Vectors 2 Question 21

21. If $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}, \overrightarrow{\mathbf{d}}$ are four distinct vectors satisfying the conditions $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{d}}$ and $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{d}}$, then prove that $\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}} \cdot \overrightarrow{\mathbf{d}} \neq \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c}}+\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{d}}$.

(2004, 2M)

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Solution:

  1. Given, $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}=\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{d}}$ and $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{d}}$

$$ \Rightarrow \quad \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{d}}-\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{d}} $$

$$ \begin{array}{lc} \Rightarrow & \overrightarrow{\mathbf{a}} \times(\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}})=(\overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{d}} \\ \Rightarrow & \overrightarrow{\mathbf{a}} \times(\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}})-(\overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{d}}=0 \\ \Rightarrow & \overrightarrow{\mathbf{a}} \times(\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}})-\overrightarrow{\mathbf{d}} \times(\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}})=0 \\ \Rightarrow & (\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{d}}) \times(\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}})=0 \Rightarrow(\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{d}}) |(\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}}) \\ \therefore & (\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{d}}) \cdot(\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{c}}) \neq 0 \\ \Rightarrow & \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{d}} \cdot \overrightarrow{\mathbf{c}} \neq \overrightarrow{\mathbf{d}} \cdot \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c}} \end{array} $$



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