Question

IF A.B=C.B, which of the following can not be a possible case : (1) Ā=Č (2) angle between A and B = 30° and angle between B and C = 150° all three vectors are of equal magnitude. (3) B is a null vector (4) A, B and C are along x,y and z-axis respectively​

Answer 1

If $\boldsymbol{\vec{A} .\vec{B}=\vec{C}. \vec{B}}$ then the option (2) is not a possible case. That is Option (2.) The angle between $\boldsymbol{\vec{A}}$ and $\boldsymbol{\vec{B}}$ = 30° and the angle between $\boldsymbol{\vec{B}}$ and $\boldsymbol{\vec{C}}$ is equal to 150° and all three vectors are of equal magnitude is not a possible case for this condition.

Explanation:

→ The scalar product or dot product of two vectors $\boldsymbol{\vec{P}}$ and $\boldsymbol{\vec{Q}}$ is given as:

   $\boldsymbol{\vec{P}=(p_{1}) \hat{\textbf{\i}}+(p_{2}) \hat{\textbf{\j}}+(p_{3} )\hat{\textbf{k}}}$

   $\boldsymbol{\vec{Q}=(q_{1}) \hat{\textbf{\i}}+(q_{2}) \hat{\textbf{\j}}+(q_{3} )\hat{\textbf{k}}}$

$\boldsymbol{\therefore Dot\:Product=\vec{P}.\vec{Q}=|\vec{P}||\vec{Q}|cos\theta=(p_{1}q_{1} +p_{2} q_{2} +p_{3} q_{3} )}$

   here 'θ' is the angle between the vectors $\boldsymbol{\vec{P}}$ and $\boldsymbol{\vec{Q}}$.

→ Some basic properties of the scalar product or the dot product are:

(i) We get the maximum value of the dot product if the angle between the two vectors is equal to zero that is when the two vectors are parallel.

(ii) We get the value of the dot product equal to zero if the angle between the two vectors is equal to 90° that is when the two vectors are perpendicular to each other.

→ In the given question:

Option (1): $\boldsymbol{\vec{A}}$ can or cannot be equal to $\boldsymbol{\vec{C}}$. Therefore $\boldsymbol{\vec{A}=\vec{C}}$ is a possible case.

$\boldsymbol{\because \vec{A}=\vec{C}}$

$\boldsymbol{\therefore \vec{A} .\vec{B}=\vec{C}. \vec{B}}$

Hence this is a possible case.

Option (2): If the angle between $\boldsymbol{\vec{A}}$ and $\boldsymbol{\vec{B}}$ = 30° and the angle between $\boldsymbol{\vec{B}}$ and $\boldsymbol{\vec{C}}$ is equal to 150° and all three vectors are of equal magnitude.

Let the magnitude of the vectors $\boldsymbol{\vec{A},\boldsymbol{\vec{B}}\:and\:\boldsymbol{\vec{C}}}$ be 'a'.

$\boldsymbol{\therefore \vec{A}.\vec{B}=|\vec{A}||\vec{B}|cos(30^{0}) =(a)(a)(\frac{\sqrt{3} }{2} )=\frac{\sqrt{3 }a^{2} }{2} }$

$\boldsymbol{\therefore \vec{C}.\vec{B}=|\vec{C}||\vec{B}|cos(150^{0}) =(a)(a)(\frac{-\sqrt{3} }{2} )=\frac{-\sqrt{3 }a^{2} }{2} }$

$\boldsymbol{\therefore \vec{A} .\vec{B}\neq \vec{C}. \vec{B}}$

Therefore $\boldsymbol{\vec{A} .\vec{B}}$ and $\boldsymbol{\vec{C} .\vec{B}}$ cannot be equal as they are opposite in sign. Hence this is not a possible case.

Option (3): $\boldsymbol{\vec{B}}$ can be a null vector making both $\boldsymbol{\vec{A} .\vec{B}}$ and $\boldsymbol{\vec{B} .\vec{C}}$ equal to zero.

$\boldsymbol{\therefore \vec{A}.\vec{B}=|\vec{A}||\vec{B}|cos(\theta) =|\vec{A}|(0)\cos(\theta) =0}$

$\boldsymbol{\therefore \vec{C}.\vec{B}=|\vec{C}||\vec{B}|cos(\theta) =|\vec{C}|(0)\cos(\theta) =0}$

$\boldsymbol{\therefore \vec{A} .\vec{B}=\vec{C}. \vec{B}}$

Hence this is a possible case.

Option (4): $\boldsymbol{\vec{A},\vec{B}\:and\:\vec{C}}$ can be along the x, y and z-axis making the angle between $\boldsymbol{\vec{A}}$ and $\boldsymbol{\vec{B}}$ as well as the angle between $\boldsymbol{\vec{B}}$ and $\boldsymbol{\vec{C}}$ equal to 90°. Thus making both $\boldsymbol{\vec{A} .\vec{B}}$ and $\boldsymbol{\vec{B} .\vec{C}}$ equal to zero.

$\boldsymbol{\therefore \vec{A}.\vec{B}=|\vec{A}||\vec{B}|cos(\theta) =|\vec{A}||\vec{B}|cos(90^{0} )=0}$

$\boldsymbol{\therefore \vec{C}.\vec{B}=|\vec{C}||\vec{B}|cos(\theta) =|\vec{C}||\vec{B}|cos(90^{0} ) =0}$

$\boldsymbol{\therefore \vec{A} .\vec{B}=\vec{C}. \vec{B}}$

Hence this is a possible case.

Therefore if $\boldsymbol{\vec{A} .\vec{B}=\vec{C}. \vec{B}}$ then the option (2) is not a possible case. That is Option (2.) The angle between $\boldsymbol{\vec{A}}$ and $\boldsymbol{\vec{B}}$ = 30° and the angle between $\boldsymbol{\vec{B}}$ and $\boldsymbol{\vec{C}}$ is equal to 150° and all three vectors are of equal magnitude is not a possible case for this condition.

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Answer 2

Answer: Only case 2 can not be possible.

Given: A.B = C.B

To Find: Which case can not be possible.

Explanation:

First of all we have to solve the given expression to check the given possible cases.

$\bold{ A.B=C.B}\\|A|.|B|\ cos\theta_1=|C||B|\ cos\theta_2\\|A|\ cos\theta_1=|C|\cos \theta_2$

Case 1: If angle between A and B is equal to angle between C and B with equal magnitude of all vectors then we can say that A=C. It can be possible with some conditions.

Case 2: If angle between A and B is 30° and angle between B and C is 150° and all vectors are equal in magnitude then

$|A|cos\theta_1=|C|\cos \theta_2\\cos30 = cos 150\\\frac{\sqrt3}{2} =-\frac{\sqrt3}{2}$

It can not possible.

Case 3: If B is a null vector. then also A.B = C.B. And it can be possible.

Case 4: If A, B, and C are along x, y, z axis then angle between all the three vectors are 90° and as we know that dot product of two mutually perpendicular vectors are zero. Then A.B = C.B. It can be possible.

For more questions please follow the given link below.

https://brainly.in/question/39679177

https://brainly.in/question/40007651

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