Question

in detail, discuss plan A, B and C and the reasons for their respective failures​

Answer 1

Answer:

Three vectors A,B andC add upto zero.Find which is false

\begin{gathered}\footnotesize{ \sf A) \: (A \times B) \times C \: is \: not \: zero \: unless \: B, C \: are \: parallel.} \\ \end{gathered}A)(A×B)×CisnotzerounlessB,Careparallel.

\begin{gathered}\footnotesize{ \sf B) \: (A \times B) .C \: is \: not \: zero \: unless \: B, C \: are \: parallel.} \\\end{gathered}B)(A×B).CisnotzerounlessB,Careparallel.

\footnotesize{ \sf C) \: If \: A,B ,C \: define \: a \: plane \: (A \times B)\times C \:is \: in \: that \: plane.}C)IfA,B,Cdefineaplane(A×B)×Cisinthatplane.

\begin{gathered}\footnotesize{ \sf D) \: (A \times B).C = |A| |B| |C| \to C^2 = A^2 + B^2 .} \\\end{gathered}D)(A×B).C=∣A∣∣B∣∣C∣→C2=A2+B2.

Things to know:

Vector product of a vector with itself is always 0, \vec{A} \times \vec{A} = A^2 sin \theta \implies A^2 sin^\circ \implies 0A×A=A2sinθ⟹A2sin∘⟹0

Similarly, the vector product of two parallel vector is always 0 as the angle b/w them will be 0.

\vec{A} \times \vec{B} = - (\vec{B} \times \vec{A})A×B=−(B×A)

\vec{A} \times \vec{B}A×B is perpendicular to both \vec{A} and BAandB . This can be proved with right hand thumb rule. The cross product of two vectors is always directed perpendicularly outwards to the plane and hence it is parallel to its operands.

Solution:

\begin{gathered} \tt Given~that~ \vec{A} + \vec{B} + \vec{C} =0 \\\\ \tt Taking~cross~product~of~ \vec{B} ~on~both~sides. We have, : \\\\ \tt \implies \vec{B} \times ( \vec{A} + \vec{B} + \vec{C}) = 0 \times \vec{B} \\\\ \tt \implies \vec{B} \times \vec{A} + \vec{B} \times \vec{B} + \vec{B} \times \vec{C} = 0 \\\\\tt \implies \vec{B} \times \vec{A} + \vec{B} \times \vec{C} = 0 \\\\ \tt\implies \vec{B} \times \vec{A} = - (\vec{B} \times \vec{C}) \\\\ \tt \implies \vec{B} \times \vec{A} = \vec{C} \times \vec{B}\end{gathered}Given that A+B+C=0Taking cross product of B on both sides.Wehave,:⟹B×(A+B+C)=0×B⟹B×A+B×B+B×C=0⟹B×A+B×C=0⟹B×A=−(B×C)⟹B×A=C×B

It can also be written as \vec{A} \times \vec{B} = \vec{B} \times \vec{C} - - - - - [i]A×B=B×C−−−−−[i]

Now, come to the first option.