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Question: For vector \(\overrightarrow{a}\), if \(\left| \overrightarrow{a} \right|=a\), then write the value ...

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

Explanation

Solution

First find the value of a in terms of x, y, z by applying dot product on ∣a→∣\left| \overrightarrow{a} \right|. After that find the values of (a→×i^)\left( \overrightarrow{a}\times \hat{i} \right), (a→×j^)\left( \overrightarrow{a}\times \hat{j} \right) and (a→×k^)\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 ∣a→∣\left| \overrightarrow{a} \right|.

Complete step by step answer:
Given: ∣a→∣=a\left| \overrightarrow{a} \right|=a
Let the vector be a→=xi^+yj^+zk^\overrightarrow{a}=x\hat{i}+y\hat{j}+z\hat{k}.
Since ∣a→∣=a\left| \overrightarrow{a} \right|=a. Then,
⇒(xi^)2+(yj^)2+(zk^)2=a\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, i^⋅j^=j^⋅k^=k^⋅i^=0\hat{i}\cdot \hat{j}=\hat{j}\cdot \hat{k}=\hat{k}\cdot \hat{i}=0. So,
⇒x2+y2+z2=a2\Rightarrow \sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}={{a}^{2}}
Square both sides of the equation,
⇒x2+y2+z2=a2\Rightarrow {{x}^{2}}+{{y}^{2}}+{{z}^{2}}={{a}^{2}} …………….….. (1)
Now,
⇒a→×i^=(xi^+yj^+zk^)×i^\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,
⇒a→×i^=xi^×i^+yj^×i^+zk^×i^\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, i^×i^=j^×j^=k^×k^=0\hat{i}\times \hat{i}=\hat{j}\times \hat{j}=\hat{k}\times \hat{k}=0.
⇒a→×i^=x(0)+y(−k^)+z(j^)\Rightarrow \overrightarrow{a}\times \hat{i}=x\left( 0 \right)+y\left( -\hat{k} \right)+z\left( {\hat{j}} \right)
Multiply the terms,
⇒a→×i^=−yk^+zj^\Rightarrow \overrightarrow{a}\times \hat{i}=-y\hat{k}+z\hat{j}
Similarly,
⇒a→×j^=xk^−zi^\Rightarrow \overrightarrow{a}\times \hat{j}=x\hat{k}-z\hat{i}
⇒a→×k^=−xj^+yi^\Rightarrow \overrightarrow{a}\times \hat{k}=-x\hat{j}+y\hat{i}
Substitute the values,
⇒(a→×i^)2+(a→×j^)2+(a→×k^)2=(−yk^+zj^)2+(xk^−zi^)2+(−xj^+yi^)2\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, i^⋅j^=j^⋅k^=k^⋅i^=0\hat{i}\cdot \hat{j}=\hat{j}\cdot \hat{k}=\hat{k}\cdot \hat{i}=0. So,
⇒(a→×i^)2+(a→×j^)2+(a→×k^)2=(−yk^)2+(zj^)2+(xk^)2+(−zi^)2+(−xj^)2+(yi^)2\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,
⇒(a→×i^)2+(a→×j^)2+(a→×k^)2=y2+z2+x2+z2+x2+y2\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,
⇒(a→×i^)2+(a→×j^)2+(a→×k^)2=2x2+2y2+2z2\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,
⇒(a→×i^)2+(a→×j^)2+(a→×k^)2=2(x2+y2+z2)\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),
⇒(a→×i^)2+(a→×j^)2+(a→×k^)2=2a2\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 (a→×i^)2+(a→×j^)2+(a→×k^)2{{\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 2a22{{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.
i^⋅i^=j^⋅j^=k^⋅k^=1\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.
i^×i^=j^×j^=k^×k^=0\hat{i}\times \hat{i}=\hat{j}\times \hat{j}=\hat{k}\times \hat{k}=0