Question

For vector $\overrightarrow{a}$, if $\left| \overrightarrow{a} \right|=a$, then write the value of ${{\left( \overrightarrow{a}\times \hat{i} \right)}^{2}}+{{\left( \overrightarrow{a}\times \hat{j} \right)}^{2}}+{{\left( \overrightarrow{a}\times \hat{k} \right)}^{2}}$.

Answer 1

Hint: First find the value of a in terms of x, y, z by applying dot product on $\left| \overrightarrow{a} \right|$. After that find the values of $\left( \overrightarrow{a}\times \hat{i} \right)$, $\left( \overrightarrow{a}\times \hat{j} \right)$ and $\left( \overrightarrow{a}\times \hat{k} \right)$ by applying cross product rule. Then square each term and add them. Now take commonly from them and substitute the values derived from the $\left| \overrightarrow{a} \right|$.

Complete step by step answer:
Given: $\left| \overrightarrow{a} \right|=a$
Let the vector be $\overrightarrow{a}=x\hat{i}+y\hat{j}+z\hat{k}$.
Since $\left| \overrightarrow{a} \right|=a$. Then,
$\Rightarrow \sqrt{{{\left( x\hat{i} \right)}^{2}}+{{\left( y\hat{j} \right)}^{2}}+{{\left( z\hat{k} \right)}^{2}}}=a$
As we know the dot product of any unit vector with any other is zero, $\hat{i}\cdot \hat{j}=\hat{j}\cdot \hat{k}=\hat{k}\cdot \hat{i}=0$. So,
$\Rightarrow \sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}={{a}^{2}}$
Square both sides of the equation,
$\Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{a}^{2}}$ …………….….. (1)
Now,
$\Rightarrow \overrightarrow{a}\times \hat{i}=\left( x\hat{i}+y\hat{j}+z\hat{k} \right)\times \hat{i}$
Multiply the terms on the right side,
$\Rightarrow \overrightarrow{a}\times \hat{i}=x\hat{i}\times \hat{i}+y\hat{j}\times \hat{i}+z\hat{k}\times \hat{i}$
Since the cross product must be perpendicular to the two-unit vectors, it must be equal to the other unit vector or the opposite of that unit vector. The cross product of any unit vector with itself is zero, $\hat{i}\times \hat{i}=\hat{j}\times \hat{j}=\hat{k}\times \hat{k}=0$.
$\Rightarrow \overrightarrow{a}\times \hat{i}=x\left( 0 \right)+y\left( -\hat{k} \right)+z\left( {\hat{j}} \right)$
Multiply the terms,
$\Rightarrow \overrightarrow{a}\times \hat{i}=-y\hat{k}+z\hat{j}$
Similarly,
$\Rightarrow \overrightarrow{a}\times \hat{j}=x\hat{k}-z\hat{i}$
$\Rightarrow \overrightarrow{a}\times \hat{k}=-x\hat{j}+y\hat{i}$
Substitute the values,
$\Rightarrow {{\left( \overrightarrow{a}\times \hat{i} \right)}^{2}}+{{\left( \overrightarrow{a}\times \hat{j} \right)}^{2}}+{{\left( \overrightarrow{a}\times \hat{k} \right)}^{2}}={{\left( -y\hat{k}+z\hat{j} \right)}^{2}}+{{\left( x\hat{k}-z\hat{i} \right)}^{2}}+{{\left( -x\hat{j}+y\hat{i} \right)}^{2}}$
As we know the dot product of any unit vector with any other is zero, $\hat{i}\cdot \hat{j}=\hat{j}\cdot \hat{k}=\hat{k}\cdot \hat{i}=0$. So,
$\Rightarrow {{\left( \overrightarrow{a}\times \hat{i} \right)}^{2}}+{{\left( \overrightarrow{a}\times \hat{j} \right)}^{2}}+{{\left( \overrightarrow{a}\times \hat{k} \right)}^{2}}={{\left( -y\hat{k} \right)}^{2}}+{{\left( z\hat{j} \right)}^{2}}+{{\left( x\hat{k} \right)}^{2}}+{{\left( -z\hat{i} \right)}^{2}}+{{\left( -x\hat{j} \right)}^{2}}+{{\left( y\hat{i} \right)}^{2}}$
Square the terms,
$\Rightarrow {{\left( \overrightarrow{a}\times \hat{i} \right)}^{2}}+{{\left( \overrightarrow{a}\times \hat{j} \right)}^{2}}+{{\left( \overrightarrow{a}\times \hat{k} \right)}^{2}}={{y}^{2}}+{{z}^{2}}+{{x}^{2}}+{{z}^{2}}+{{x}^{2}}+{{y}^{2}}$
Add the lime terms,
$\Rightarrow {{\left( \overrightarrow{a}\times \hat{i} \right)}^{2}}+{{\left( \overrightarrow{a}\times \hat{j} \right)}^{2}}+{{\left( \overrightarrow{a}\times \hat{k} \right)}^{2}}=2{{x}^{2}}+2{{y}^{2}}+2{{z}^{2}}$
Take 2 commons from the right side,
$\Rightarrow {{\left( \overrightarrow{a}\times \hat{i} \right)}^{2}}+{{\left( \overrightarrow{a}\times \hat{j} \right)}^{2}}+{{\left( \overrightarrow{a}\times \hat{k} \right)}^{2}}=2\left( {{x}^{2}}+{{y}^{2}}+{{z}^{2}} \right)$
Substitute the value from equation (1),
$\Rightarrow {{\left( \overrightarrow{a}\times \hat{i} \right)}^{2}}+{{\left( \overrightarrow{a}\times \hat{j} \right)}^{2}}+{{\left( \overrightarrow{a}\times \hat{k} \right)}^{2}}=2{{a}^{2}}$

Hence, the value of ${{\left( \overrightarrow{a}\times \hat{i} \right)}^{2}}+{{\left( \overrightarrow{a}\times \hat{j} \right)}^{2}}+{{\left( \overrightarrow{a}\times \hat{k} \right)}^{2}}$ is $2{{a}^{2}}$.

Note:
The vector can be multiplied by,
Dot product:- The dot product of two vectors is the magnitude of one time the projection of the second onto the first.
Since a vector has no projection perpendicular to itself, the dot product of any unit vector with any other is zero.
$\hat{i}\cdot \hat{i}=\hat{j}\cdot \hat{j}=\hat{k}\cdot \hat{k}=1$
Cross product:- The cross product of two vectors is the area of the parallelogram between them.
The cross product of any unit vector with itself is zero.
$\hat{i}\times \hat{i}=\hat{j}\times \hat{j}=\hat{k}\times \hat{k}=0$