Kinematics Question 146

Question: The unit vector parallel to the resultant of the vectors $ \vec{A}=4\hat{i}+3\hat{j}+6\hat{k} $ and $ \vec{B}=-\hat{i}+3\hat{j}-8\hat{k} $ it’s [EAMCET 2000]

Options:

A) $ \frac{1}{7}(3\hat{i}+6\hat{j}-2\hat{k}) $

B) $ \frac{1}{7}(3\hat{i}+6\hat{j}+2\hat{k}) $

C) $ \frac{1}{49}(3\hat{i}+6\hat{j}-2\hat{k}) $

D) $ \frac{1}{49}(3\hat{i}-6\hat{j}+2\hat{k}) $

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Answer:

Correct Answer: A

Solution:

Resultant of vectors $ \overrightarrow{A} $ and $ \overrightarrow{B} $

$ \overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}=4\hat{i}+3\hat{j}+6\hat{k}-\hat{i}+3\hat{j}-8\hat{k} $

$ \overrightarrow{R}=3\hat{i}+6\hat{j}-2\hat{k} $

$ \hat{R}=\frac{\overrightarrow{R}}{|\vec{R}|}=\frac{3\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{3^{2}+6^{2}+{{(-2)}^{2}}}}=\frac{3\hat{i}+6\hat{j}-2\hat{k}}{7} $



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