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Find the distance from the point (1,-2,4) to the plane 3x+2y+6z = 5

User Justlike
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Final answer:

The distance from the point (1, -2, 4) to the plane 3x+2y+6z = 5 is calculated using the point-to-plane distance formula, resulting in a distance of 18/7 units.

Step-by-step explanation:

To find the distance from the point (1,-2,4) to the plane 3x+2y+6z = 5, we can use the point-to-plane distance formula. This formula states that the distance D from a point (x1,y1,z1) to the plane Ax+By+Cz+D=0 is given by:

D = |Ax1+By1+Cz1+D| / √(A2+B2+C2)

In this case, A=3, B=2, C=6, and D=-5. Substituting the point (1, -2, 4), we get:

D = |3(1) + 2(-2) + 6(4) - 5| / √(32 + 22 + 62)

D = |3 - 4 + 24 - 5| / √(9 + 4 + 36)

D = |18| / √49

D = 18 / 7

Thus, the distance from the point to the plane is 18/7 units.

User PmanAce
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