Vector Algebra Question 235
Question: A, B, C, D are any four points, then $ \overrightarrow{AB}.\overrightarrow{CD}+,\overrightarrow{,BC}.\overrightarrow{AD}+\overrightarrow{CA}.\overrightarrow{BD}= $
[MNR 1986]
Options:
A) $ 2\overrightarrow{AB}.\overrightarrow{BC}.\overrightarrow{CD} $
B) $ \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD} $
C) $ 5\sqrt{3} $
D) 0
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Answer:
Correct Answer: D
Solution:
- $ \overrightarrow{AD}=\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}=\mathbf{a}+\mathbf{b}+\mathbf{c} $ $ \overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}=\mathbf{a}+\mathbf{b} $ or $ \overrightarrow{CA}=-(\mathbf{a}+\mathbf{b}) $ $ \overrightarrow{BD}=\overrightarrow{BC}+\overrightarrow{CD}=\mathbf{b}+\mathbf{c} $ Therefore, $ \overrightarrow{AB},.,\overrightarrow{CD}+\overrightarrow{BC},.,\overrightarrow{AD}+\overrightarrow{CA},.,\overrightarrow{BD} $ $ =\mathbf{a},.,\mathbf{c}+\mathbf{b},.(\mathbf{a}+\mathbf{b}+\mathbf{c})+(-\mathbf{a}-\mathbf{b}),.,(\mathbf{b}+\mathbf{c}) $ $ =\mathbf{a},.,\mathbf{c}+\mathbf{b},.,\mathbf{a}+\mathbf{b},.,\mathbf{b}+\mathbf{b},.,\mathbf{c}-\mathbf{a},.,\mathbf{b}-\mathbf{a},.,\mathbf{c}-\mathbf{b},.,\mathbf{b}-\mathbf{b},.,\mathbf{c} $ $ =0 $ .
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