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  4. If a , b , c , d are four distinct vectors satisfying the conditions a × b = c × d and a × c = b × d , then prove that a ⋅ b + c ⋅ d ≠ a ⋅ c + b ⋅ d

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

IIT JEEIIT JEE 2004Vector Algebra

Solution:

Given, $\overrightarrow{ a } \times \overrightarrow{ b }=\overrightarrow{ c } \times \overrightarrow{ d }$ and $\overrightarrow{ a } \times \overrightarrow{ c }=\overrightarrow{ b } \times \overrightarrow{ d }$
$\Rightarrow \overrightarrow{ a } \times \overrightarrow{ b }-\overrightarrow{ a } \times \overrightarrow{ c }=\overrightarrow{ c } \times \overrightarrow{ d }-\overrightarrow{ b } \times \overrightarrow{ d }$
$\Rightarrow \overrightarrow{ a } \times(\overrightarrow{ b }-\overrightarrow{ c })=(\overrightarrow{ c }-\overrightarrow{ b }) \times \overrightarrow{ d }$
$\Rightarrow \overrightarrow{ a } \times(\overrightarrow{ b }-\overrightarrow{ c })-(\overrightarrow{ c }-\overrightarrow{ b }) \times \overrightarrow{ d }=0$
$\Rightarrow \overrightarrow{ a } \times(\overrightarrow{ b }-\overrightarrow{ c })-\overrightarrow{ d } \times(\overrightarrow{ b }-\overrightarrow{ c })=0$
$\Rightarrow (\overrightarrow{ a }-\overrightarrow{ d }) \times(\overrightarrow{ b }-\overrightarrow{ c })=0 \Rightarrow(\overrightarrow{ a }-\overrightarrow{ d }) \|(\overrightarrow{ b }-\overrightarrow{ c })$
$\therefore (\overrightarrow{ a }-\overrightarrow{ d }) \cdot(\overrightarrow{ b }-\overrightarrow{ c }) \neq 0$
$\Rightarrow \overrightarrow{ a } \cdot \overrightarrow{ b }+\overrightarrow{ d } \cdot \overrightarrow{ c } \neq \overrightarrow{ d } \cdot \overrightarrow{ b }+\overrightarrow{ a } \cdot \overrightarrow{ c }$