Show that the vectors a b c are coplanar if and only if a + b, b + c, c plus are coplanar
Answers
Answer:
The proof is given below.
Step-by-step explanation:
Given the vectors a b c . we have to prove that if a b c are coplanar then
As we know, if the three vectors are coplanar then their scalar triple product equals to 0.
Therefore,
Now,
⇔
⇔
⇔
⇔
⇔
Hence, a + b, b + c, c plus are coplanar.
By reversing steps we can prove that if a + b, b + c, c plus are coplanar then a, b , c are coplanar.
Hence, a b c are coplanar if and only if a + b, b + c, c plus are coplanar
Answer:
The three vectors are coplanar if their scalar triple product is zero.
i.e. a,b and c are coplanar that means:
a·(b×c)=0. ( This expression is same as: [a,b,c] )
Now:
[a+b,b+c,c+a]=(a+b)·[(b+c)×(c+a)]
=(a+b)·[b×c+b×a+c×c+c×a]
= (a+b)··[b×c+b×a+0+c×a]
(since cross product of same vector is zero.
as c×c=|c|.|c| sin θ
and θ=0 for same vectors.
so, sin θ=0 )
Hence,
[a+b,b+c,c+a]=(a+b).(b×c+b×a+c×a)
=a·(b×c)+a·(b×a)+a·(c×a)+b·(b×c)+b·(b×a)+b·(c×a)
=[a,b,c]+[a,b,a]+[a,c,a]+[b,b,c]+[b,b,a]+[b,c,a]
Also,
[a,b,a]=[a,c,a]=[b,b,c]=[b,b,a]=0
⇒ [a+b,b+c,c+a]=[a,b,c]+[b,c,a]
⇒ [a+b,b+c,c+a]=2[a,b,c]
Hence, [a+b,b+c,c+a]=0 if and only if [a,b,c]=0.
Hence, we can say that:
the vectors a,b,c are coplanar if and only if a + b, b + c, c plus are coplanar.