Collinear vectors
Fig. 1 |
Condition of vectors collinearity
Two vectors are collinear, if any of these conditions done:a = n · b
N.B. Condition 2 is not valid if one of the components of the vector is zero.
N.B. Condition 3 applies only to three-dimensional (spatial) problems.
The proof of the condition of collinearity 3
Let there are two collinear vectors a = {a_{x}; a_{y}; a_{z}} and b = {na_{x}; na_{y}; na_{z}}. We find their cross product
a × b = | i | j | k | = i (a_{y}b_{z} - a_{z}b_{y}) - j (a_{x}b_{z} - a_{z}b_{x}) + k (a_{x}b_{y} - a_{y}b_{x}) = |
a_{x} | a_{y} | a_{z} | ||
b_{x} | b_{y} | b_{z} |
= i (a_{y}na_{z} - a_{z}na_{y}) - j (a_{x}na_{z} - a_{z}na_{x}) + k (a_{x}na_{y} - a_{y}na_{x}) = 0i + 0j + 0k = 0
Examples of tasks
Examples of plane tasks
Solution: Since the vectors does not contain a components equal to zero, then use the condition of collinearity 2, which in the case of the plane problem for vectors a and b will view:
a_{x} | = | a_{y} | . |
b_{x} | b_{y} |
Means:
Vectors a and b are collinear because | 1 | = | 2 | . |
4 | 8 |
Vectors a and с are not collinear because | 1 | ≠ | 2 | . |
5 | 9 |
Vectors с and b are not collinear because | 5 | ≠ | 9 | . |
4 | 8 |
Solution: Since the vector components contain zero, then use the condition of collinearity 1, we find there is a number n for which:
For this we find a nonzero component of vector a in this case this is a_{y}. If the vectors are collinear then
n = | b_{y} | = | 6 | = 2 |
a_{y} | 3 |
Calculate the value of na:
na = {2 · 0; 2 · 3} = {0; 6}Since b = 2a, the vectors a and b are collinear.
Solution: Since the vectors does not contain a components equal to zero, then use the condition of collinearity 2
a_{x} | = | a_{y} | . |
b_{x} | b_{y} |
Means:
3 | = | 2 | . |
9 | n |
Solve this equation:
n = | 2 · 9 | = 6 |
3 |
Answer: vectors a and b are collinear when n = 6.
Examples of spatial tasks
Solution: Since the vectors does not contain a components equal to zero, then use the condition of collinearity 2, which in the case of the plane problem for vectors a and b will view:
a_{x} | = | a_{y} | = | a_{z} | . |
b_{x} | b_{y} | b_{z} |
Means:
Vectors a and b are collinear because | 1 | = | 2 | = | 3 | . |
4 | 8 | 12 |
Vectors a and с are not collinear because | 1 | = | 2 | ≠ | 3 | . |
5 | 10 | 12 |
Vectors с and b are not collinear because | 5 | = | 10 | ≠ | 12 | . |
4 | 8 | 12 |
Solution: Since the vector components contain zero, then use the condition of collinearity 1, we find there is a number n for which:
For this we find a nonzero component of vector a in this case this is a_{y}. If the vectors are collinear then
n = | b_{y} | = | 6 | = 2 |
a_{y} | 3 |
Calculate the value of na:
na = {2 · 0; 2 · 3; 2 · 1} = {0; 6; 2}Since b = 2a, the vectors a and b are collinear.
Solution: Since the vectors does not contain a components equal to zero, then use the condition of collinearity 2
a_{x} | = | a_{y} | = | a_{z} | . |
b_{x} | b_{y} | b_{z} |
Means:
3 | = | 2 | = | m |
9 | n | 12 |
From this relations we obtain two equations:
3 | = | 2 |
9 | n |
3 | = | m |
9 | 12 |
Solve this equations:
n = | 2 · 9 | = 6 |
3 |
m = | 3 · 12 | = 4 |
9 |
Answer: vectors a and b are collinear when n = 6 and m = 4.
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