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In three-dimensional space, coplanar vectors are vectors that are on the same plane. These are also referred to as vectors that run parallel to each other. Any two random vectors that are coplanar may always be found in a plane. A plane can be understood as a two-dimensional figure that goes up to infinity in a three-dimensional space. In this article, we will learn about the coplanarity of two lines in a three-dimensional space, as well as how to represent it in vector form.
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Key terms: Coplanar vectors, vector equations, scalar product, non-trivial combination, Three-dimensional space, Two-dimensional figure, plane
What is a Coplanar Vector?
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In a three-dimensional plane, coplanar vectors are defined as vectors that are parallel to each other. The vectors are perpendicular to each other in the same plane. Any two random vectors on a plane might be always coplanar. The coplanarity of two lines exists in a three-dimensional space that may be expressed in vector form. Three vectors are said to be planar when the scalar product of these three vectors is equal to zero.
Coplanar Vectors
Coplanarity in Theory
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In three-dimensional geometry, the coplanar line is a popular topic. The coplanarity of three vectors is referred to as a condition in mathematical theory, where three lines located on the same plane are said to be coplanar. A plane is a two-dimensional shape that extends into infinity in three-dimensional space, although straight lines have been utilized as vector equations.
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| Vectors | Types of Vectors | Signum Function |
| Addition of Vectors | Negative of a Vector | Conditional Probability |
Conditions of Coplanar Vectors
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- If three vectors are coplanar, their scalar product is zero, and these vectors exist in a three-dimensional space.
- If the three vectors are in 3d space and are linearly independent, they are also referred to as coplanar.
- All vectors are coplanar if more than two vectors are linearly independent.
- So, if the scalar product of the vectors is 0 and they exist in three dimensions; then these vectors are coplanar.
A vector is a linear combination of vectors v1, …, vn with coefficients a1, …, an, such that;
a1v1 + … + anvn
A linear combination a1v1 + … + anvn is considered trivial if all of the coefficients a1,..., an are zero, and non-trivial if at least one of the coefficients is not zero.
A non-trivial solution is an equation system in which the determinant of the coefficient is zero. A trivial solution is an equation system with a non-zero determinant of the coefficient matrix but solutions of x = y = z = 0.
What are Linearly Independent Vectors?
When no non-trivial combination of the vectors is a zero vector, the vectors v1,......vn are linearly independent.
a1v1 +... + a vn = 0, where a1=0,....., and an=0 are the coefficients.
What are Linearly Dependent Vectors and How Do They Work?
When there is at least one non-trivial combination of these vectors that equals zero vector, the vectors v1,......,vn are linearly dependent.
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| Integrals | Methods of Integration | Degree of polynomial |
| Integration by Parts | Fundamental Theorem of Calculus | Vector Formula |
Things to Remember
- Any three vectors are said to be coplanar if their scalar triple product is zero.
- If any three vectors are linearly dependent and no more than two vectors are linearly independent, the vectors are coplanar.
- A plane is a two-dimensional shape that extends into infinity in three-dimensional space.
- All vectors are coplanar if more than two vectors are linearly independent.
Also Read: Applications of the Integrals
Previous Year Questions
- If a is a unit vector, then… [TS EAMCET 2017]
- Let \(\overrightarrow{a}, \overrightarrow{b}, \& \overrightarrow{c}\) be non-coplanar unit vectors equally inclined to one… [BITSAT 2017]
- From a point A with position vector… [JEE Main 2018]
- If a is a unit vector, then….[TS EAMCET 2017]
- Let a,b & c be non-coplanar unit vectors equally inclined to one another at an acute angle \thetaθ. Then |∣[abc]∣ in terms of θ is equal to…..[BITSAT 2017]
- Find the third vector C, if A+3B−C=0….[JKCET 2007]
- The relationship between the angular velocity ω of the disc and number of rotations (n) made by the disc is governed by…
- From a point A with position vector...[JEE MAIN 2018]
- Then its angular momentum, about the origin is perpendicular to...[KEAM]
- If the volume of a parallelepiped whose coterminous edges are….[JEE MAIN 2020]
- If \(\overrightarrow{a} and \overrightarrow{b}\) are unit vectors and… [KCET 2008]
- f a and b are vectors such that… [KCET 2007]
- If \(\overrightarrow{a}= 2\hat{i}+3\hat{j}-\hat{k}, \overrightarrow{b}= \hat{i}+2\hat{j}-5\hat{k}, \overrightarrow{c}= 3\hat{i}+5\hat{j}-\hat{k} \)… [KCET 2007]
- OA and BO are two vectors of magnitudes 5 and 6 respectively… [KCET 2007]
- If a,b and c are three non-coplanar vectors and p,q and r are vectors defined by... [KCET 2012]
- If a|(b+c) and a|(b−c) where a,b,c ∈ N then [KCET 2006]
- The vector equation of the symmetrical form of equation of straight line… [VITEEE 2018]
Sample Questions
Ques. What do you mean by coplanar vectors? (1 mark)
Ans. Vectors that are coplanar in a three-dimensional plane are known as coplanar vectors. These axes are always perpendicular to the plane.
Ques. Determine if the vectors x = {1; 2; 3}, y = {1; 1; 1}, and z = {1; 2; 1} are coplanar. (3 marks)
Ans. We must compute the scalar triple product to see if the three vectors x, y, and z are coplanar or not:
x.[y × z] = (1)·(1)·(1) + (1)·(1)·(2) + (1)·(2)·(3) – (1)·(1)·(3) – (1)·(1)·(2) – (1)·(1)·(2)
= 1+2+6–3–2–2
= 2
As can be seen, the scalar triple product is not zero, therefore the vectors x, y, and z are not coplanar.
Ques. Prove that x = {1; 1; 1}, y = {1; 3; 1}, and z = {2; 2; 2} are three vectors that are coplanar. (3 marks)
Ans. Calculate a scalar triple product of vectors
x·[y × z] = (1)·(2)·(3) + (1)·(1)·(2) + (1)·(1)·(2) – (1)·(2)·(3) – (1)·(1)·(2) – (1)·(1)·(2)
= 6 + 2 + 2 – 6 – 2 – 2
= 0
As can be seen, the scalar triple product equals zero, indicating that the vectors x, y, and z are coplanar.
Ques. When points A (3,2,1), B (4, x,5), C (4,2, -2), and D (6,5, -1) are coplanar, solve for the value of x. (5 marks)
Ans. The points are as follows: A (3, 2, 2), B (2, x, 5), C (4, 2, -2) and D (4, 2, -2). (6, 5, -1)
Now, AB = B's position vector minus A's position vector
=> AB = (4i + xj + 5k) - (3i + 2j + 2k) = i + (x - 2) j + 3k
AC = C's position vector minus A's position vector
=> AC = (4i + 2j - 2k) - (3i + 2j + 2k) = i + 0j - 4k
Now, AD = D's position vector minus A's position vector
=> AD = (6i + 5j - k) - (3i + 2j + 2k) = 3i + 3j - k
Because A, B, C, and D are all coplanar,
|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
Ques. Find Scalar Triple Product of the given Vectors which are (i + 2j + 3k), (- i - 2j + k) and (i + k). (3 marks)
Ans. |1 2 3|
|-1 -2 1| = 0
Scalar triple product = 1(-2-1)- 2(-1-1) + 3(0+2)
= 1(-3)- 2 (-2) + 3(2)
= -3+4+6 =7
Ques. Prove that a+b, b+c, c+a = 2(a, b, c) (3 marks)
Ans. 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 = (2a, b,c)
Ques. If the vectors i+j+2k, yi-j+k, and 3i-2j-k are coplanar, find the value of y. (3 marks)
Ans. Given that vectors are coplanar
|1 1 2|
|y -1 1|
=> 1 (1 + 2) - 1 (- y - 3) + 2 (- 2 y + 3) = 0
=> 3 + y + 3 - 4y + 6 = 0
=> -3y + 12= 0
=> y = 4
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