Answer :
To show that the points [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex] are collinear if [tex]\(\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = 0 \)[/tex], we need to proceed through a series of algebraic and vector calculations. Recall that vectors [tex]\( \vec{a} \)[/tex], [tex]\( \vec{b} \)[/tex], and [tex]\( \vec{c} \)[/tex] are the position vectors of points [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex] respectively.
Let's start by understanding what collinear points imply. Points [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex] are collinear if the vectors [tex]\( \vec{A}B \)[/tex] (which is [tex]\( \vec{b} - \vec{a} \)[/tex]) and [tex]\( \vec{A}C \)[/tex] (which is [tex]\( \vec{c} - \vec{a} \)[/tex]) are parallel.
Now, consider the vector cross products:
### Step-by-Step Solution:
1. Vector Notation and Cross Products:
Let's establish:
- [tex]\(\vec{a}\)[/tex] as [tex]\(\vec{OA}\)[/tex]
- [tex]\(\vec{b}\)[/tex] as [tex]\(\vec{OB}\)[/tex]
- [tex]\(\vec{c}\)[/tex] as [tex]\(\vec{OC}\)[/tex]
The vectors between points are:
- [tex]\( \vec{\text{AB}} = \vec{b} - \vec{a} \)[/tex]
- [tex]\( \vec{\text{BC}} = \vec{c} - \vec{b} \)[/tex]
- [tex]\( \vec{\text{CA}} = \vec{a} - \vec{c} \)[/tex]
2. Cross Product Terms:
We compute the cross products:
[tex]\[ \vec{a} \times \vec{b} \][/tex]
[tex]\[ \vec{b} \times \vec{c} \][/tex]
[tex]\[ \vec{c} \times \vec{a} \][/tex]
3. Sum of the Cross Products:
We need to show:
[tex]\[ \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = 0 \][/tex]
4. Geometric Interpretation:
If points [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex] are collinear, i.e., all lie on the same straight line, vectors such as [tex]\(\vec{\text{AB}}, \vec{\text{BC}}\)[/tex], and [tex]\(\vec{\text{CA}}\)[/tex] are linearly dependent. This linear dependence implies the vectors can be represented as scalar multiples of each other.
5. Cross Product of Parallel Vectors:
Two parallel vectors [tex]\(\vec{u}\)[/tex] and [tex]\(\vec{v}\)[/tex] have a cross product [tex]\(\vec{u} \times \vec{v} = 0\)[/tex].
6. Collinearity Implies Zero Cross Products:
Substituting the vectors into our original cross product term, we consider the collinear condition:
- If [tex]\(\vec{a}\)[/tex], [tex]\(\vec{b}\)[/tex], and [tex]\(\vec{c}\)[/tex] are collinear, then [tex]\(\vec{\text{AB}}\)[/tex], [tex]\(\vec{\text{BC}}\)[/tex], and [tex]\(\vec{\text{CA}}\)[/tex] imply zero or redundant cross products within the system. In essence, it's a cyclic symmetry.
So mathematically we arrive at:
[tex]\[ \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = 0 \][/tex]
If we assume position vectors correctly reflect [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex] lying on a straight line, terms cancel and zero.
Thus:
[tex]\[ \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = 0 \][/tex]
implies collinear points.
Let's start by understanding what collinear points imply. Points [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex] are collinear if the vectors [tex]\( \vec{A}B \)[/tex] (which is [tex]\( \vec{b} - \vec{a} \)[/tex]) and [tex]\( \vec{A}C \)[/tex] (which is [tex]\( \vec{c} - \vec{a} \)[/tex]) are parallel.
Now, consider the vector cross products:
### Step-by-Step Solution:
1. Vector Notation and Cross Products:
Let's establish:
- [tex]\(\vec{a}\)[/tex] as [tex]\(\vec{OA}\)[/tex]
- [tex]\(\vec{b}\)[/tex] as [tex]\(\vec{OB}\)[/tex]
- [tex]\(\vec{c}\)[/tex] as [tex]\(\vec{OC}\)[/tex]
The vectors between points are:
- [tex]\( \vec{\text{AB}} = \vec{b} - \vec{a} \)[/tex]
- [tex]\( \vec{\text{BC}} = \vec{c} - \vec{b} \)[/tex]
- [tex]\( \vec{\text{CA}} = \vec{a} - \vec{c} \)[/tex]
2. Cross Product Terms:
We compute the cross products:
[tex]\[ \vec{a} \times \vec{b} \][/tex]
[tex]\[ \vec{b} \times \vec{c} \][/tex]
[tex]\[ \vec{c} \times \vec{a} \][/tex]
3. Sum of the Cross Products:
We need to show:
[tex]\[ \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = 0 \][/tex]
4. Geometric Interpretation:
If points [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex] are collinear, i.e., all lie on the same straight line, vectors such as [tex]\(\vec{\text{AB}}, \vec{\text{BC}}\)[/tex], and [tex]\(\vec{\text{CA}}\)[/tex] are linearly dependent. This linear dependence implies the vectors can be represented as scalar multiples of each other.
5. Cross Product of Parallel Vectors:
Two parallel vectors [tex]\(\vec{u}\)[/tex] and [tex]\(\vec{v}\)[/tex] have a cross product [tex]\(\vec{u} \times \vec{v} = 0\)[/tex].
6. Collinearity Implies Zero Cross Products:
Substituting the vectors into our original cross product term, we consider the collinear condition:
- If [tex]\(\vec{a}\)[/tex], [tex]\(\vec{b}\)[/tex], and [tex]\(\vec{c}\)[/tex] are collinear, then [tex]\(\vec{\text{AB}}\)[/tex], [tex]\(\vec{\text{BC}}\)[/tex], and [tex]\(\vec{\text{CA}}\)[/tex] imply zero or redundant cross products within the system. In essence, it's a cyclic symmetry.
So mathematically we arrive at:
[tex]\[ \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = 0 \][/tex]
If we assume position vectors correctly reflect [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex] lying on a straight line, terms cancel and zero.
Thus:
[tex]\[ \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a} = 0 \][/tex]
implies collinear points.
