Vectors 4 Question 14

14. If the vectors $\overrightarrow{\mathbf{b}}, \overrightarrow{\mathbf{c}}, \overrightarrow{\mathbf{d}}$ are not coplanar, then prove that the vector $(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times(\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{d}})+(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}}) \times(\overrightarrow{\mathbf{d}} \times \overrightarrow{\mathbf{b}})$

$$ +(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{d}}) \times(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}) \text { is parallel to } \overrightarrow{\mathbf{a}} $$

$(1994,4 M)$

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Answer:

Correct Answer: 14. 5

Solution:

  1. Considering first part $(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times(\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{d}})$

Let

$$ \overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{d}}=\overrightarrow{\mathbf{e}} $$

$(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{e}}=(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{e}}) \overrightarrow{\mathbf{b}}-(\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{e}}) \overrightarrow{\mathbf{a}}$ $[\because(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times \overrightarrow{\mathbf{c}}=(\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c}}) \overrightarrow{\mathbf{b}}-(\overrightarrow{\mathbf{b}} \cdot \overrightarrow{\mathbf{c}}) \overrightarrow{\mathbf{a}}]$

$={\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{d}})} \overrightarrow{\mathbf{b}}-{\overrightarrow{\mathbf{b}} \cdot(\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{d}})} \overrightarrow{\mathbf{d}}$

$=[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{c}} \overrightarrow{\mathbf{d}}] \overrightarrow{\mathbf{b}}-[\overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}} \overrightarrow{\mathbf{d}}] \overrightarrow{\mathbf{a}}$

Similarly,

$$ \begin{aligned} (\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}}) \times(\overrightarrow{\mathbf{d}} \times \overrightarrow{\mathbf{b}}) & =[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{d}} \overrightarrow{\mathbf{b}}] \overrightarrow{\mathbf{c}}-[\overrightarrow{\mathbf{c}} \overrightarrow{\mathbf{d}} \overrightarrow{\mathbf{b}}] \overrightarrow{\mathbf{a}} \\ & =[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{d}} \overrightarrow{\mathbf{b}}] \overrightarrow{\mathbf{c}}-[\overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}} \overrightarrow{\mathbf{d}}] \overrightarrow{\mathbf{a}} \end{aligned} $$

Also, $\quad(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{d}}) \times(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})=-(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}) \times(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{d}})$

$$ \begin{aligned} =(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}) \times(\overrightarrow{\mathbf{d}} \times \overrightarrow{\mathbf{a}}) & =[\overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{d}} \overrightarrow{\mathbf{a}}] \overrightarrow{\mathbf{c}}-[\overrightarrow{\mathbf{c}} \overrightarrow{\mathbf{d}} \overrightarrow{\mathbf{a}}] \overrightarrow{\mathbf{b}} \\ & =[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{d}} \overrightarrow{\mathbf{b}}] \overrightarrow{\mathbf{c}}-[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{c}} \overrightarrow{\mathbf{d}}] \overrightarrow{\mathbf{b}} \end{aligned} $$

From Eqs. (i), (ii) and (iii),

$(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \times(\overrightarrow{\mathbf{c}} \times \overrightarrow{\mathbf{d}})+(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}}) \times(\overrightarrow{\mathbf{d}} \times \overrightarrow{\mathbf{b}})+(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{d}}) \times(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}})$

$=[\overrightarrow{\mathbf{a}} \mathbf{c} \overrightarrow{\mathbf{d}}] \overrightarrow{\mathbf{b}}-[\overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}} \overrightarrow{\mathbf{d}}] \overrightarrow{\mathbf{a}}+[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{d}} \overrightarrow{\mathbf{b}}] \overrightarrow{\mathbf{c}}-[\overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}} \overrightarrow{\mathbf{d}}] \overrightarrow{\mathbf{a}}-[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{d}} \overrightarrow{\mathbf{b}}] \overrightarrow{\mathbf{c}}$

$$ -[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{c}} \overrightarrow{\mathbf{d}}] \overrightarrow{\mathbf{b}}=-2[\overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}} \overrightarrow{\mathbf{d}}] \overrightarrow{\mathbf{a}} $$

$\therefore$ Parallel to $\overrightarrow{\mathbf{a}}$.



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