Question

find direction cosines of vector A which is equal to 5i+6j-4k?
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Answer 1

you can understand from the above example

Answer 2

Answer:

The direction cosines of given vector are

$\frac{5}{\sqrt{77}},\frac{6}{\sqrt{77}}\:and\:\frac{-4}{\sqrt{77}}$

Step-by-step explanation:

Let the given vector be $\vec p=5\hat i+6\hat j-4\hat k$ and we are required to find the direction cosines of the given vector.

For any vector $\vec p=x\hat i+y\hat j+z\hat k$ the direction cosine are given by the relation as follows

$cos\alpha=\frac{x}{\sqrt{x^{2}+y^{2}+z^{2}} } \\\\cos\beta=\frac{y}{\sqrt{x^{2}+y^{2}+z^{2}} } \\\\cos\gamma=\frac{z}{\sqrt{x^{2}+y^{2}+z^{2}} }$

Hence the direction cosines of the given vectors are calculated as follows

$cos\alpha=\frac{5}{\sqrt{5^{2}+6^{2}+(-4)^{2}} }=\frac{5}{\sqrt{77} } \\\\cos\beta=\frac{6}{\sqrt{5^{2}+6^{2}+(-4)^{2}} }=\frac{6}{\sqrt{77}}\\\\cos\gamma=\frac{-4}{\sqrt{5^{2}+6^{2}+(-4)^{2}} }=-\frac{4}{\sqrt{77}}$

Therefore,

The direction cosines of given vector are

$\frac{5}{\sqrt{77}},\frac{6}{\sqrt{77}}\:and\:\frac{-4}{\sqrt{77}}$

#SPJ2