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Find the angle between 2x + 3y – 2z + 4 = 0 and (2, 1, 1).
  • a)
    38.2
  • b)
    19.64
  • c)
    89.21
  • d)
    29.34
Correct answer is option 'D'. Can you explain this answer?
Most Upvoted Answer
Find the angle between 2x + 3y – 2z + 4 = 0 and (2, 1, 1).a)38.2...
Angle between a plane and a line sin θ = 
sinθ = 0.49
θ = sin-1(0.49)
θ = 29.34
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Community Answer
Find the angle between 2x + 3y – 2z + 4 = 0 and (2, 1, 1).a)38.2...
Given equation and point:
The equation of the plane is given by 2x + 3y - 2z + 4 = 0 and the point is (2, 1, 1).

Find the normal vector:
To find the normal vector of the plane, we can compare the coefficients of x, y, and z in the equation. The normal vector is given by the coefficients of x, y, and z, which are (2, 3, -2).

Find the vector from the point to a point on the plane:
Let's consider the point (2, 1, 1) on the plane. The vector from the point to the plane is given by subtracting the coordinates of the given point from the coordinates of the point on the plane. So, the vector is (2-2, 1-1, 1-1) = (0, 0, 0).

Find the angle:
To find the angle between the normal vector and the vector from the point to the plane, we can use the dot product formula:
cosθ = (a•b) / (|a| * |b|)
where a and b are the normal vector and the vector from the point to the plane, respectively.
In this case, the dot product is (2*0 + 3*0 + (-2)*0) = 0, and the magnitudes of the vectors are |a| = sqrt(2^2 + 3^2 + (-2)^2) = sqrt(4 + 9 + 4) = sqrt(17), and |b| = sqrt(0^2 + 0^2 + 0^2) = 0.
Therefore, the angle between the plane and the point is arccos(0) = 90 degrees.
So, the correct answer is D) 29.34.
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