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If $$\mathbf{D} = (6.00 \hat{i} + 3.00 \hat{j} - 1.00 \hat{k})m$$ $$\mathbf{E} = (4.00 \hat{i} - 5.00 \hat{j} + 8.00 \hat{k})m$$

Find $2\mathbf{D}-\mathbf{E}$ and $|2\mathbf{D}-\mathbf{E}|$.

The number 2 has infinite number of significant figures. Each of the components has only 3 significant figures, so I keep only 3 sig figs when multiplying by 2:

$$2\mathbf{D}-\mathbf{E} = 12.0\hat{i} - 4.00\hat{i} + 6.00\hat{j} + 5.00\hat{j} - 2.00\hat{k}-8.00\hat{k}$$ When adding or subtracting, the sum has only the least precise decimal place: $$2\mathbf{D}-\mathbf{E} = (8.0\hat{i}+11.00\hat{j}-10.00\hat{k})m$$ And $$|2\mathbf{D}-\mathbf{E}| = \sqrt{8.0^2+11.00^2+10.00^2}$$ $$= \sqrt{64+121.0+100.0}$$ $$= \sqrt{285}=16.9m$$ University Physics gives the following answer for $2\mathbf{D}-\mathbf{E}$: $$2\mathbf{D}-\mathbf{E} = (8.00\hat{i} + 11.00\hat{j} - 10.00\hat{k})m$$

I don't understand how they arrive at that answer using the rules of significant figures and even if this is according to the rules, the magnitude calculated using these numbers would be:

$$|2\mathbf{D}-\mathbf{E}| = \sqrt{64.0+121.0+100.0}$$ $$=\sqrt{285.0} = 16.88m,$$

which is the not the answer they give in the book. It is $16.9m$.

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  • $\begingroup$ Related, possible duplicates: physics.stackexchange.com/q/14533/124, physics.stackexchange.com/q/63328/124 $\endgroup$ – David Z Feb 11 '18 at 11:04
  • $\begingroup$ If $\mathbf{D} = (6.00 \hat{i} + 3.00 \hat{j} - 1.00 \hat{k})\,\rm m$ then $2\mathbf{D} = (2\times 6.00 \hat{i} + 2\times 3.00 \hat{j} - 2\times 1.00 \hat{k})\,\rm m = (12.00 \hat{i} + 6.00 \hat{j} - 2.00 \hat{k})\,\rm m$ as you are finding a vector which is twice the size of the original vector. $\endgroup$ – Farcher Feb 11 '18 at 11:17
  • $\begingroup$ @Farcher What's the point? $\endgroup$ – Samama Fahim Feb 11 '18 at 11:34

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