Answer
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Hint: 1) The vector form coordinates as written as $x\hat i + y\hat j + z\hat k$.
2) If two vectors are perpendicular, then their dot product is zero.
3) The magnitude of the vector is calculated as $\left| A \right| = \sqrt {{x^2} \times {y^2} \times {z^2}} $.
4) A unit vector for a given vector in its direction is calculated as \[\vec A = \dfrac{{x\hat i + y\hat j + z\hat k}}{{\sqrt {{x^2} + {y^2} + {z^2}} }}\].
Complete step by step solution:
We know that the electric field vector ${\vec E_A}$can be written as,
\[
{{\vec E}_A} = \dfrac{{kq}}{{{{\left| {{{\vec r}_{OA}}} \right|}^2}}} \times {{\hat r}_{OA}} \\
\therefore {{\vec E}_A} = \dfrac{{kq}}{{{{\left| {\sqrt {{1^2} + {2^2} + {3^2}} } \right|}^2}}} \times \dfrac{{1\hat i + 2\hat j + 3\hat k}}{{\sqrt {{1^2} + {2^2} + {3^2}} }} \\
\Rightarrow {{\vec E}_A} = \dfrac{{kq}}{{{{\left| {\sqrt {14} } \right|}^2}}} \times \dfrac{{1\hat i + 2\hat j + 3\hat k}}{{\sqrt {14} }} \\
\Rightarrow {{\vec E}_A} = \dfrac{{kq\left( {1\hat i + 2\hat j + 3\hat k} \right)}}{{14\sqrt {14} }} \\
\]
Now similarly we will calculate for vector${\vec E_B}$
\[
{{\vec E}_B} = \dfrac{{kq}}{{{{\left| {{{\vec r}_{OB}}} \right|}^2}}} \times {{\hat r}_{OB}} \\
\therefore {{\vec E}_B} = \dfrac{{kq}}{{{{\left| {\sqrt {{1^2} + {1^2} + {{\left( { - 1} \right)}^2}} } \right|}^2}}} \times \dfrac{{1\hat i + 1\hat j - 1\hat k}}{{\sqrt {{1^2} + {1^2} + {{\left( { - 1} \right)}^2}} }} \\
\Rightarrow {{\vec E}_B} = \dfrac{{kq}}{{{{\left| {\sqrt 3 } \right|}^2}}} \times \dfrac{{1\hat i + 1\hat j - 1\hat k}}{{\sqrt 3 }} \\
\Rightarrow {{\vec E}_B} = \dfrac{{kq\left( {1\hat i + 2\hat j + 3\hat k} \right)}}{{3\sqrt 3 }} \\
\]
similarly, we will calculate for vector ${\vec E_C}$
\[
{{\vec E}_C} = \dfrac{{kq}}{{{{\left| {{{\vec r}_{OC}}} \right|}^2}}} \times {{\hat r}_{OC}} \\
\therefore {{\vec E}_C} = \dfrac{{kq}}{{{{\left| {\sqrt {{2^2} + {2^2} + {2^2}} } \right|}^2}}} \times \dfrac{{2\hat i + 2\hat j + 2\hat k}}{{\sqrt {{2^2} + {2^2} + {2^2}} }} \\
\Rightarrow {{\vec E}_C} = \dfrac{{kq}}{{{{\left| {\sqrt {12} } \right|}^2}}} \times \dfrac{{2\hat i + 2\hat j + 2\hat k}}{{\sqrt {12} }} \\
\Rightarrow {{\vec E}_C} = \dfrac{{kq\left( {2\hat i + 2\hat j + 2\hat k} \right)}}{{12 \times 2\sqrt 3 }} \\
\]
Taking 2 OUT and canceling it by the 2 in the denominator the above equation can be written as
\[ \Rightarrow {\vec E_C} = \dfrac{{kq\left( {\hat i + \hat j + \hat k} \right)}}{{12\sqrt 3 }}\]
Now for the first option $\left( 1 \right)$:
We know that if two vectors are perpendicular to each other their dot product will be zero
If ${\vec E_A} \bot {\vec E_B}$
$\therefore {\vec E_A}.{\vec E_B} = 0$
Substituting the values of both vectors we get
\[
{{\vec E}_A}.{{\vec E}_B} = \left( {\dfrac{{kq}}{{14\sqrt {14} }}} \right)\left( {1\hat i + 2\hat j + 3\hat k} \right).\left( {\dfrac{{kq}}{{3\sqrt 3 }}} \right)\left( {1\hat i + 1\hat j - 1\hat k} \right) \\
\Rightarrow {{\vec E}_A}.{{\vec E}_B} = \left( {\dfrac{{kq}}{{14\sqrt {14} }}} \right)\left( {\dfrac{{kq}}{{3\sqrt 3 }}} \right)\left( {\left( {1\hat i.1\hat i} \right) + \left( {2\hat j.1\hat j} \right) + \left( {3\hat k.\left( { - 1\hat k} \right)} \right)} \right) \\
\Rightarrow {{\vec E}_A}.{{\vec E}_B} = \left( {\dfrac{{kq}}{{14\sqrt {14} }}} \right)\left( {\dfrac{{kq}}{{3\sqrt 3 }}} \right)\left( {1 + 2 - 3} \right) \\
\Rightarrow {{\vec E}_A}.{{\vec E}_B} = 0 \\
\]
Here we have seen that the dot product of both the vectors is zero hence the required condition is satisfied so we can say that ${\vec E_A} \bot {\vec E_B}$.
So, the option $\left( 1 \right)$is correct.
Now for the first option $\left( 2 \right)$:
We know that
$
\left| {{{\vec E}_B}} \right| = \dfrac{{kq}}{{{{\left| {{{\vec r}_{OB}}} \right|}^2}}} \\
\therefore \left| {{{\vec E}_B}} \right| = \dfrac{{kq}}{{{{\left| {\sqrt {{1^2} + {1^2} + {{\left( { - 1} \right)}^2}} } \right|}^2}}} \\
$
$ \Rightarrow \left| {{{\vec E}_B}} \right| = \dfrac{{kq}}{3}$______________________$\left( a \right)$
Now similarly,
$
\left| {{{\vec E}_C}} \right| = \dfrac{{kq}}{{{{\left| {{{\vec r}_{OC}}} \right|}^2}}} \\
\therefore \left| {{{\vec E}_C}} \right| = \dfrac{{kq}}{{{{\left| {\sqrt {{2^2} + {2^2} + {2^2}} } \right|}^2}}} \\
\\
$
$ \Rightarrow \left| {{{\vec E}_C}} \right| = \dfrac{{kq}}{{12}}$_________________________$\left( b \right)$
On dividing the equation $\left( a \right)$by$\left( b \right)$ we get
\[
\dfrac{{\left| {{{\vec E}_B}} \right|}}{{\left| {{{\vec E}_C}} \right|}} = \dfrac{{\left( {\dfrac{{kq}}{3}} \right)}}{{\left( {\dfrac{{kq}}{{12}}} \right)}} \\
\Rightarrow \dfrac{{\left| {{{\vec E}_B}} \right|}}{{\left| {{{\vec E}_C}} \right|}} = 4 \\
\\
\]
So, we can say that
\[\left| {{{\vec E}_B}} \right| = 4\left| {{{\vec E}_C}} \right|\] Hence option $\left( 2 \right)$ is also correct.
Both $\left( 1 \right)$ and $\left( 2 \right)$ are correct
Note: The unit vector in the direction vector is the vector divided by its mod
In the dot multiplication, $\left( {\hat i} \right)$is always multiplied with $\left( {\hat i} \right)$similarly $\left( {\hat j} \right)$is multiplied with $\left( {\hat j} \right)$and $\left( {\hat k} \right)$is multiplied with $\left( {\hat k} \right)$, and they all are summed together.
In dot product \[\left( {\hat i.\hat i} \right) = 1,\left( {\hat j.\hat j = 1} \right)\]and $\left( {\hat k.\hat k} \right) = 1$
In vector calculation direction of the vector is very important and must be taken care of.
If two vectors are perpendicular, their dot product will be zero.
2) If two vectors are perpendicular, then their dot product is zero.
3) The magnitude of the vector is calculated as $\left| A \right| = \sqrt {{x^2} \times {y^2} \times {z^2}} $.
4) A unit vector for a given vector in its direction is calculated as \[\vec A = \dfrac{{x\hat i + y\hat j + z\hat k}}{{\sqrt {{x^2} + {y^2} + {z^2}} }}\].
Complete step by step solution:
We know that the electric field vector ${\vec E_A}$can be written as,
\[
{{\vec E}_A} = \dfrac{{kq}}{{{{\left| {{{\vec r}_{OA}}} \right|}^2}}} \times {{\hat r}_{OA}} \\
\therefore {{\vec E}_A} = \dfrac{{kq}}{{{{\left| {\sqrt {{1^2} + {2^2} + {3^2}} } \right|}^2}}} \times \dfrac{{1\hat i + 2\hat j + 3\hat k}}{{\sqrt {{1^2} + {2^2} + {3^2}} }} \\
\Rightarrow {{\vec E}_A} = \dfrac{{kq}}{{{{\left| {\sqrt {14} } \right|}^2}}} \times \dfrac{{1\hat i + 2\hat j + 3\hat k}}{{\sqrt {14} }} \\
\Rightarrow {{\vec E}_A} = \dfrac{{kq\left( {1\hat i + 2\hat j + 3\hat k} \right)}}{{14\sqrt {14} }} \\
\]
Now similarly we will calculate for vector${\vec E_B}$
\[
{{\vec E}_B} = \dfrac{{kq}}{{{{\left| {{{\vec r}_{OB}}} \right|}^2}}} \times {{\hat r}_{OB}} \\
\therefore {{\vec E}_B} = \dfrac{{kq}}{{{{\left| {\sqrt {{1^2} + {1^2} + {{\left( { - 1} \right)}^2}} } \right|}^2}}} \times \dfrac{{1\hat i + 1\hat j - 1\hat k}}{{\sqrt {{1^2} + {1^2} + {{\left( { - 1} \right)}^2}} }} \\
\Rightarrow {{\vec E}_B} = \dfrac{{kq}}{{{{\left| {\sqrt 3 } \right|}^2}}} \times \dfrac{{1\hat i + 1\hat j - 1\hat k}}{{\sqrt 3 }} \\
\Rightarrow {{\vec E}_B} = \dfrac{{kq\left( {1\hat i + 2\hat j + 3\hat k} \right)}}{{3\sqrt 3 }} \\
\]
similarly, we will calculate for vector ${\vec E_C}$
\[
{{\vec E}_C} = \dfrac{{kq}}{{{{\left| {{{\vec r}_{OC}}} \right|}^2}}} \times {{\hat r}_{OC}} \\
\therefore {{\vec E}_C} = \dfrac{{kq}}{{{{\left| {\sqrt {{2^2} + {2^2} + {2^2}} } \right|}^2}}} \times \dfrac{{2\hat i + 2\hat j + 2\hat k}}{{\sqrt {{2^2} + {2^2} + {2^2}} }} \\
\Rightarrow {{\vec E}_C} = \dfrac{{kq}}{{{{\left| {\sqrt {12} } \right|}^2}}} \times \dfrac{{2\hat i + 2\hat j + 2\hat k}}{{\sqrt {12} }} \\
\Rightarrow {{\vec E}_C} = \dfrac{{kq\left( {2\hat i + 2\hat j + 2\hat k} \right)}}{{12 \times 2\sqrt 3 }} \\
\]
Taking 2 OUT and canceling it by the 2 in the denominator the above equation can be written as
\[ \Rightarrow {\vec E_C} = \dfrac{{kq\left( {\hat i + \hat j + \hat k} \right)}}{{12\sqrt 3 }}\]
Now for the first option $\left( 1 \right)$:
We know that if two vectors are perpendicular to each other their dot product will be zero
If ${\vec E_A} \bot {\vec E_B}$
$\therefore {\vec E_A}.{\vec E_B} = 0$
Substituting the values of both vectors we get
\[
{{\vec E}_A}.{{\vec E}_B} = \left( {\dfrac{{kq}}{{14\sqrt {14} }}} \right)\left( {1\hat i + 2\hat j + 3\hat k} \right).\left( {\dfrac{{kq}}{{3\sqrt 3 }}} \right)\left( {1\hat i + 1\hat j - 1\hat k} \right) \\
\Rightarrow {{\vec E}_A}.{{\vec E}_B} = \left( {\dfrac{{kq}}{{14\sqrt {14} }}} \right)\left( {\dfrac{{kq}}{{3\sqrt 3 }}} \right)\left( {\left( {1\hat i.1\hat i} \right) + \left( {2\hat j.1\hat j} \right) + \left( {3\hat k.\left( { - 1\hat k} \right)} \right)} \right) \\
\Rightarrow {{\vec E}_A}.{{\vec E}_B} = \left( {\dfrac{{kq}}{{14\sqrt {14} }}} \right)\left( {\dfrac{{kq}}{{3\sqrt 3 }}} \right)\left( {1 + 2 - 3} \right) \\
\Rightarrow {{\vec E}_A}.{{\vec E}_B} = 0 \\
\]
Here we have seen that the dot product of both the vectors is zero hence the required condition is satisfied so we can say that ${\vec E_A} \bot {\vec E_B}$.
So, the option $\left( 1 \right)$is correct.
Now for the first option $\left( 2 \right)$:
We know that
$
\left| {{{\vec E}_B}} \right| = \dfrac{{kq}}{{{{\left| {{{\vec r}_{OB}}} \right|}^2}}} \\
\therefore \left| {{{\vec E}_B}} \right| = \dfrac{{kq}}{{{{\left| {\sqrt {{1^2} + {1^2} + {{\left( { - 1} \right)}^2}} } \right|}^2}}} \\
$
$ \Rightarrow \left| {{{\vec E}_B}} \right| = \dfrac{{kq}}{3}$______________________$\left( a \right)$
Now similarly,
$
\left| {{{\vec E}_C}} \right| = \dfrac{{kq}}{{{{\left| {{{\vec r}_{OC}}} \right|}^2}}} \\
\therefore \left| {{{\vec E}_C}} \right| = \dfrac{{kq}}{{{{\left| {\sqrt {{2^2} + {2^2} + {2^2}} } \right|}^2}}} \\
\\
$
$ \Rightarrow \left| {{{\vec E}_C}} \right| = \dfrac{{kq}}{{12}}$_________________________$\left( b \right)$
On dividing the equation $\left( a \right)$by$\left( b \right)$ we get
\[
\dfrac{{\left| {{{\vec E}_B}} \right|}}{{\left| {{{\vec E}_C}} \right|}} = \dfrac{{\left( {\dfrac{{kq}}{3}} \right)}}{{\left( {\dfrac{{kq}}{{12}}} \right)}} \\
\Rightarrow \dfrac{{\left| {{{\vec E}_B}} \right|}}{{\left| {{{\vec E}_C}} \right|}} = 4 \\
\\
\]
So, we can say that
\[\left| {{{\vec E}_B}} \right| = 4\left| {{{\vec E}_C}} \right|\] Hence option $\left( 2 \right)$ is also correct.
Both $\left( 1 \right)$ and $\left( 2 \right)$ are correct
Note: The unit vector in the direction vector is the vector divided by its mod
In the dot multiplication, $\left( {\hat i} \right)$is always multiplied with $\left( {\hat i} \right)$similarly $\left( {\hat j} \right)$is multiplied with $\left( {\hat j} \right)$and $\left( {\hat k} \right)$is multiplied with $\left( {\hat k} \right)$, and they all are summed together.
In dot product \[\left( {\hat i.\hat i} \right) = 1,\left( {\hat j.\hat j = 1} \right)\]and $\left( {\hat k.\hat k} \right) = 1$
In vector calculation direction of the vector is very important and must be taken care of.
If two vectors are perpendicular, their dot product will be zero.
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