Question

Two lines \(2x=3y=-z\) and \(6z=-y=-4z\) intersect at a point, then find the angle between them.

Answer 1

The direction vectors of the two lines can be found by looking at the coefficients of x, y, and z in their respective equations. For the first line, the direction vector is <2, 3, -1>, and for the second line, it is <6, -1, -4>. 
The angle between two lines with direction vectors a and b can be found using the dot product formula: 
\(cos\theta=\frac{(a.b)}{(\left |a  \right |\left |b  \right |)}\) 
where θ is the angle between the two lines. 
Substituting the direction vectors for the two lines, we get: 
cosθ = (<2, 3, -1> · <6, -1, -4>) / (|<2, 3, -1>| |<6, -1, -4>|) 
Evaluating the dot product and magnitudes, we get: 
\(cos\theta=\frac{(-6-3+4)}{\sqrt14\times \sqrt53}\)
Simplifying, we get: 
\(cos\theta=\frac{-5}{\sqrt14\times\sqrt53}\)
Using a calculator, we can find the angle θ to be approximately 128.9 degrees. 
Therefore, the angle between the two lines is approximately 128.9 degrees.