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Question

Question: If \(\overset{\rightarrow}{A} = 2\widehat{i} + 4\widehat{j} - 5\widehat{k}\) the direction of cosine...

If A=2i^+4j^5k^\overset{\rightarrow}{A} = 2\widehat{i} + 4\widehat{j} - 5\widehat{k} the direction of cosines of the vector A\overset{\rightarrow}{A} are

A

245,445and545\frac{2}{\sqrt{45}},\frac{4}{\sqrt{45}}\text{and}\frac{- 5}{\sqrt{\text{45}}}

B

145,245and345\frac{1}{\sqrt{45}},\frac{2}{\sqrt{45}}\text{and}\frac{3}{\sqrt{\text{45}}}

C

445,0and445\frac{4}{\sqrt{45}},0\text{and}\frac{4}{\sqrt{45}}

D

345,245and545\frac{3}{\sqrt{45}},\frac{2}{\sqrt{45}}\text{and}\frac{5}{\sqrt{\text{45}}}

Answer

245,445and545\frac{2}{\sqrt{45}},\frac{4}{\sqrt{45}}\text{and}\frac{- 5}{\sqrt{\text{45}}}

Explanation

Solution

A=(2)2+(4)2+(5)2=45|\overset{\rightarrow}{A}| = \sqrt{(2)^{2} + (4)^{2} + ( - 5)^{2}} = \sqrt{45}

cosα=245,cosβ=445,cosγ=545\cos\alpha = \frac{2}{\sqrt{45}},\cos\beta = \frac{4}{\sqrt{45}},\cos\gamma = \frac{- 5}{\sqrt{45}}