# Answer to Question #2179 in Linear Algebra for biju

Question #2179

Find the direction cosines of the perpendicular from the origin to the plane 3r.(2i − 3j+ k) + 7 = 0.

Expert's answer

When the plane is defined by he equation Ai - Bj+Ck + D =0, the directional cosines can be found by the following formulae:

<img src="/cgi-bin/mimetex.cgi?l%20=%20%5Cfrac%7BA%7D%7B%5Csqrt%7BA%5E2+B%5E2+C%5E2%7D%7D%20=%20%5Cfrac%7B2%7D%7B%5Csqrt%7B2%5E2+%28-1%29%5E2+1%5E2%7D%7D%20=%20%5Cfrac%7B2%7D%7B%5Csqrt%7B6%7D%7D%20%5C%5C%20m%20=%20%5Cfrac%7BB%7D%7B%5Csqrt%7BA%5E2+B%5E2+C%5E2%7D%7D%20=%20-%20%5Cfrac%7B1%7D%7B%5Csqrt%7B6%7D%7D%20%5C%5C%20n%20=%20%5Cfrac%7BC%7D%7B%5Csqrt%7BA%5E2+B%5E2+C%5E2%7D%7D%20=%20%5Cfrac%7B1%7D%7B%5Csqrt%7B6%7D%7D" title="l = \frac{A}{\sqrt{A^2+B^2+C^2}} = \frac{2}{\sqrt{2^2+(-1)^2+1^2}} = \frac{2}{\sqrt{6}} \\ m = \frac{B}{\sqrt{A^2+B^2+C^2}} = - \frac{1}{\sqrt{6}} \\ n = \frac{C}{\sqrt{A^2+B^2+C^2}} = \frac{1}{\sqrt{6}}">

<img src="/cgi-bin/mimetex.cgi?l%20=%20%5Cfrac%7BA%7D%7B%5Csqrt%7BA%5E2+B%5E2+C%5E2%7D%7D%20=%20%5Cfrac%7B2%7D%7B%5Csqrt%7B2%5E2+%28-1%29%5E2+1%5E2%7D%7D%20=%20%5Cfrac%7B2%7D%7B%5Csqrt%7B6%7D%7D%20%5C%5C%20m%20=%20%5Cfrac%7BB%7D%7B%5Csqrt%7BA%5E2+B%5E2+C%5E2%7D%7D%20=%20-%20%5Cfrac%7B1%7D%7B%5Csqrt%7B6%7D%7D%20%5C%5C%20n%20=%20%5Cfrac%7BC%7D%7B%5Csqrt%7BA%5E2+B%5E2+C%5E2%7D%7D%20=%20%5Cfrac%7B1%7D%7B%5Csqrt%7B6%7D%7D" title="l = \frac{A}{\sqrt{A^2+B^2+C^2}} = \frac{2}{\sqrt{2^2+(-1)^2+1^2}} = \frac{2}{\sqrt{6}} \\ m = \frac{B}{\sqrt{A^2+B^2+C^2}} = - \frac{1}{\sqrt{6}} \\ n = \frac{C}{\sqrt{A^2+B^2+C^2}} = \frac{1}{\sqrt{6}}">

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