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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

Vector A=3i4j+5k\overset{\rightarrow}{A} = 3i - 4j + 5k. We know that direction cosines of

A=332+42+52,432+42+52,532+42+52\overset{\rightarrow}{A} = \frac{3}{\sqrt{3^{2} + 4^{2} + 5^{2}}},\frac{- 4}{\sqrt{3^{2} + 4^{2} + 5^{2}}},\frac{5}{\sqrt{3^{2} + 4^{2} + 5^{2}}}

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