Q 58. The distance between the point 4i - 3j - 12k and the plane r. (8i - 6j) = 20 is? Ops: A. 10 B. 6 C. 2 D. 3
Question
Q 58. The distance between the point 4i - 3j - 12k and the plane r . (8i - 6j) = 20 is?
Ops:
- A. 10
- B. 6
- C. 2
- D. 3
Solution
The distance (d) between a point and a plane is given by the formula:
d = |Ax1 + By1 + Cz1 + D| / sqrt(A^2 + B^2 + C^2)
Where (x1, y1, z1) are the coordinates of the point and Ax + By + Cz + D = 0 is the equation of the plane.
In this case, the point is (4, -3, -12) and the plane equation can be written as 8x - 6y + 0z - 20 = 0, so A = 8, B = -6, C = 0 and D = -20.
Substituting these values into the formula gives:
d = |84 - 6(-3) - 20| / sqrt(8^2 + (-6)^2 + 0^2) d = |32 + 18 - 20| / sqrt(64 + 36) d = |30| / sqrt(100) d = 30 / 10 d = 3
So, the distance between the point and the plane is 3 units. Therefore, the correct answer is D. 3.
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