Question

determine whether the lines intersect, and if so, find the point of intersection. if the lines intersect, find the angle between the lines. (round your answer to
$\frac{x}{3}=\frac{y - 2}{-1}=z + 1$, $\frac{x - 1}{4}=y+2=\frac{z + 3}{-3}$
$(x,y,z)=(square)$
$\theta=square^{circ}$

Explanation:

Step1: Write the symmetric - form of lines in vector - form

The first line is \(\frac{x}{3}=\frac{y - 2}{-1}=\frac{z + 1}{1}\), its vector - form is \(\vec{r_1}=\lambda\vec{b_1}\), where \(\vec{b_1}=3\vec{i}-\vec{j}+\vec{k}\) and we can assume a point on it \(P_1=(0,2, - 1)\) (when \(\lambda = 0\)). The second line is \(\frac{x - 1}{4}=\frac{y+2}{1}=\frac{z + 3}{-3}\), its vector - form is \(\vec{r_2}=\vec{a_2}+\mu\vec{b_2}\), where \(\vec{a_2}=\vec{i}-2\vec{j}-3\vec{k}\) and \(\vec{b_2}=4\vec{i}+\vec{j}-3\vec{k}\).

Step2: Check for intersection

If the two lines intersect, then \(\vec{r_1}=\vec{r_2}\), so \(0 + 3\lambda=1 + 4\mu\), \(2-\lambda=-2+\mu\), \(-1+\lambda=-3 - 3\mu\). From \(2-\lambda=-2+\mu\), we get \(\lambda=4 - \mu\). Substitute \(\lambda = 4 - \mu\) into \(0 + 3\lambda=1 + 4\mu\): \(3(4 - \mu)=1 + 4\mu\), \(12-3\mu=1 + 4\mu\), \(7\mu = 11\), \(\mu=\frac{11}{7}\). Then \(\lambda=4-\frac{11}{7}=\frac{28 - 11}{7}=\frac{17}{7}\). Substitute \(\lambda=\frac{17}{7}\) and \(\mu=\frac{11}{7}\) into the third equation \(-1+\lambda=-3 - 3\mu\). Left - hand side: \(-1+\frac{17}{7}=\frac{-7 + 17}{7}=\frac{10}{7}\). Right - hand side: \(-3-3\times\frac{11}{7}=\frac{-21 - 33}{7}=-\frac{54}{7}\). Since the left - hand side is not equal to the right - hand side, the two lines do not intersect.

Step3: Calculate the angle between the lines

The angle \(\theta\) between two lines with direction vectors \(\vec{b_1}\) and \(\vec{b_2}\) is given by \(\cos\theta=\frac{\vert\vec{b_1}\cdot\vec{b_2}\vert}{\vert\vec{b_1}\vert\vert\vec{b_2}\vert}\). \(\vec{b_1}\cdot\vec{b_2}=(3\times4)+(-1\times1)+(1\times(-3))=12 - 1-3 = 8\). \(\vert\vec{b_1}\vert=\sqrt{3^{2}+(-1)^{2}+1^{2}}=\sqrt{9 + 1+1}=\sqrt{11}\), \(\vert\vec{b_2}\vert=\sqrt{4^{2}+1^{2}+(-3)^{2}}=\sqrt{16 + 1 + 9}=\sqrt{26}\). So \(\cos\theta=\frac{\vert8\vert}{\sqrt{11}\sqrt{26}}=\frac{8}{\sqrt{286}}\), and \(\theta=\cos^{-1}(\frac{8}{\sqrt{286}})\approx64.9^{\circ}\)

Answer:

The lines do not intersect and the angle between them is approximately \(64.9^{\circ}\)