Three vectors A, B, and C add up to zero. Find which is false, a) vector (A \times B) C is not zero unless vectors B, C are parallel d) vector (A \times B) . C is not zero unless vectors B, C are parallel c) if vectors A, B, C define a plane, (A \times B) C is in that plane d) (A \times B) \cdot C= |A||B||C| \quad such that C ^{2}= A ^{2}+ B ^{2}
Three vectors A, B, and C add up to zero. Find which is false, a) vector (A \times B) C is not zero unless vectors B, C are parallel d) vector (A \times B) . C is not zero unless vectors B, C are parallel c) if vectors A, B, C define a plane, (A \times B) C is in that plane d) (A \times B) \cdot C= |A||B||C| \quad such that C ^{2}= A ^{2}+ B ^{2}

Answer: The correct answer is c ) if vectors A, B, C define a plane, (A \times B) C is in that plane and d) (A \times B) \cdot C= |A \| B||C| \quad such that C ^{2}= A ^{2}+ B ^{2}

Given that \overrightarrow{ A }+\overrightarrow{ B }+\overrightarrow{ C }=0
Now, if the vector triple product of A, B, and C is true, then the vector will always lie on the plane that will be produced by the three variables A, B, and C, respectively. It entails \vec{A}+\vec{B}+\vec{C}=0 will always lie in a single plane forming sides of the triangle.
Intially take,

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\[\overrightarrow{ A } \times \overrightarrow{ B }=\overrightarrow{ B } \times \overrightarrow{ C\]

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Dot product with C on both sides of the above equation is taken into consideration here.

(A×B)·C=(B×C)·C(\overrightarrow{ A } \times \overrightarrow{ B }) \cdot \overrightarrow{ C }=(\overrightarrow{ B } \times \overrightarrow{ C }) \cdot \overrightarrow{ C }

Now it’s zero on two conditions. First, B and C are parallel. But C need not be parallel to B. We can use any vector perpendicular (say, P) to both B and C to compute the cross product. The dot product of P and C is also zero because the angle between them is always 90°. So B is untrue.