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Coplanar vectors are defined as vectors which are lying on the same in a three-dimensional plane. The vectors are parallel to the same plane. It is always easy to find any two random vectors in a plane, which are coplanar. Coplanarity of two lines lies in a three-dimensional space, which is represented in vector form. The coplanarity of three vectors is defined when their scalar product is zero.

Coplanar lines are a very common topic in three-dimensional geometry. In mathematical theory, the coplanarity of three vectors is called a condition where three lines lying on the same plane are referred to as coplanar.

A plane is a two-dimensional figure going into infinity in the three-dimensional space, while we have used the straight lines as vector equations.

If three vectors are coplanar then their scalar product is zero, and if these vectors are existing in a 3d- space.

The three vectors are also coplanar if the vectors are in 3d and are linearly independent.

If more than two vectors are linearly independent; then all the vectors are coplanar.

So, the condition for vectors to be coplanar is that their scalar product should be 0, and they should exist on 3d; then these vectors are coplanar.

The equation system that has the determinant of the coefficient as zero is called a non-trivial solution. The equation system that has a determinant of the coefficient matrix as non zero, but the solutions are x=y=z=0 is called a trivial solution.

The vectors, v1,……vn, are linearly independent when the no non-trivial combination of the vectors is zero vector.

a1v1 + … + an vn = 0 where the coefficients a1= 0 ….. an=0.

The vectors, v1,……vn, are linearly dependent when there is at least one non-trivial combination of these vectors, which is equal to zero vector.

If the scalar triple product of any three vectors is 0, then they are called coplanar. The vectors are coplanar if any three vectors are linearly dependent, and if among them not more than two vectors are linearly independent.

FAQ (Frequently Asked Questions)

1. Solve for the Value of x When Point A(3,2,1) B (4,x,5) C (4,2,-2) and D (6,5,-1) are Coplanar.

Given, points are: A(3, 2, 2), B(2, x, 5), C (4, 2, -2) and D (6, 5, -1)

Now, AB = position vector of B - position vector of A

=> AB = (4i + xj + 5k) - (3i + 2j + 2k) = i + (x - 2)j + 3k

AC = position vector of C - position vector of A

=> AC = (4i + 2j - 2k) - (3i + 2j + 2k) = i + 0j -4k

Now, AD = position vector of D - position vector of A

=> AD = (6i + 5j - k) - (3i + 2j + 2k) = 3i + 3j - k

As A, B, C and D are coplanar, then

|AB AC AD| = 0

=> |1 x-2 3|

|1 0 -4| = 0

|3 3 -1|

=> 1(0 +12) - (x - 2)*(-1 + 12) + 3(3 - 0) = 0

=> 12 - 11(x - 2) + 3 * 3 = 0

=> 9 - 11x + 22 + 9 = 0

=> 39 - 7x = 0

=> 7x = 39

=> x = 39/7

2. Find the Scalar Triple Product of Vectors i + 2j + 3k, - i - 2j + k and i + k

|1 2 3|

|-1 -2 1| = 0

|1 0 1|.

Scalar triple product = 1(-2-1)- 2 (-1-1)+ 3(0+2)

=1(-3)- 2 (-2)+ 3(2)=-3+4+6 =7

3. Prove that [a+b,b+c,c+a]=2[a,b,c]

LHS=(a+b).{(b+c)X(c+a)}

=(a+b).{bxc+bxa+cxc+cxa}

=(a+b).{bxc-axb+cxa}

=a.(bxc)-a.(axb)+a.(cxa)+b(bxc)-b(axb)+b.(cxa)

=a.(bxc)-0+0+0-0+b.(cxa)

=a.(bxc)+a(bxc) = 2a.(bxc)=2[a,b,c]

RHS = 2[a,b,c]

4. Search the Value of y if the Vectors i+j+2k,yi-j+k and 3i-2j-k are Coplanar.

Ans: Given that vectors are coplanar

|1 1 2|

|y -1 1|

|3 -2 - 1|.

= 1 (1 + 2 ) - 1 ( - y - 3 ) + 2 ( - 2 y + 3 ) = 0

3 + y + 3 - 4y + 6 = 0

-3y + 12= 0

y = 4

5. Define Coplanar Vectors

Ans. Coplanar vectors are defined as vectors which exist on the same in a three-dimensional plane. These vectors are always parallel to the plane.