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:
 $cos\theta=\frac{(12-3+4)}{\sqrt{14}\times \sqrt{53}}$
Simplifying, we get:
 

$cos\theta=\frac{13}{\sqrt{14}\times\sqrt{53}}$
Using a calculator, we can find the angle θ to be approximately 54.5 degrees.
Therefore, the angle between the two lines is approximately 54.5 degrees.