Vectors 2 Question 23
23. If $A, B, C, D$ are any four points in space, then prove that $|\overrightarrow{\mathbf{A B}} \times \overrightarrow{\mathbf{C D}}+\overrightarrow{\mathbf{B C}} \times \overrightarrow{\mathbf{A D}}+\overrightarrow{\mathbf{C A}} \times \overrightarrow{\mathbf{B D}}|$
$=4$ (area of $\triangle A B C)$.
(1987, 2M)
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Solution:
- Let the position vectors of points $A, B, C, D$ be $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}$ and $\overrightarrow{\mathbf{d}}$, respectively.
Then, $\quad \overrightarrow{AB}=\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{B C}}=\overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{b}}, \overrightarrow{AD}=\overrightarrow{\mathbf{d}}-\overrightarrow{\mathbf{a}}$,
$$ \overrightarrow{BD}=\overrightarrow{\mathbf{d}}-\overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{C A}}=\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{c}}, \overrightarrow{\mathbf{C D}}=\overrightarrow{\mathbf{d}}-\overrightarrow{\mathbf{c}} $$
Now, $|\overrightarrow{\mathbf{A B}} \times \overrightarrow{\mathbf{C D}}+\overrightarrow{\mathbf{B C}} \times \overrightarrow{\mathbf{A D}}+\overrightarrow{\mathbf{C A}} \times \overrightarrow{\mathbf{B D}}|$
$$ \begin{aligned} & =|(\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}}) \times(\overrightarrow{\mathbf{d}}-\overrightarrow{\mathbf{c}})+(\overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{b}}) \times(\overrightarrow{\mathbf{d}}-\overrightarrow{\mathbf{a}})+(\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{c}}) \times(\overrightarrow{\mathbf{d}}-\overrightarrow{\mathbf{b}})| \\ & =\mid \overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{d}}-\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{d}}-\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{d}}-\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{d}} \\ & \quad+\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{d}}-\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{d}}+\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{b}} \mid \\ & =2 \mid \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}) \end{aligned} $$
Also, area of $\triangle A B C$
$$ \begin{aligned} & =\frac{1}{2}|\overrightarrow{\mathbf{A B}} \times \overrightarrow{\mathbf{A C}}|=\frac{1}{2}|(\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}}) \times(\overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{a}})| \\ & =\frac{1}{2}|\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{a}}| \\ & =\frac{1}{2}|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{a}}| \end{aligned} $$
From Eqs. (i) and (ii),
$|\overrightarrow{\mathbf{A B}} \times \overrightarrow{\mathbf{C D}}+\overrightarrow{\mathbf{B C}} \times \overrightarrow{\mathbf{A D}}+\overrightarrow{\mathbf{C A}} \times \overrightarrow{\mathbf{B D}}| 2(2$ area of $\triangle A B C)$
$$ =4(\text { area of } \triangle A B C) $$
