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find the angle theta between the lines whose direction cosines are given by the equations 3l+m+5n=0 and 6mn-2nl+5lm=0.


9 years ago

							Dear student,
The direction cosines of a vector are the cosines of the angles between the vector and the three coordinate axes.
If v is a vector ${\mathbf v}= v_1 \boldsymbol{\hat{x}} + v_2 \boldsymbol{\hat{y}} + v_3 \boldsymbol{\hat{z}}$
where $\boldsymbol{\hat{x}}, \boldsymbol{\hat{y}}, \boldsymbol{\hat{z}}$ is a basis. Then the direction cosines are \begin{align} \alpha & = \cos a = \frac{{\mathbf v} \cdot \boldsymbol{\hat{x}} }{ \left \Vert {\mathbf v} \right \Vert } & = \frac{v_1}{\sqrt{v_1^2 + v_2^2 + v_3^2}} ,\\ \beta & = \cos b = \frac{{\mathbf v} \cdot \boldsymbol{\hat{y}} }{ \left \Vert {\mathbf v} \right \Vert } & = \frac{v_2}{\sqrt{v_1^2 + v_2^2 + v_3^2}} ,\\ \gamma &= \cos c = \frac{{\mathbf v} \cdot \boldsymbol{\hat{z}} }{ \left \Vert {\mathbf v} \right \Vert } & = \frac{v_3}{\sqrt{v_1^2 + v_2^2 + v_3^2}}. \end{align}
Note that
α2 + β2 + γ2 = 1
and
(α, β, γ) is the Cartesian coordinates of the unit vector $\boldsymbol{\hat{v}}$

9 years ago
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