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3-1.Vectors
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If $\overrightarrow A = 2\hat i + 4\hat j - 5\hat k$ the direction of cosines of the vector $\overrightarrow A $ are
A$\frac{2}{{\sqrt {45} }},\frac{4}{{\sqrt {45} }}\,{\rm{and}}\,\frac{{ - \,{\rm{5}}}}{{\sqrt {{\rm{45}}} }}$
B$\frac{1}{{\sqrt {45} }},\frac{2}{{\sqrt {45} }}\,{\rm{and}}\,\frac{{\rm{3}}}{{\sqrt {{\rm{45}}} }}$
C$\frac{4}{{\sqrt {45} }},\,0\,{\rm{and}}\,\frac{{\rm{4}}}{{\sqrt {45} }}$
D$\frac{3}{{\sqrt {45} }},\frac{2}{{\sqrt {45} }}\,{\rm{and}}\,\frac{{\rm{5}}}{{\sqrt {{\rm{45}}} }}$
Solution
(a)$\vec A = 2\hat i + 4\hat j – 5\hat k$
$|\overrightarrow A |\, = \sqrt {{{(2)}^2} + {{(4)}^2} + {{( – 5)}^2}} \, = \,\sqrt {45} $
$\cos \alpha = \frac{2}{{\sqrt {45} }},\,\,\,\,\,\cos \beta = \frac{4}{{\sqrt {45} }},\,\,\,\,\cos \gamma = \frac{{ – 5}}{{\sqrt {45} }}$
$|\overrightarrow A |\, = \sqrt {{{(2)}^2} + {{(4)}^2} + {{( – 5)}^2}} \, = \,\sqrt {45} $
$\cos \alpha = \frac{2}{{\sqrt {45} }},\,\,\,\,\,\cos \beta = \frac{4}{{\sqrt {45} }},\,\,\,\,\cos \gamma = \frac{{ – 5}}{{\sqrt {45} }}$
Standard 11
Physics
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