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 byx + y + 3z = 0and the line given byx = 4 + ty = 1 − 3tz = 2 + t. substituting the parametric equations for the line into the equation for the plane and solving for t givest =correct: your answer is correct. ,so that the plane and the line intersect at the point(x, y, z) =correct: your answer is correct..what is the acute angle o

  angle of intersection: 5.2°

Step-by-step explanation:

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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The attached graph shows the line and plane meet at a shallow angle, consistent with the above answer.

i believe the correct answer is b.

3 is greater then both of those numbers