Question

Prove that \[\vec{i}\times \left( \vec{a}\times \hat{i} \right)+\hat{j}\left( \vec{a}\times \hat{j} \right)+\hat{k}\times \left( \vec{a}\times \hat{k} \right)=2\vec{a}\]

Answer 1

Hint: First use the property of the orthogonal unit vectors to find $\vec{i}\cdot \vec{i}=\vec{j}\cdot \vec{j}=\vec{k}\cdot \vec{k}=1$. Use the conversion formula from cross product to dot product $\vec{a}\times \left( \vec{b}\times \vec{c} \right)=\vec{b}\left( \vec{a}\cdot \vec{c} \right)-\vec{c}\left( \vec{a}\cdot \vec{b} \right)$ to convert the vector triple products at the left hand side of the statement. Use the fact that a vector is sum of component vectors along the direction of unit vectors $\left( \vec{a}={{a}_{1}}\vec{i}+{{a}_{2}}\vec{j}+{{a}_{3}}\vec{k} \right)$.\[\]

Complete step-by-step solution:
We know that the dot product of two vectors $\vec{a}$ and $\vec{b}$ is denoted as $\vec{a}\cdot \vec{b}$ and is given by $\vec{a}\cdot \vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\cos \theta $ where $\theta $ is the angle between the vectors $\vec{a}$ and $\vec{b}$. The cross product between two vectors is denoted as $\vec{a}\times \vec{b}$ and is given by $\vec{a}\times \vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\sin \theta \hat{n}$ where $\hat{n}$ is a vector perpendicular to both $\vec{a}$ and $\vec{b}$ and in a direction according to right hand rule.
We also know that $\hat{i}$,$\hat{j}$ and $\hat{k}$ are unit vectors(vectors with magnitude 1) along $x,y$ and $z$ axes respectively. So the magnitude of these vectors \[\left| {\hat{i}} \right|=\left| {\hat{j}} \right|=\left| {\hat{k}} \right|=1\]. The vectors just like their axes are perpendicular to each other which means any angle among $\hat{i}$,$\hat{j}$ and $\hat{k}$is ${{90}^{\circ }}.$ So $\hat i \cdot \hat i = \hat j \cdot \hat j = \hat k \cdot \hat k = 1 \cdot 1 \cdot \cos {0^ \circ } = 1$
We know the formula to convert cross product of three vectors to dot product $\vec{a}\times \left( \vec{b}\times \vec{c} \right)=\vec{b}\left( \vec{a}\cdot \vec{c} \right)-\vec{c}\left( \vec{a}\cdot \vec{b} \right)$ \[\]
We are asked to prove the statement
\[\vec{i}\times \left( \vec{a}\times \hat{i} \right)+\hat{j}\left( \vec{a}\times \hat{j} \right)+\hat{k}\times \left( \vec{a}\times \hat{k} \right)=2\vec{a}\]
Let us use the formula and covert into dot products
\[\begin{align}
  & \vec{i}\times \left( \vec{a}\times \hat{i} \right)+\hat{j}\left( \vec{a}\times \hat{j} \right)+\hat{k}\times \left( \vec{a}\times \hat{k} \right)= \\
 & =\left( \hat{i}\cdot \hat{i} \right)\vec{a}-\left( \hat{i}\cdot \vec{a} \right)\hat{i}+\left( \hat{j}\cdot \hat{j} \right)\vec{a}-\left( \hat{j}\cdot \vec{a} \right)\hat{j}+\left( \hat{k}\cdot \hat{k} \right)\vec{a}-\left( \hat{k}\cdot \vec{a} \right)\hat{k} \\
 & =\vec{a}-\left( \hat{i}\cdot \vec{a} \right)\hat{i}+\vec{a}-\left( \hat{j}\cdot \vec{a} \right)\hat{j}+\vec{a}-\left( \hat{k}\cdot \vec{a} \right)\hat{k} \\
 & =3\vec{a}-\left\{ \left( \hat{i}\cdot \vec{a} \right)\hat{i}+\left( \hat{j}\cdot \vec{a} \right)\hat{j}+\left( \hat{k}\cdot \vec{a} \right)\hat{k} \right\} \\
\end{align}\]
We can express any vector as a combination of orthogonal unit vectors. Let us assume that ${a_1}\hat i + {a_2}\hat j + {a_3}\hat k$ where \[\left| {\vec{a}} \right|=\sqrt{{{a}_{1}}^{2}+{{a}_{2}}^{2}+{{a}_{3}}^{2}}\] and ${{a}_{1}},{{a}_{2}},{{a}_{3}}$ are the magnitude of the vector $\vec{a}$ along the direction of unit orthogonal vectors $\hat{i}$,$\hat{j}$ and $\hat{k}$. So we can express $\hat{i}\cdot \vec{a}=a\cos \alpha ={{a}_{1}},\hat{j}\cdot \vec{a}=a\cos \beta ={{a}_{2}},\hat{k}\cdot \vec{a}=a\cos \gamma ={{a}_{3}}$ where $\alpha ,\beta ,\gamma $ are the angles the vector $\vec{a}$ makes with $\hat{i}$,$\hat{j}$ and $\hat{k}$.
\[\begin{align}
  & =3\vec{a}-\left\{ \left( \hat{i}\cdot \vec{a} \right)\hat{i}+\left( \hat{j}\cdot \vec{a} \right)\hat{j}+\left( \hat{k}\cdot \vec{a} \right)\hat{k} \right\} \\
 & =3\vec{a}-\left( {{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k} \right) \\
 & =3\vec{a}-\vec{a}=2\vec{a} \\
\end{align}\]
Which is right hand side of the proof. Hence proved.\[\]

Note: The vector triple products of the forms like $\vec{a}\times \left( \vec{b}\times \vec{c} \right)$ are also called vector volume. The formula $\vec{a}\times \left( \vec{b}\times \vec{c} \right)=\vec{b}\left( \vec{a}\cdot \vec{c} \right)-\vec{c}\left( \vec{a}\cdot \vec{b} \right)$ is also called Lagrange’s formula or expansion of vector triple product. We note another useful identity called Jacobi’s identity $\left( \vec{a}\times \vec{b} \right)\times \vec{c}=\vec{a}\times \left( \vec{b}\times \vec{c} \right)-\vec{b}\times \left( \vec{a}\times \vec{c} \right).$