MHT CET · Physics · Mathematics in Physics
A unit vector in the direction of resultant vector of \(\vec{A}=-2 \hat{i}+3 \hat{j}+\hat{k}\) and \(\vec{B}=\hat{i}+2 \hat{j}-4 \hat{k}\) is
- A \(\frac{-3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+5 \hat{\mathrm{k}}}{\sqrt{35}}\)
- B \(\frac{\hat{i}+2 \hat{j}-4 \hat{k}}{\sqrt{35}}\)
- C \(\frac{-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{35}}\)
- D \(\frac{-\hat{\mathrm{i}}+5 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}}{\sqrt{35}}\)
Answer & Solution
Correct Answer
(D) \(\frac{-\hat{\mathrm{i}}+5 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}}{\sqrt{35}}\)
Step-by-step Solution
Detailed explanation
\(\vec{R} = \vec{A} + \vec{B} = (-2 \hat{i}+3 \hat{j}+\hat{k}) + (\hat{i}+2 \hat{j}-4 \hat{k}) = -\hat{i}+5 \hat{j}-3 \hat{k}\) \(\hat{R} = \frac{\vec{R}}{|\vec{R}|} = \frac{-\hat{i}+5 \hat{j}-3 \hat{k}}{\sqrt{(-1)^2+5^2+(-3)^2}} = \frac{-\hat{i}+5 \hat{j}-3 \hat{k}}{\sqrt{1+25+9}} = \frac{-\hat{i}+5 \hat{j}-3 \hat{k}}{\sqrt{35}}\)
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