Vector Algebra Question 99
Question: If ABCD is a parallelogram, $ \overrightarrow{AB}=2,\mathbf{i}+4,\mathbf{j}-5,\mathbf{k} $ and $ \overrightarrow{AD}=,\mathbf{i}+2,\mathbf{j}+3,\mathbf{k}, $ then the unit vector in the direction of BD is
[Roorkee 1976]
Options:
A) $ \frac{1}{\sqrt{69}},(\mathbf{i}+2\mathbf{j}-8\mathbf{k}) $
B) $ \frac{1}{69},(\mathbf{i}+2\mathbf{j}-8,\mathbf{k}) $
C) $ \frac{1}{\sqrt{69}},(-\mathbf{i}-2\mathbf{j}+8\mathbf{k}) $
D) $ \frac{1}{69},(-\mathbf{i}-2\mathbf{j}+8,\mathbf{k}) $
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Answer:
Correct Answer: C
Solution:
- Since $ \overrightarrow{AB}+\overrightarrow{BD}=\overrightarrow{AD}\Rightarrow \overrightarrow{BD}=\overrightarrow{AD}-\overrightarrow{AB} $ $ =(\mathbf{i}+2\mathbf{j}+3\mathbf{k})-(2\mathbf{i}+4\mathbf{j}-5\mathbf{k})=-\mathbf{i}-2\mathbf{j}+8\mathbf{k} $ Hence unit vector in the direction of $ \overrightarrow{BD} $ is $ \frac{-\mathbf{i}-2\mathbf{j}+8\mathbf{k}}{|-\mathbf{i}-2\mathbf{j}+8\mathbf{k}|}=\frac{-\mathbf{i}-2\mathbf{j}+8\mathbf{k}}{\sqrt{69}}. $
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