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.