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Determine whether the planes are parallel, perpendicular, or neither. If neither, find the angle between them. (Round to one decimal place.)5x12y13z−2, y−4x26z

User Anselma
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To determine the relationship between two planes, we need to find their normal vectors and then check their dot product. Let's write the equations of the planes in standard form:

Plane 1: 5x + 12y + 13z - 2 = 0 (normal vector = <5, 12, 13>)
Plane 2: -4x + 2y + 6z = 0 (normal vector = <-4, 2, 6>)

Now let's take the dot product of the two normal vectors:

<5, 12, 13> · <-4, 2, 6> = (5)(-4) + (12)(2) + (13)(6) = -20 + 24 + 78 = 82

Since the dot product is not zero, the planes are not perpendicular. To determine if they are parallel, we need to check if one normal vector is a scalar multiple of the other. Let's divide the dot product by the length of each normal vector:

|<5, 12, 13>| = sqrt(5^2 + 12^2 + 13^2) = sqrt(338)
|<-4, 2, 6>| = sqrt((-4)^2 + 2^2 + 6^2) = sqrt(56)

(82) / (sqrt(338) * sqrt(56)) ≈ 0.998

Since the result is very close to 1, we can conclude that the planes are almost parallel. The angle between them is approximately:

cos⁻¹(0.998) ≈ 0.06 radians ≈ 3.4 degrees
User Lyd
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