Solution
Three vectors are coplanar if their scalar triple product is zero. This can be represented by a determinant of their coefficients.
Step 1: Set up the determinant.
a11amp;1amp;bamp;1amp;1amp;1amp;c=0
Step 2: Apply row operations to simplify.
To create terms like (1−a), (1−b), and (1−c), we perform the following operations:
R2→R2−R1
R3→R3−R1
a1−a1−aamp;1amp;b−1amp;0amp;1amp;0amp;c−1=0
Step 3: Expand the determinant.
Expanding along the first row:
a[(b−1)(c−1)−0]−1[(1−a)(c−1)−0]+1[0−(1−a)(b−1)]=0
a(b−1)(c−1)−(1−a)(c−1)−(1−a)(b−1)=0
Step 4: Divide by (1−a)(1−b)(1−c).
Note that (b−1)=−(1−b) and (c−1)=−(1−c).
Substituting these:
a[−(1−b)][−(1−c)]−(1−a)[−(1−c)]−(1−a)[−(1−b)]=0
a(1−b)(1−c)+(1−a)(1−c)+(1−a)(1−b)=0
Now, divide the entire equation by (1−a)(1−b)(1−c):
(1−a)(1−b)(1−c)a(1−b)(1−c)+(1−a)(1−b)(1−c)(1−a)(1−c)+(1−a)(1−b)(1−c)(1−a)(1−b)=0
Step 5: Final Adjustment.
The question asks for 1−a1+1−b1+1−c1.
We can rewrite 1−aa as follows:
1−aa=1−aa−1+1=1−a−(1−a)+1=−1+1−a1
Substituting this back into our result:
Final Answer:
