Question Type 4: Finding the angle between two lines using the direction vectors Exercises

Use the cross product method to find the acute angle between $l_1: (2, 1, 4) + k(5, 9, 1)$ and $l_2: (1, 0, 1) + t(8, 7, 2)$.

If $\mathbf{d}_1'=3\mathbf{d}_1$ and $\mathbf{d}_2'=-2\mathbf{d}_2$ are scaled direction vectors of lines $l_1$ and $l_2$, show that the angle between the lines with direction vectors $\mathbf{d}_1'$ and $\mathbf{d}_2'$ is the same as the angle between the lines with direction vectors $\mathbf{d}_1$ and $\mathbf{d}_2$.

Calculate the acute angle between the lines $l_1$ and $l_2$ defined by:

$l_1: \mathbf{r} = \begin{pmatrix} 2 \\ 1 \\ 4 \end{pmatrix} + k \begin{pmatrix} 5 \\ 9 \\ 1 \end{pmatrix}$

$l_2: \mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + t \begin{pmatrix} 8 \\ 7 \\ 2 \end{pmatrix}$

Give your answer in degrees.

Find the angle between the line $l_1$ and the line with direction $d_3$.

Express the acute angle $\theta$ between the lines $l_1: (2,1,4)+k(5,9,1)$ and $l_2: (1,0,1)+t(8,7,2)$ in radians.

Find $\tan\theta$ where $\theta$ is the acute angle between the lines $l_1: \mathbf{r} = \begin{pmatrix} 2 \\ 1 \\ 4 \end{pmatrix} + k \begin{pmatrix} 5 \\ 9 \\ 1 \end{pmatrix}$ and $l_2: \mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} + t \begin{pmatrix} 8 \\ 7 \\ 2 \end{pmatrix}$.