Question

$\dag \: \footnotesize{ \bf Three \: vectors \: A , B \: and \: C \: add \: upto \: zero. } \\ \footnotesize{ \dag \: \bf Find \: which \: is \: false }$
$\footnotesize{ \sf A) \: (A \times B) \times C \: is \: not \: zero \: unless \: B, C \: are \: parallel.}$
$\footnotesize{ \sf B) \: (A \times B) .C \: is \: not \: zero \: unless \: B, C \: are \: parallel.}$
$\footnotesize{ \sf C) \: If \: A,B ,C \: define \: a \: plane \: (A \times B)\times C \:is \: in \: that \: plane.}$
$\footnotesize{ \sf D) \: (A \times B).C = |A| |B| |C| \to C^2 = A^2 + B^2 .}$
Need Explaination for each option. ​

Answer 1

Question:

$\footnotesize{ \sf Three \: vectors \: A , B \: and \: C \: add \: upto \: zero. } \footnotesize{ \: \sf Find \: which \: is \: false }$

• $\footnotesize{ \sf A) \: (A \times B) \times C \: is \: not \: zero \: unless \: B, C \: are \: parallel.}$
• $\footnotesize{ \sf B) \: (A \times B) .C \: is \: not \: zero \: unless \: B, C \: are \: parallel.}$
• $\footnotesize{ \sf C) \: If \: A,B ,C \: define \: a \: plane \: (A \times B)\times C \:is \: in \: that \: plane.}$
• $\footnotesize{ \sf D) \: (A \times B).C = |A| |B| |C| \to C^2 = A^2 + B^2 .}$

Things to know:

• Vector product of a vector with itself is always 0, $\vec{A} \times \vec{A} = A^2 sin \theta \implies A^2 sin^\circ \implies 0$
• Similarly, the vector product of two parallel vector is always 0 as the angle b/w them will be 0.
• $\vec{A} \times \vec{B} = - (\vec{B} \times \vec{A})$
• $\vec{A} \times \vec{B}$ is perpendicular to both $\vec{A} and B$. 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:

$\tt Given~that~ \vec{A} + \vec{B} + \vec{C} =0 \\\\ \tt Taking~cross~product~of~ \vec{B} ~on~both~sides. We have, : \\\\ \tt \implies \vec{B} \times ( \vec{A} + \vec{B} + \vec{C}) = 0 \times \vec{B} \\\\ \tt \implies \vec{B} \times \vec{A} + \vec{B} \times \vec{B} + \vec{B} \times \vec{C} = 0 \\\\\tt \implies \vec{B} \times \vec{A} + \vec{B} \times \vec{C} = 0 \\\\ \tt\implies \vec{B} \times \vec{A} = - (\vec{B} \times \vec{C}) \\\\ \tt \implies \vec{B} \times \vec{A} = \vec{C} \times \vec{B}$

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

Now, come to the first option.

$\footnotesize{ \sf A) \: (A \times B) \times C \: is \: not \: zero \: unless \: B, C \: are \: parallel.}$

From [i] we have, $\vec{A} \times \vec{B} = \vec{B} \times \vec{C}$

Taking vector product of C on both sides, we have:

$(\vec{A} \times \vec{B}) \times \vec{C} =( \vec{B} \times \vec{C}) \times \vec{C}$

$(\vec{A} \times \vec{B}) \times \vec{C}$ can only be zero if $\vec{B} ~and~\vec{C}$ are parallel I.e their vector product is zero

Hence, it is true.

Now, come to 2nd option.

$\footnotesize{ \sf B) \: (A \times B) .C \: is \: not \: zero \: unless \: B, C \: are \: parallel.}$

Similarly, Taking dot product of C vector on both sides in [i], we have :

$\tt (\vec{A} \times \vec{B}). \vec{C} =( \vec{B} \times \vec{C}). \vec{C}$

$(\vec{B} \times \vec{C}). \vec{C}$ will always be zero. It doesn't require $\vec{A} and \vec{B}$ to be parallel. It will gonna be zero in any case and so will be $(\vec{A} \times \vec{B}).\vec{C}$

Hence, Option B is incorrect as $(\vec{A} \times \vec{B}).\vec{C}$ can be zero even when B and C are not parallel.

Now, come to 3rd option.

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

It is given that A, B and C defines a plane.

Now $\vec{A} \times \vec{B}$lies outside that plane and will be perpendicular to $\vec{A}~and~ \vec{B}$ For an instânce, let $\vec{A}~and~ \vec{B}$ be $\vec{K}$.

In the option, we have,

$(\vec{A} \times \vec{B}) \times \vec{C} \\\\ \implies \vec{K} \times \vec{C}.$

This cross product lies perpendicular to vector K, Perpendicular to vector K means it will lie in the plane defined by A, B and C. (See the attachment).

Hence, we conclude that option C is corect.

Now come to 4th option.

$\footnotesize{ \sf D) \: (A \times B).C = |A| |B| |C| \to C^2 = A^2 + B^2 .}$

$( \vec{A} \times \vec{B}). \vec{C} \implies (|A| |B| sin \theta). \vec{C} \implies |(|A| |B| sin \theta)|~|C| cos \theta$ We can bring this $|(|A| |B| sin \theta)|~|C| cos \theta$ to |A| |B| |C| only when the angle b/w A and B is 90° as sin 90=1.

And, it is already given that, A+B+C=0, that means they are the sides of a triangle. And, hence they will satisfy, the equation $A^2+B^2=C^2$

Hence, D is also correct.

The correct option is B

Answer 2

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.

C option is correct