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# A point charge $\left( q \right)$ is placed at the origin. Let ${\vec E_A}$,${\vec E_B}$${\vec E_C}$ be the electric fields at three points $A\left( {1,2,3} \right), B\left( {1,1, - 1} \right),C\left( {2,2,2} \right)$respectively due to the charge $\left( q \right)$. Then ,$\left( 1 \right){{\vec E}_A} \bot {{\vec E}_B} \\ \left( 2 \right)\left| {{{\vec E}_b}} \right| = 4\left| {{{\vec E}_c}} \right| \\$Select the correct alternativea) Only $\left( 1 \right)$ is correctb) Only$\left( 2 \right)$ is correctc) Both $\left( 1 \right)$ and $\left( 2 \right)$ are correctd) Both $\left( 1 \right)$ and $\left( 2 \right)$ are wrong

Last updated date: 04th Mar 2024
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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.