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

SAGAR SINGH - IIT DELHI
878 Points
13 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 $\boldsymbol{\hat{v}}$