we have.

(c) Now, we will evaluate the vector

Taking the scalar common, here, we have (a₁, a₂, a₃) = (3, –6, 2) and (b₁, b₂, b₃) = (6, 2, –3)

Hence, we have.

(d) Now, we will evaluate the vector

Taking the scalar common, here, we have (a₁, a₂, a₃) = (6, 2, –3) and (b₁, b₂, b₃) = (2, 3, 6)

Hence, we have.

Thus,,, is also another right handed orthogonal system of unit vectors.

Given, and

We know the dot product of two vectors and forming an angle θ is

⇒ 65 cos θ = 60

We also know the cross product of two vectors and forming an angle θ is

where is a unit vector perpendicular to and

But, we have sin²θ + cos²θ = 1

is a unit vector

Thus,

Given.

Let the angle between vectors and be θ.

We know the cross product of two vectors and forming an angle θ is

where is a unit vector perpendicular to and

is a unit vector

We also have the dot product of two vectors and forming an angle θ is

But, it is given that

⇒ sin θ = cos θ

⇒ tan θ = 1

Thus, the angle between two vectors is.

Given.

We have

Using distributive property of vectors, we have

We know that if the cross product of two vectors is the null vector, then the vectors are parallel.

Here,

So, vector is parallel to.

Thus, for some scalar m.

Given, and

Let the angle between vectors and be θ.

We know the cross product of two vectors and forming an angle θ is

where is a unit vector perpendicular to and

is a unit vector

Recall the magnitude of the vector is

⇒ 14 sin θ = 7

Thus, the angle between two vectors is.

Given and.

To draw inferences from this, we shall analyze these two equations one at a time.

First, let us consider.

We know the cross product of two vectors and forming an angle θ is

where is a unit vector perpendicular to and .

So, if, we have at least one of the following true –

(a)

(b)

(c) and

(d) is parallel to

Now, let us consider.

We have the dot product of two vectors and forming an angle θ is

So, if, we have at least one of the following true –

(a)

(b)

(c) and

(d) is perpendicular to

Given both these conditions are true.

Hence, the possibility (d) cannot be true as can’t be both parallel and perpendicular to at the same time.

Thus, either one or both of and are zero vectors if we have as well as.

Given, and.

Considering the first equation, is the cross product of the vectors and.

By the definition of the cross product of two vectors, we have perpendicular to both and.

Similarly, considering the second equation, we have perpendicular to both and.

Once again, considering the third equation, we have perpendicular to both and.

From the above three statements, we can observe that the vectors, and are mutually perpendicular.

It is also said that, and are three unit vectors.

Thus,,, form an orthonormal right handed triad of unit vectors.

Given points A(3, –1, 2), B(1, –1, –3) and C(4, –3, 1)

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

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

Similarly, we have and

Plane ABC contains the two vectors and.

So, a vector perpendicular to this plane is also perpendicular to both of these vectors.

Recall the vector is given by

Similarly, the vector is given by

We need to find a unit vector perpendicular to and.

Recall a vector that is perpendicular to two vectors and is

Here, we have (a₁, a₂, a₃) = (–2, 0, –5) and (b₁, b₂, b₃) = (1, –2, –1)

Let the unit vector in the direction of be.

We know unit vector in the direction of a vector is given by .

Recall the magnitude of the vector is

Now, we find.

So, we have

Thus, the required unit vector that is perpendicular to plane ABC is.

Given ABC is a triangle with BC = a, CA = b and AB = c.

Firstly, we need to prove.

From the triangle law of vector addition, we have

But, we know

Let, and

By taking cross product with, we get

We know the cross product of two vectors and forming an angle θ is

where is a unit vector perpendicular to and

Here, all the vectors are coplanar. So, the unit vector perpendicular to and is same as that of and.

Consider equation (I) again.

We have

By taking cross product with, we get

From (II) and (III), we get

Thus, and in ΔABC.

Given and

Recall the cross product of two vectors and is

Here, we have (a₁, a₂, a₃) = (1, –2, 3) and (b₁, b₂, b₃) = (2, 3, –5)

We need to prove and are perpendicular to each other.

We know that two vectors are perpendicular if their dot product is zero.

So, we will evaluate.

But,, and are mutually perpendicular.

Thus and it is perpendicular to.

Given two unit vectors and forming an angle of 30°.

We know the cross product of two vectors and forming an angle θ is

where is a unit vector perpendicular to and .

Given two diagonals of parallelogram and

Recall the area of the parallelogram whose diagonals are given by the two vectors and is.

We have

We have

But, we found .

is a unit vector

Thus, area of the parallelogram is square units.

Let the angle between vectors and be θ.

We know the cross product of two vectors and forming an angle θ is

where is a unit vector perpendicular to and

is a unit vector

Now, consider the LHS of the given expression.

But, we have sin²θ + cos²θ = 1

We know, and

But as dot product is commutative

Thus,

Cross Product: The vector or cross product of two non-zero vectors and, denoted by, is defined as

where θ is the angle between and , 0≤θ≤π and is a unit vector perpendicular to both and, such that, and form a right handed system.

We have

is a unit vector

But, we have the dot product of two vectors and forming an angle θ as

Now, we divide these two equations.

Thus,

Given, and

We know the cross product of two vectors and forming an angle θ is

where is a unit vector perpendicular to and

is a unit vector

We also have the dot product of two vectors and forming an angle θ is

But, we have sin²θ + cos²θ = 1

Thus,

Given and are two adjacent sides of a triangle.

Recall the area of the triangle whose adjacent sides are given by the two vectors and is where

Here, we have (a₁, a₂, a₃) = (1, 2, 3) and (b₁, b₂, b₃) = (–3, –2, 1)

Recall the magnitude of the vector is

Now, we find.

Thus, area of the triangle is square units.

Given, and

We need to find a vector perpendicular to and such that.

Recall a vector that is perpendicular to two vectors and is

Here, we have (a₁, a₂, a₃) = (1, 4, 2) and (b₁, b₂, b₃) = (3, –2, 7)

So, is a vector parallel to.

Let for some scalar λ.

We have.

⇒ λ[(2)(32) + (–1)( –1) + (4)(–14)] = 15

⇒ λ(64 + 1 – 56) = 15

⇒ 9λ = 15

So, we have.

Thus,

Given and

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

Recall a vector that is perpendicular to two vectors and is

Here, we have (a₁, a₂, a₃) = (4, 4, 0) and (b₁, b₂, b₃) = (2, 0, 4)

Let the unit vector in the direction of be.

We know unit vector in the direction of a vector is given by .

Recall the magnitude of the vector is

Now, we find.

So, we have

Thus, the required unit vector that is perpendicular to both and is.

Given three points A(2, 3, 5), B(3, 5, 8) and C(2, 7, 8) forming a triangle.

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

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

Similarly, we have and

To find area of ΔABC, we need to find at least two sides of the triangle. So, we will find vectors and.

Recall the vector is given by

Similarly, the vector is given by

Recall the area of the triangle whose adjacent sides are given by the two vectors and is where

Here, we have (a₁, a₂, a₃) = (1, 2, 3) and (b₁, b₂, b₃) = (0, 4, 3)

Recall the magnitude of the vector is

Now, we find.

Thus, area of the triangle is square units.

Given, and

We need to find area of the parallelogram with vectors and as diagonals.

Recall the area of the parallelogram whose diagonals are given by the two vectors and is where

Here, we have (a₁, a₂, a₃) = (1, –3, 2) and (b₁, b₂, b₃) = (–1, 2, 0)

Recall the magnitude of the vector is

Now, we find.

Thus, area of the parallelogram is square units.

Let ABCD be a parallelogram with sides AB and AC given.

We have and

We need to find unit vector parallel to diagonal.

From the triangle law of vector addition, we have

Let the unit vector in the direction of be.

We know unit vector in the direction of a vector is given by .

Recall the magnitude of the vector is

Now, we find.

So, we have

Thus, the required unit vector that is parallel to diaonal is.

Now, we have to find the area of parallelogram ABCD.

Recall the area of the parallelogram whose adjacent sides are given by the two vectors and is where

Here, we have (a₁, a₂, a₃) = (2, –4, 5) and (b₁, b₂, b₃) = (1, –2, –3)

Recall the magnitude of the vector is

Now, we find.

Thus, area of the parallelogram is square units.

We know if either or.

To verify if the converse is true, we suppose

We know the cross product of two vectors and forming an angle θ is

where is a unit vector perpendicular to and .

So, if, we have at least one of the following true –

(a)

(b)

(c) and

(d) is parallel to

The first three possibilities mean that either or or both of them are true.

However, there is another possibility that when the two vectors are parallel. Thus, the converse is not true.

We will justify this using an example.

Given and

Recall the cross product of two vectors and is

Here, we have (a₁, a₂, a₃) = (1, 3, –2) and (b₁, b₂, b₃) = (2, 6, –4)

Hence, we have even when and.

Thus, the converse of the given statement is not true.

Given, and

We need to verify that

First, we will find.

Recall the cross product of two vectors and is

Now, we will find.

We have

Finally, we will find.

We have

So,

Observe that that RHS of both and are the same.

Thus,

Given three points A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5) forming a triangle.

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

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

Similarly, we have and

To find area of ΔABC, we need to find at least two sides of the triangle. So, we will find vectors and.

Recall the vector is given by

Similarly, the vector is given by

Recall the area of the triangle whose adjacent sides are given by the two vectors and is where

Here, we have (a₁, a₂, a₃) = (1, 2, 3) and (b₁, b₂, b₃) = (0, 4, 3)

Recall the magnitude of the vector is

Now, we find.

Thus, area of the triangle is square units.

Given three points A(1, 2, 3), B(2, –1, 4) and C(4, 5, –1) forming a triangle.

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

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

Similarly, we have and

To find area of ΔABC, we need to find at least two sides of the triangle. So, we will find vectors and.

Recall the vector is given by

Similarly, the vector is given by

Recall the area of the triangle whose adjacent sides are given by the two vectors and is where

Here, we have (a₁, a₂, a₃) = (1, –3, 1) and (b₁, b₂, b₃) = (3, 3, –4)

Recall the magnitude of the vector is

Now, we find.

Thus, area of the triangle is square units.

Given two vectors and

We need to find vectors of magnitude perpendicular to and.

Recall a vector that is perpendicular to two vectors and is

Here, we have (a₁, a₂, a₃) = (1, 2, 1) and (b₁, b₂, b₃) = (–1, 3, 4)

Let the unit vector in the direction of be.

We know unit vector in the direction of a vector is given by .

Recall the magnitude of the vector is

Now, we find.

So, we have

So, a vector of magnitude in the direction of is

Observe that is also a unit vector perpendicular to the same plane. This vector is along the direction opposite to the direction of vector.

Thus, the vectors of magnitude that are perpendicular to plane of both and are.

We need to find a unit vector parallel to

Now from the Parallel law of vector Addition, we know that,

Therefore,

Now we need to find the unit vector parallel to

Any unit vector is given by,

Therefore,

Now, we need to find Area of parallelogram. From the figure above it can be easily found by the cross product of adjacent sides.

Therefore, Area of Parallelogram =

If and

Here, we have,

(a₁, a₂, a₃) = (2, -4, -5) and (b₁, b₂, b₃) = (2, 3, 3)

Area of Parallelogram = 21 sq units.

Given and

We know the dot product of two vectors and forming an angle θ is

We also know the cross product of two vectors and forming an angle θ is

where is a unit vector perpendicular to and

is a unit vector

We have

But, we know sin²θ + cos²θ = 1

Thus,

Definition: VECTOR PRODUCT: When multiplication of two vectors yields another vector then it is called vector product of two vectors.

Example:

Figure 1: Vector Product

[where is a unit vector perpendicular to the plane containing and (referred to the figure provided)]

.

We know, , and are 3 unit vectors along x, y and z axis whose magnitudes are unity.

[where is a unit vector perpendicular to the plane containing and ]

[here is , as is perpendicular to both and ]

And, .

So,

[∵ is an unit vector].

.

We know, , and are 3 unit vectors along x, y and z axis whose magnitudes are unity.

We have,

and

And, ,

and

.

=1+1+(-1)

=1.

.

We know, , and are 3 unit vectors along x, y and z axis whose magnitudes are unity.

We have,

,

and

And, ,

and

.

=1+1+1

=3

We know, , and are 3 unit vectors along x, y and z axis whose magnitudes are unity.

We have, ,

and

.

=0

Area of parallelogram

From the figure, it is clear that, and

i.e. .

Now,

.

Now, we know, area of parallelogram .

So, Area of parallelogram .

.

We know,

and .

So,

.

Angle between and .

Given,

Also given, is a unit vector

i.e. .

∴Angle between and

.

Given, and

=12

.

We know,

is perpendicular to both and .

So, [∵ and are perpendicular to each other]

The angle between and is 60 ̊ .

We have, and .

………………… (1)

and ………………….. (2)

Dividing equation (1) by equation (2),

∴ The angle between and is 60 ̊ .

.

= 0

We know, is perpendicular to both and .

So, [∵ and are perpendicular to each other]

.

We know, , and are 3 unit vectors along x, y and z axis whose magnitudes are unity.

.

NOTE: The product of and is not mentioned here.

and .

We know, , and are 3 unit vectors along x, y and z axis whose magnitudes are unity.

Given, and .

.

“FOR CROSS PRODUCT”

.

We know that cross product of two vectors gives us a vector which is perpendicular to both the vectors.

Let and be the vector perpendicular to vectors and .

Inserting the given values we get,

Now, as we know unit vector can be obtained by dividing the given vector by its magnitude.

Unit vector in the direction of

∴Desired unit vector is

We know that,

Now,

→ sin²θ + cos²θ = 1

So we have and in order to find we need to work out the problem by finding cross product through determinant.

→

Now then,

→ From (1)

Let’s see what all things we know from the given question.

→ Unit Vectors

1 = (1)(1) sinθ

sinθ=1

Equations we already have –

Now,

sin θ = cos θ

Let’s have a look at everything we have before proceeding to solve the question.

→ Given (Unit Vectors)

Now then,

= 2(1)(1)sin²θ

= 2 sin²θ

In case, the question asks for

=(1)(1)

= 1

We know that →

Now,

→ Given and (3)

On comparing LHS and RHS we get :

→ From (5)

→ From (4)

We know that cross product of two vectors gives us a vector which is perpendicular to both the vectors. And keeping in mind that is a Unit vector we get the equation –

→ (Vector divided its magnitude gives unit vector)

is perpendicular to and

Thus, is another unit vector perpendicular to

Alternative Solution –

Since is perpendicular to , any unit vector parallel/anti-parallel to will be perpendicular to .

Given

sinθ

sinθ

sinθ

Let’s have a look at everything given in the problem.

We can use the basic cross product formula to solve the question –

We need to solve the problem by finding cross product through determinant.

Let , also (Given)

Inserting the given values we get,

On comparing LHS and RHS we get,

42+14λ=0 and-2λ-6=0

14λ= -42and-2λ=6

λ= -3andλ=-3

Area of the parallelogram is give by

Let,

→ ( is an unit vector)

sq. units.

We know that,

Now,

→ (From 1)

= 1+1+0 → (From 4 and 5)

= 2

We know that cross product of two vectors gives us a vector which is perpendicular to both the vectors. If we can find an unit vector

perpendicular to the given vectors, we can easily get the answer by multiplying the unit vector.

Unit vectors perpendicular to the given vectors

Now,

∵ Unit vectors perpendicular to and

Vectors of magnitude which are perpendicular to and →

As we know, for vectors and unit vectors perpendicular to them is give by

Unit vector can be either in positive or negative direction.

Hence, the number of vectors of unit length perpendicular to both the vectors is 2.

Given question gives us two same vectors so the angle is

In case, it asks write the angle between the vectors

The angle between the vectors will be 180^° as they are equal in magnitude and opposite in direction.

Let

=

_(B)

…(1)

…(2)

Out of the four options the only option that satisfies both (1) and (2) is

_(A)

Let

Since

Since β is perpendicular to a

3-γ-γ=0

(A)

The equations of the plane is given by

A(x-x₁)+B(y-y₁)+C(z-z₁)=0

Where A,B and C are the drs of the normal to the plane.

Putting the first point,

=A(x-1)+B(y+1)+C(z-2)=0 …(1)

Putting the second point in Eqn (1)

=A(2-1)+B(0+1)+C(-1-2)=0

A+B-3C=0 …(a)

Putting the third point in Eqn (1)

=A(0-1)+B(2+1)+C(1-2)=0

= -A+3B-C=0 …(b)

Solving (a) and (b) using cross multiplication method

A+B-3C=0

-A+3B-C=0

Put these in Eqn(1)

=8α(x-1)+4α(y+1)+4α(z-2)=0

=2(x-1)+(y+1)+(z-2)=0

=2x+2+y+1+z-2=0

2x+y+z+1=0

Now the vector perpendicular to this plane is

Now the unit vector of is given by

((C)

The diagnols of a rhombus are always perpendicular

It means

Q=90°

_(B)

(A)

Let be perpendicular to both of these vectors

Now the unit vector of is given by

(A)

Let and

A vector perpendicular to both of them is given by

= =

Now the unit vector of is given by

(D)

_(B)

are unit vectors and angle between each of them is 90°

So,

So (A) is false ∵

Option (B) is true because angle between them is 0°

So, cosQ=cos0=1

∵

(C) False as

(D) is False as

And then

(B)

Let

We know

|

(B)

We know,

Let Q be the angle between vectors a and b

₌

sin²Q=1-cos²Q

(B) ∵

We know,

Using them,

We know,

(C)