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Question: The direction cosines of the vector \(3\mathbf{i} - 4\mathbf{j} + 5\mathbf{k}\) are...

The direction cosines of the vector 3i4j+5k3\mathbf{i} - 4\mathbf{j} + 5\mathbf{k} are

A

35,45,15\frac{3}{5},\frac{- 4}{5},\frac{1}{5}

B

352,452,12\frac{3}{5\sqrt{2}},\frac{- 4}{5\sqrt{2}},\frac{1}{\sqrt{2}}

C

32,42,12\frac{3}{\sqrt{2}},\frac{- 4}{\sqrt{2}},\frac{1}{\sqrt{2}}

D

352,452,12\frac{3}{5\sqrt{2}},\frac{4}{5\sqrt{2}},\frac{1}{\sqrt{2}}

Answer

352,452,12\frac{3}{5\sqrt{2}},\frac{- 4}{5\sqrt{2}},\frac{1}{\sqrt{2}}

Explanation

Solution

r=3i4j+5k\mathbf{r} = 3\mathbf{i} - 4\mathbf{j} + 5\mathbf{k}; r=32+(4)2+52=52|\mathbf{r}| = \sqrt{3^{2} + ( - 4)^{2} + 5^{2}} = 5\sqrt{2}

Hence, direction cosines are 352,452,552\frac{3}{5\sqrt{2}},\frac{- 4}{5\sqrt{2}},\frac{5}{5\sqrt{2}} i.e., 352,452,12\frac{3}{5\sqrt{2}},\frac{- 4}{5\sqrt{2}},\frac{1}{\sqrt{2}}