A) \[(\overrightarrow{a}-\overrightarrow{d})=\lambda (\overrightarrow{b}-\overrightarrow{c})\]
B) \[(\overrightarrow{a}+\overrightarrow{d})=\lambda (\overrightarrow{b}+\overrightarrow{c})\]
C) \[(\overrightarrow{a}-\overrightarrow{d})=\lambda (\overrightarrow{c}+\overrightarrow{d})\]
D) \[(\overrightarrow{a}+\overrightarrow{d})=\lambda (\overrightarrow{c}-\overrightarrow{d})\]
E) None of the above
Correct Answer: A
Solution :
Now, \[(\overrightarrow{a}-\overrightarrow{d})\times (\overrightarrow{b}-\overrightarrow{c})\] \[=\overrightarrow{a}\times \overrightarrow{b}-\overrightarrow{a}\times \overrightarrow{c}-\overrightarrow{d}\times \overrightarrow{b}+\overrightarrow{d}\times \overrightarrow{c}\] \[=\overrightarrow{c}\times \overrightarrow{d}-\overrightarrow{b}\times \overrightarrow{d}-\overrightarrow{d}\times \overrightarrow{b}+\overrightarrow{d}\times \overrightarrow{c}\] (\[\because \]\[\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{c}\times \overrightarrow{d},\overrightarrow{a}\times \overrightarrow{c}=\overrightarrow{b}\times \overrightarrow{d}\] given) \[=\overrightarrow{0}\] \[\Rightarrow \] \[(\overrightarrow{a}-\overrightarrow{b})||(\overrightarrow{b}-\overrightarrow{c})\] \[\Rightarrow \] \[\vec{a}-\vec{d}\,=\lambda (\vec{b}-\vec{c})\]
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