IB Mathematics Analysis and Approaches 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 ${\mathbf{d}}_1^{\prime }=3{\mathbf{d}}_1$ and ${\mathbf{d}}_2^{\prime }=-2{\mathbf{d}}_2$ are scaled direction vectors of lines $l_1$ and $l_2$, show that the angle between the lines with direction vectors ${\mathbf{d}}_1^{\prime }$ and ${\mathbf{d}}_2^{\prime }$ 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:

$$l_1:\mathbf{r}=\left(\begin{array}{c}2\\ 1\\ 4\end{array}\right)+k\left(\begin{array}{c}5\\ 9\\ 1\end{array}\right)$$

$$l_2:\mathbf{r}=\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right)+t\left(\begin{array}{c}8\\ 7\\ 2\end{array}\right)$$

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 $l_1:\mathbf{r}=\left(\begin{array}{c}2\\ 1\\ 4\end{array}\right)+k\left(\begin{array}{c}5\\ 9\\ 1\end{array}\right)$ and $l_2:\mathbf{r}=\left(\begin{array}{c}1\\ 0\\ 1\end{array}\right)+t\left(\begin{array}{c}8\\ 7\\ 2\end{array}\right)$.