Lagrange's Identity - Practice Questions & MCQ

$
\begin{aligned}
(\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d}) & =\left|\begin{array}{ll}
\vec{a} \cdot \vec{c} & \vec{a} \cdot \vec{d} \\
\vec{b} \cdot \vec{c} & \vec{b} \cdot \vec{d}
\end{array}\right| \\
& =(\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d})-(\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c})
\end{aligned}
$

Proof:
Let $\quad(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \cdot(\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{d}})=\overrightarrow{\mathbf{u}} \cdot(\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{d}})$ (where $(\vec{a} \times \vec{b})$ )
$
\begin{aligned}
& =\overrightarrow{\mathbf{u}}=(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{c}}) \cdot \overrightarrow{\mathbf{d}} \\
& =((\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{c}}) \cdot \overrightarrow{\mathbf{d}} \\
& =((\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}}) \overrightarrow{\mathbf{b}}-(\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{b}}) \overrightarrow{\mathbf{a}}) \cdot \overrightarrow{\mathbf{d}} \\
=(\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}})(\vec{b} \cdot \overrightarrow{\mathbf{d}})- & (\overrightarrow{\mathbf{c}} \cdot \vec{b})(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{d}}) \\
=(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c}})(\vec{b} \cdot \overrightarrow{\mathbf{d}})- & (\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{d}})(\vec{b} \cdot \overrightarrow{\mathbf{c}})
\end{aligned}
$

NOTE:
$
\begin{aligned}
& (\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})=[(\vec{a} \times \vec{b}) \cdot \vec{d}] \vec{c}-[(\vec{a} \times \vec{b}) \cdot \vec{c}] \vec{d} \\
& =\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{d}
\end{array}\right] \vec{c}-\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right] \vec{d}
\end{aligned}
$
$\Rightarrow$ Thus, vector $\overrightarrow{(a} \times \vec{b}) \times(\vec{c} \times \vec{d})$ lies in the plane of $\vec{c}$ and $\vec{d}$

If we take the dot product of two system of vectors and get unity, then the system is called reciprocal system of vectors.

Thus if $\tilde{\mathrm{a}}, \tilde{\mathrm{b}}$ and $\tilde{\mathrm{c}}$ are three non - coplanar vectors, and if $\overrightarrow{a^{\prime}}=\frac{\vec{b} \times \vec{c}}{\left[\begin{array}{ll}\vec{a} & \vec{b}\end{array}\right]}$, $\overrightarrow{b^{\prime}}=\frac{\vec{c} \times \vec{a}}{\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]}$ and $\overrightarrow{c^{\prime}}=\frac{\vec{a} \times \vec{b}}{\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]}$ then $\overrightarrow{a^{\prime}}, \overrightarrow{b^{\prime}}, \vec{c}$ are said to be the reciprocal systems of vectors for vectors $\vec{a}, \vec{b}$ and $\vec{c}$.

Properties

1. If $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ and $\overrightarrow{\mathbf{a}^{\prime}}, \overrightarrow{\mathbf{b}^{\prime}}$ and $\overrightarrow{\mathbf{c}^{\prime}}$ are reciprocal system of vectors, then $\left.\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{a}^{\prime}}=\frac{\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})}{[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]}=\frac{\left[\begin{array}{lll}\overrightarrow{\mathbf{a}} & \overrightarrow{\mathbf{b}} & \overrightarrow{\mathbf{c}}\end{array}\right]}{[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}}} \overrightarrow{\mathbf{c}}\right]$.

Similarly,
$
\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{b}^{\prime}}=\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{c}^{\prime}}=1
$

Due to the above property, the two systems of vectors are called reciprocal systems.
2. $\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}^{\prime}}=\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c}^{\prime}}=\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{a}^{\prime}}=\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}^{\prime}}=\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{a}^{\prime}}=\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{b}^{\prime}}=\mathbf{0}$
3. $\left[\begin{array}{lll}\overrightarrow{\mathbf{a}} & \overrightarrow{\mathrm{b}} & \overrightarrow{\mathbf{c}}\end{array}\right]\left[\begin{array}{lll}\overrightarrow{\mathbf{a}^{\prime}} & \overrightarrow{\mathbf{b}^{\prime}} & \overrightarrow{\mathbf{c}^{\prime}}\end{array}\right]=1$

 Proof:

 We have,
$
\begin{aligned}
{\left[\begin{array}{lll}
\overrightarrow{a^{\prime}} & \overrightarrow{b^{\prime}} & \overrightarrow{c^{\prime}}
\end{array}\right]=\left[\begin{array}{ll}
\vec{b} \times \vec{c} \\
{\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right] \frac{\vec{c} \times \vec{a}}{\left.\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right] \frac{\vec{a} \times \vec{b}}{\left[\begin{array}{ll}
\vec{a} & \vec{b} \\
\vec{c}
\end{array}\right]}\right]}} \\
& =\frac{1}{\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]^3}\left[\begin{array}{lll}
\vec{b} & \vec{c} & \vec{c} \times \vec{a} \\
& \vec{a} \times \vec{b}
\end{array}\right] \\
& =\frac{1}{\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]^3}\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]^2 \\
& =\frac{1}{\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]} \\
\Rightarrow\left[\begin{array}{lll}
\overrightarrow{a^{\prime}} & \overrightarrow{b^{\prime}} & \vec{c}
\end{array}\right]\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]=1
\end{array}\right.}
\end{aligned}
$
4. The orthogonal triad of vectors $\hat{\mathbf{i}}, \hat{\mathbf{j}}$ and $\hat{\mathbf{k}}$ is self reciprocal.

5. $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ are non-coplanar iff $\overrightarrow{\mathbf{a}^{\prime}}, \overrightarrow{\mathbf{b}^{\prime}}$ and $\overrightarrow{\mathbf{c}^{\prime}}$ are non-coplanar.
$
\text { As }\left[\begin{array}{lll}
\overrightarrow{a^{\prime}} & \overrightarrow{b^{\prime}} & \vec{c}
\end{array}\right]\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]=1_{\text {and }}\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right] \neq 0 \text { are non-coplanar } \quad \Leftrightarrow \frac{1}{\left[\begin{array}{lll}
\vec{a} & \vec{b} & \vec{c}
\end{array}\right]} \neq 0 \Leftrightarrow\left[\begin{array}{lll}
\overrightarrow{a^{\prime}} & \overrightarrow{b^{\prime}} & \overrightarrow{c^{\prime}}
\end{array}\right]
$
are non-coplanar.