# Collinear Vectors When two vectors A y B appear on the 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 = (Ax, TOy, TOz) Y B = (Bx, By, Bz), 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 = (Ax, TOy, TOz) Y B = (Bx, By, Bz) are collinear if:

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

( TOx / Bx ) = (Ay / By ) = (Az / Bz )

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

A × B = (AyBz - TOzBy) î + (AxBz - TOzBx) ĵ + (AxBy   AyBx) k = 0

OR

A × B = |A| |B| sinθ = 0

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

## Exercises

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

A y B they will be collinear if the relations of their coordinates are the same:

( TOx / Bx ) = (Ay / By ) = (Az / Bz )

(1 / -1) ≠ (-2 / 3) ≠ (1/1)

Vectors are not collinear.

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

A y B they will be collinear if their coordinate relationships are the same.

( TOx / Bx ) = (Ay / By ) = (Az / Bz )

( 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 × B = (AyBz - TOzBy) î + (AxBz - TOzBx) ĵ + (AxBy   AyBx) 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 AB y CD they are collinear.

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

(ABx / CDx ) = (ABy / CDy ) = (ABz / CDz )

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

AB = (Bx - TOx, By - TOy, Bz - TOz) = (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 = (Dx - Cx, Dy - Cy, Dz - Cz) = (-3, 6, -12)

In consecuense:

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

– 2/3 = – 2/3 = – 2/3

Vectors are collinear.

AB × CD = (AByCDz - ABzCDy) î + (ABxCDz - ABzCDx) ĵ + (ABxCDy   AByCDx) 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