Before starting the concept of how to find perpendicular vectors first we need to know about the vectors. The physical quantities which have both magnitude and direction are termed as vectors.

**If to illustrate the concept of a vector first we need to take a vector quantity in consideration. For example force is a vector. Let us the assume that the weight of a body is 5N,it means that the magnitude of the weight is 5N and it is acting in downward direction. If we talk about two vectors, then fulfilling the condition that their dot product is zero, it can be said that they are perpendicular to each other.**

For example, if we consider the dot product of vectors A and B, then-

A.B= AB cos x

Here x is defined as the angle between two vectors A and B. If we take x= 90 degree, then the value of dot product of A and B is= ABcos90=0 (as cos 90=0). Conversely if the dot product of two vectors is 0 then they can be said to be perpendicular to each other.

A.B=0

AB cos x=0

cos x=0

cos x= cos 90

x=90

The concept of perpendicular vectors can be illustrated with some numerical problems written below-

**Example**

The values of two vectors A and B are (5i+3j-8k) and (i+j+k) respectively. Are these two vectors perpendicular or not?

A.B

=(5i+3j-8k).(i+j+k)

=5i.i+3j.j-8k.k [as i.i=1,j.j=1 and k.k=1]

=5+3-8

=8-8

=0

As per the rule derived earlier when the dot product of two vectors is zero then they are said to be perpendicular to each other. Hence A and B vectors are perpendicular to each other.

Two vectors (3i+7j+7k) and (-7i-aj+7k) are perpendicular to each other. Find the value of a.

First we need to calculate the dot product of these two vectors.

(3i+7j+7k).(-7i-aj+7k)

=(-21-7a+49)

As we know that in case of perpendicular vectors the dot product of the two vectors must be 0. So –

(-21-7a+49)=0

Or, -7a+28=0

Or, 7a=28

Or, a=28/7

Or, a=4

The similar concept is used in no work force. No work force is a force that is acting perpendicularly with the direction of displacement of a body.

W=F.S

= FScosx

Here, W= work done

F= the force

S= the displacement

x=90 degree

W= FS cos90

=0

Here,the work done is 0 that means no work is done by the force. Hence it is named as no work force.

**Example**

A person is walking on a horizontal ground carrying a luggage on his head. Here the work done is 0 as the force due gravity and his displacement on the ground are perpendicular to each other.

**How to find a vector perpendicular to two vectors?**

**Here a condition is applied that both the vectors on which the third vector is perpendicular should be non parallel vectors. By calculating the cross product of two non-parallel vectors we will be able to get a vector that is perpendicular to both of them individually. **For example, A and B are two non parallel vectors where A= (a1i+a2j+a3k) and B= (b1i+b2j+b3k) then their perpendicular vector will be= cross product of A and B

p = (a1i+a2j+a3k)*(b1i+b2j+b3k)

Or, p = (a1b2k-a1b3j-a2b1k+a2b3i+a3b1j-a3b2i)

[As i*i=0,i*j=k,i*k=-j,j*j=0,j*k=i,j*i=-k,k*k=0,k*i=j,k*j=-i]

Or, p= i(a2b3-a3b2)+j(a3b1-a1b3)+k(a1b2-a2b1)

So the required perpendicular vector is- i(a2b3-a3b2)+j(a3b1-a1b3)+k(a1b2-a2b1)

Now if we want to cross check whether the vector is perpendicular or not we need to consider the individual dot products of p.A and p.B.

Now, p.A= { i(a2b3-a3b2)+j(a3b1-a1b3)+k(a1b2-a2b1)}.(a1i+a2j+a3k)

= a1a2b3-a1b2a3+b1a2a3-a1a2b3+a1b2a3-b1a2a3

= 0

And p.B= {i(a2b3-a3b2)+j(a3b1-a1b3)+k(a1b2-a2b1)}. (b1i+b2j+b3k)

= b1a2b3-b1b2a3+b1b2a3-a1b2b3+a1b2b3-b1a2b3

= 0

So from the above calculations it can be seen that the perpendicular vector p is perpendicular to both A and B vectors individually. Hence it is proved that the cross product of two non parallel vectors gives the perpendicular vector of them.

**How** **to find** **a unit** **vector** **perpendicular** **to** **two** **vectors?**

**Let us take two vectors X and Y.**

**The cross product of X and Y is= X*Y**

**The magnitude of X and Y is= |X*Y|**

**So the required unit vector that is perpendicular to both X and Y is=( X*Y)/ |X*Y|**

**Example**

A and B are two vectors where A= (5i+5j+7k) and B= (i+j+k). Find the unit vector that is perpendicular to both A and B.

A*B=(5i+5j+7k)*(i+j+k)

= 5k-5j-5k+5i+7j-7i

=-2i+2j

=2(-i+j)

|A*B|=[(-2)^2+(2)^2]^(½)

=2.(2)^(½)

Therefore, unit vector that is perpendicular to A and B=2(-i+j)/2.(2)^(½)

=(-i+j)/(2)^(½)

**How to find if two vectors are perpendicular?**

**To find whether two vectors are perpendicular to each other or not, we should calculate their dot product first. From the result we will be able to conclude whether the vectors are perpendicular or not. The 0 value of the dot product signifies that the vectors are perpendicular to each other whereas the non zero value signifies that the vectors are not mutually perpendicular.**

**Example**

The weight of a car F=3i+3j+3k is acting downwards. Its displacement in the horizontal plane is given by D=-3i+3j. What can be said about the work done on the car?

Work done=F.D

=(3i+3j+3k).(-3i+3j+0k)

= -9+9

= 0

Here in the above problem the dot product is 0. So it can be concluded that F and D are perpendicular to each other.

**How to find the resultant of two perpendicular vectors?**

** A+B=R**

**Let there be two vectors A and B having an angle C between them. If their resultant is R then the magnitude of R will be,**

**R= (A^2+B^2-2ABcosC)^(½)**

**Example**

Two forces F1 and F2 are acting on a body where the value of F1 is 8 N and value of F2 is 6 N and they are mutually perpendicular. What will be the value of the resultant force?

F1=8 N

F2=6 N

C= 90 degree

R= (F1^2+F2^2-2F1F2cosC)^(½)

={(8)^2+(6)^2-2.8.6.cos 90}^(½)

={64+36-0}^(½)

=100^(½)

=10 N

**How to find a vector perpendicular to three vectors?**

**Let us take three different points lying on the same plane but not on the same straight line. These points have three different position vectors that are x,y,z respectively. Let r be the position vector of another point on the same plane where the other three points lie. So the vectors (r-x), (y-x) and (z -x) are coplanar**. Using the formula for coplanar vectors it can be said that

(r-x). (y-x)* (z-x)=0

Or,(r-x).(x*y+y*z+z*x)=0

Hence, (x*y+y*z+z*x) is perpendicular to (r-x) and is therefore perpendicular to the plane of the three different points.

How to find perpendicular distance between two vectors?

Perpendicular distance is the shortest distance between two vectors. Formula for the shortest distance between two vectors is-

Shortest distance=|(b1*b2).(a1-a2)|/|b1*b2|

**Example**

r1=(i+j+k)+k1(i-j-k)

r2=(2i+2j+2k)+k2(2i-2j-2k)

a1=i+j+k

a2=2i+2j+2k

b1=i-j-k

b2=2i-2j-2k

a2-a1=(2i+2j+2k)-(i+j+k)

=2i+2j+2k-i-j-k

=i+j+k

b1*b2=(i-j-k)*(2i-2j-2k)

=-2k+2j+2k+2i+2j-2i

=4j

Shortest distance=|(b1*b2).(a1-a2)|/|b1*b2|

=|4j.(i+j+k)/4|

=|1|

=1 unit

**How to find parallel and perpendicular vectors?**

**We can use the following formula to find the parallel vectors-**

**k.a=b**

**Here k is a constant and a and b are two vectors.**

**Example**

There are two vectors a and b whose values are (3i+6j) and (4i+8j) respectively. Are these two vectors parallel to each other or not?

According to the formula,written above k.a=b

k.(3 6)=(4 8)

3k=4 6k=8

Or, k=4/3 or,k=8/6=4/3

As the values of k are same in both the cases so vectors a and b are parallel to each other.

Similarly we can identify whether any two vectors are perpendicular or not by calculating their slopes. If the value of the product of these slopes is -1 then it can be said that the vectors are perpendicular or orthogonal vectors.

There are two vectors (3i-5j) and (5i+3j). Identify whether they are perpendicular or not?

Let a=(3i-5j)

and b=(5i+3j)

Therefore slope of a=ma=- 5/3 and slope of b=mb= ⅗

Here, ma.mb=(-5/3)*(⅗)

=-1

So,vectors a and b are perpendicular vectors.

**How to find perpendicular vectors in 3 D?**

**Let us take an example to get the answer.**

**p=(1 2 -2) and q=(-2 2 1). Are these two 3d vectors perpendicular or not?**

**p.q= (-2+4-2)**

**=0**

Therefore p and q are two perpendicular 3d vectors.

Before starting the concept of perpendicular vectors first we need to know about the vectors. The physical quantities which have both magnitude and direction are termed as vectors.

If to illustrate the concept of a vector first we need to take a vector quantity in consideration. For example force is a vector. Let us the assume that the weight of a body is 5N,it means that the magnitude of the weight is 5N and it is acting in downward direction. If we talk about two vectors, then fulfilling the condition that their dot product is zero, it can be said that they are perpendicular to each other. For example, if we consider the dot product of vectors A and B, then-

A.B= AB cos x

Here x is defined as the angle between two vectors A and B. If we take x= 90 degree, then the value of dot product of A and B is= ABcos90=0 (as cos 90=0). Conversely if the dot product of two vectors is 0 then they can be said to be perpendicular to each other.

A.B=0

AB cos x=0

cos x=0

cos x= cos 90

x=90

The concept of perpendicular vectors can be illustrated with some numerical problems written below-

**Example**

The values of two vectors A and B are (5i+3j-8k) and (i+j+k) respectively. Are these two vectors perpendicular or not?

A.B

=(5i+3j-8k).(i+j+k)

=5i.i+3j.j-8k.k [as i.i=1,j.j=1 and k.k=1]

=5+3-8

=8-8

=0

As per the rule derived earlier when the dot product of two vectors is zero then they are said to be perpendicular to each other. Hence A and B vectors are perpendicular to each other.

2) Two vectors (3i+7j+7k) and (-7i-aj+7k) are perpendicular to each other. Find the value of a.

First we need to calculate the dot product of these two vectors.

(3i+7j+7k).(-7i-aj+7k)

=(-21-7a+49)

As we know that in case of perpendicular vectors the dot product of the two vectors must be 0. So –

(-21-7a+49)=0

Or, -7a+28=0

Or, 7a=28

Or, a=28/7

Or, a=4

The similar concept is used in no work force. No work force is a force that is acting perpendicularly with the direction of displacement of a body.

W=F.S

= FScosx

Here, W= work done

F= the force

S= the displacement

x=90 degree

W= FS cos90

=0

Here,the work done is 0 that means no work is done by the force. Hence it is named as no work force.

**Example**

A person is walking on a horizontal ground carrying a luggage on his head. Here the work done is 0 as the force due gravity and his displacement on the ground are perpendicular to each other.

**How to find a vector perpendicular to two vectors?**

**Here a condition is applied that both the vectors on which the third vector is perpendicular should be non parallel vectors. By calculating the cross product of two non-parallel vectors we will be able to get a vector that is perpendicular to both of them individually.**

For example, A and B are two non parallel vectors where A= (a1i+a2j+a3k) and B= (b1i+b2j+b3k) then their perpendicular vector will be= cross product of A and B

p = (a1i+a2j+a3k)*(b1i+b2j+b3k)

Or, p = (a1b2k-a1b3j-a2b1k+a2b3i+a3b1j-a3b2i)

[As i*i=0,i*j=k,i*k=-j,j*j=0,j*k=i,j*i=-k,k*k=0,k*i=j,k*j=-i]

Or, p= i(a2b3-a3b2)+j(a3b1-a1b3)+k(a1b2-a2b1)

So the required perpendicular vector is- i(a2b3-a3b2)+j(a3b1-a1b3)+k(a1b2-a2b1)

Now if we want to cross check whether the vector is perpendicular or not we need to consider the individual dot products of p.A and p.B.

Now, p.A= { i(a2b3-a3b2)+j(a3b1-a1b3)+k(a1b2-a2b1)}.(a1i+a2j+a3k)

= a1a2b3-a1b2a3+b1a2a3-a1a2b3+a1b2a3-b1a2a3

= 0

And p.B= {i(a2b3-a3b2)+j(a3b1-a1b3)+k(a1b2-a2b1)}. (b1i+b2j+b3k)

= b1a2b3-b1b2a3+b1b2a3-a1b2b3+a1b2b3-b1a2b3

= 0

So from the above calculations it can be seen that the perpendicular vector p is perpendicular to both A and B vectors individually. Hence it is proved that the cross product of two non parallel vectors gives the perpendicular vector of them.

How to find a unit vector perpendicular to two vectors?

Let us take two vectors X and Y.

The cross product of X and Y is= X*Y

The magnitude of X and Y is= |X*Y|

So the required unit vector that is perpendicular to both X and Y is=( X*Y)/ |X*Y|

**Example**

A and B are two vectors where A= (5i+5j+7k) and B= (i+j+k). Find the unit vector that is perpendicular to both A and B.

A*B=(5i+5j+7k)*(i+j+k)

= 5k-5j-5k+5i+7j-7i

=-2i+2j

=2(-i+j)

|A*B|=[(-2)^2+(2)^2]^(½)

=2.(2)^(½)

Therefore, unit vector that is perpendicular to A and B=2(-i+j)/2.(2)^(½)

=(-i+j)/(2)^(½)

**How to find if two vectors are perpendicular?**

**To find whether two vectors are perpendicular to each other or not, we should calculate their dot product first. From the result we will be able to conclude whether the vectors are perpendicular or not. The 0 value of the dot product signifies that the vectors are perpendicular to each other whereas the non zero value signifies that the vectors are not mutually perpendicular.**

**Example**

The weight of a car F=3i+3j+3k is acting downwards. Its displacement in the horizontal plane is given by D=-3i+3j. What can be said about the work done on the car?

Work done=F.D

=(3i+3j+3k).(-3i+3j+0k)

= -9+9

= 0

Here in the above problem the dot product is 0. So it can be concluded that F and D are perpendicular to each other.

**How to find the resultant of two perpendicular vectors?**

**Let there be two vectors A and B having an angle C between them. If their resultant is R then the magnitude of R will be,**

**R= (A^2+B^2-2ABcosC)^(½)**

**Example**

Two forces F1 and F2 are acting on a body where the value of F1 is 8 N and value of F2 is 6 N and they are mutually perpendicular. What will be the value of the resultant force?

F1=8 N

F2=6 N

C= 90 degree

R= (F1^2+F2^2-2F1F2cosC)^(½)

={(8)^2+(6)^2-2.8.6.cos 90}^(½)

={64+36-0}^(½)

=100^(½)

=10 N

**How to find a vector perpendicular to three vectors?**

**Let us take three different points lying on the same plane but not on the same straight line. These points have three different position vectors that are x,y,z respectively. Let r be the position vector of another point on the same plane where the other three points lie.**

So the vectors (r-x), (y-x) and (z -x) are coplanar. Using the formula for coplanar vectors it can be said that

(r-x). (y-x)* (z-x)=0

Or,(r-x).(x*y+y*z+z*x)=0

Hence, (x*y+y*z+z*x) is perpendicular to (r-x) and is therefore perpendicular to the plane of the three different points.

How to find perpendicular distance between two vectors?

Perpendicular distance is the shortest distance between two vectors. Formula for the shortest distance between two vectors is-

Shortest distance=|(b1*b2).(a1-a2)|/|b1*b2|

**Example**

r1=(i+j+k)+k1(i-j-k)

r2=(2i+2j+2k)+k2(2i-2j-2k)

a1=i+j+k

a2=2i+2j+2k

b1=i-j-k

b2=2i-2j-2k

a2-a1=(2i+2j+2k)-(i+j+k)

=2i+2j+2k-i-j-k

=i+j+k

b1*b2=(i-j-k)*(2i-2j-2k)

=-2k+2j+2k+2i+2j-2i

=4j

Shortest distance=|(b1*b2).(a1-a2)|/|b1*b2|

=|4j.(i+j+k)/4|

=|1|

=1 unit

**How to find parallel and perpendicular vectors?**

**We can use the following formula to find the parallel vectors-**

**k.a=b**

**Here k is a constant and a and b are two vectors.**

**Example**

There are two vectors a and b whose values are (3i+6j) and (4i+8j) respectively. Are these two vectors parallel to each other or not?

According to the formula,written above k.a=b

k.(3 6)=(4 8)

3k=4 6k=8

Or, k=4/3 or,k=8/6=4/3

As the values of k are same in both the cases so vectors a and b are parallel to each other.

Similarly we can identify whether any two vectors are perpendicular or not by calculating their slopes. If the value of the product of these slopes is -1 then it can be said that the vectors are perpendicular or orthogonal vectors.

**Example**

There are two vectors (3i-5j) and (5i+3j). Identify whether they are perpendicular or not?

Let a=(3i-5j)

and b=(5i+3j)

Therefore slope of a=ma=- 5/3 and slope of b=mb= ⅗

Here, ma.mb=(-5/3)*(⅗)

=-1

So,vectors a and b are perpendicular vectors.

**How to find perpendicular vectors in 3 d?**

**Let us take an example to get the answer.**

**p=(1 2 -2) and q=(-2 2 1). Are these two 3d vectors perpendicular or not**?

**p.q= (-2+4-2)**

**=0**

Therefore p and q are two perpendicular 3d vectors.