Question

The angle between the straight lines 𝑥+12=𝑦−24=𝑧+35 and 𝑥−11=𝑦+22=𝑧−3−4

Answer 1

arccos(10/(3√105))

Answer 2

The angle between two straight lines in 3D space can be found using their direction vectors. The general symmetric form of a straight line is given by:

$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$

where $(x_1, y_1, z_1)$ is a point on the line and $\vec{d} = a\hat{i} + b\hat{j} + c\hat{k}$ is the direction vector of the line.

Given the first line:

$L_1: \frac{x+1}{2} = \frac{y-2}{4} = \frac{z+3}{5}$

The direction vector for $L_1$ is $\vec{d_1} = 2\hat{i} + 4\hat{j} + 5\hat{k}$.

Given the second line:

$L_2: \frac{x-1}{1} = \frac{y+2}{2} = \frac{z-3}{-4}$

The direction vector for $L_2$ is $\vec{d_2} = 1\hat{i} + 2\hat{j} - 4\hat{k}$.

The angle $\theta$ between two lines with direction vectors $\vec{d_1}$ and $\vec{d_2}$ is given by the formula:

$\cos \theta = \frac{|\vec{d_1} \cdot \vec{d_2}|}{||\vec{d_1}|| \cdot ||\vec{d_2}||}$

First, calculate the dot product $\vec{d_1} \cdot \vec{d_2}$:

$\vec{d_1} \cdot \vec{d_2} = (2)(1) + (4)(2) + (5)(-4) = 2 + 8 - 20 = -10$

Next, calculate the magnitudes of the direction vectors:

$||\vec{d_1}|| = \sqrt{2^2 + 4^2 + 5^2} = \sqrt{4 + 16 + 25} = \sqrt{45} = 3\sqrt{5}$

$||\vec{d_2}|| = \sqrt{1^2 + 2^2 + (-4)^2} = \sqrt{1 + 4 + 16} = \sqrt{21}$

Now, substitute these values into the formula for $\cos \theta$:

$\cos \theta = \frac{|-10|}{(3\sqrt{5})(\sqrt{21})} = \frac{10}{3\sqrt{5 \times 21}} = \frac{10}{3\sqrt{105}}$

Therefore, the angle $\theta$ between the lines is:

$\theta = \arccos\left(\frac{10}{3\sqrt{105}}\right)$