Show that the plane and line with the given equations intersect, and then find the acute angle of intersection between them. (Give the angle in degrees and round to one decimal place.) The plane given by x + y + 2z = 0 and the line given by x = 2 + t y = 1 - 2t z = 3 + t. Substituting the parametric equations for the line into the equation for the plane and solving for t gives t = so that the plane and the line intersect at the point (x, y, z) = (). What is the acute angle of intersection? theta = degree

Less than 90 degrees is the acute angle measurement. Right angles are 90 degrees in length. Angles that are obtuse are more than 90 degrees. Discover the various sorts of angles and examples of each.

The direction vector normal to the plane is ...

n = (1, 1, 3)

The direction vector of the line is ...

m = (1, -3, 1)

Then the angle θ between them can be found from the dot product:

n•m = |n|·|m|·cos(θ)

(1·1 +1(-3) +3·1) = 1 -3 +3 = 1 = √(1²+1²+3²)·√(1²+(-3)²+1²)·cos(θ)

1 = 11·cos(θ)

θ = arccos(1/11) ≈ 84.8°

This is the angle between the line and the normal to the plane, so the angle between the line and the plane will be the complement of this. Since this angle is not 90°, the line and plane must intersect.

acute angle = 90° -84.8° = 5.2°

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