Question

by what number should (-6)¹ be divided so that the quotient is 27¹ ? please solve this right answer gets brainliest, follow, and thanks to all of his or her answers.
$( - 6) { - }^{4} \times x = 27 - {}^{1}$
please solve this​

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.