When two vectors ** A** y

**appear on the**

*B***same straight**or are

**parallels**to a certain line, they are said to be

**collinear vectors - example**.

On the other hand, when the relations that maintain their coordinates are **equal** I the **vector product is null** (zero), they are also said to be **collinear vectors - example**.

**Definition**

Sean ** A** = (A

_{x}, TO

_{y}, TO

_{z}) Y

**= (B**

*B*_{x}, B

_{y}, B

_{z}), if these are parallel to a line or are on the same line, then they will be

**collinear vectors - example**. It doesn't matter if they have a different meaning.

**Vector collinearity conditions**

Two vectors ** A** = (A

_{x}, TO

_{y}, TO

_{z}) Y

**= (B**

*B*_{x}, B

_{y}, B

_{z}) are collinear if:

- The relationships of its coordinates are the same, that is:

( TO_{x} / B_{x} ) = (A_{y} / B_{y} ) = (A_{z} / B_{z} )

- The vector product of both vectors is null (zero), that is:

** A** ×

**= (A**

*B*_{y}B

_{z }- TO

_{z}B

_{y}) î + (A

_{x}B

_{z }- TO

_{z}B

_{x}) ĵ + (A

_{x}B

_{y }–

_{ }A

_{y}B

_{x}) k = 0

OR

** A** ×

**= |**

*B***| |**

*A***| sinθ = 0**

*B*Since θ = 0 or θ = 180 ⁰ (remember that θ is the angle between both vectors).

**Exercises**

- Determine if the vectors
= (1, -2, 1) and*A*= (-1, 3, 1) are collinear.*B*

** A** y

**they will be collinear if the relations of their coordinates are the same:**

*B*( TO_{x} / B_{x} ) = (A_{y} / B_{y} ) = (A_{z} / B_{z} )

(1 / -1) ≠ (-2 / 3) ≠ (1/1)

Vectors are not collinear.

- Determine if the vectors
= (2, 4, 12) and*A*= (1, 2, 6) are collinear.*B*

** A** y

**they will be collinear if their coordinate relationships are the same.**

*B*( TO_{x} / B_{x} ) = (A_{y} / B_{y} ) = (A_{z} / B_{z} )

( 2 / 1 ) = ( 4 / 2 ) = ( 12 / 6 )

2 = 2 = 2

Vectors are collinear.

We can also determine if they are collinear, if when calculating its vector product, it gives us zero:

** A** ×

**= (A**

*B*_{y}B

_{z }- TO

_{z}B

_{y}) î + (A

_{x}B

_{z }- TO

_{z}B

_{x}) ĵ + (A

_{x}B

_{y }–

_{ }A

_{y}B

_{x}) k =

= (4 · 6 - 12 · 2) î + (2 · 6 - 12 · 1) ĵ + (2 · 2 - 4 · 1) k =

= (24 - 24) î + (12 - 12 ĵ + (4 - 4) k =

= 0 î + 0 ĵ + 0 k

- Since the end points of two vectors are A (-1, 5, -3), B (1, 1, 5), C (-2, 1, 2) and D (-5, 7, -10). Determine if the vectors
y*AB*they are collinear.*CD*

** AB** y

**they will be collinear if the relations of their coordinates are equal and / or their vector product is null:**

*CD*(AB_{x} / CD_{x} ) = (AB_{y} / CD_{y} ) = (AB_{z} / CD_{z} )

Now, we have two points A (-1, 5, -3) and B (1, 1, 5) that correspond to the ends of the vector ** AB**Therefore, the coordinates of said vector will be:

** AB** = (B

_{x }- TO

_{x}, B

_{y }- TO

_{y}, B

_{z }- TO

_{z}) = (2, -4, 8)

We also have two points C (-2, 1, 2) and D (-5, 7, -10) that correspond to the ends of the vector ** CD**; therefore, the coordinates of said vector will be:

** CD** = (D

_{x }- C

_{x}, D

_{y }- C

_{y}, D

_{z }- C

_{z}) = (-3, 6, -12)

In consecuense:

( 2 / -3 ) = ( -4 / 6 ) = ( 8 / -12 )

– 2/3 = – 2/3 = – 2/3

Vectors are collinear.

** AB** ×

**= (AB**

*CD*_{y}CD

_{z }- AB

_{z}CD

_{y}) î + (AB

_{x}CD

_{z }- AB

_{z}CD

_{x}) ĵ + (AB

_{x}CD

_{y }–

_{ }AB

_{y}CD

_{x}) k =

= [(-4) · (-12) - 8 · 6) î + [2 · (-12) - 8 · (-3)] ĵ + [2 · 6 - (- 4) · (-3)] k =

= (48 - 48) î + (-24 - 24) ĵ + (12 - 12) k =

= 0 î + 0 ĵ + 0 k