Answer:
The area of a parallelogram with adjacent sides u and v is given by the magnitude of the cross product of u and v:
A = |u x v|
We can find the cross product as follows:
u x v = (8)(-1) - (0)(-2), -(9)(-2) - (-3)(1), (9)(2) - (-3)(8) = (-8, -15, 42)
So the area of the parallelogram is:
A = |(-8, -15, 42)| = √(8^2 + 15^2 + 42^2) ≈ 46.09 square units
(a) The triple scalar product of three vectors u, v, and w is given by:
u . (v x w)
We can find the cross product v x w as follows:
v x w = (2)(1) - (-2)(1), (-2)(3) - (0)(1), (0)(1) - (2)(1) = (4, -6, -2)
So the triple scalar product is:
u . (v x w) = (-3)(4) - (4)(-6) - (-1)(-2) = -12 + 24 - 2 = 10
(b) The volume of a parallelepiped with adjacent edges u, v, and w is given by the scalar triple product of u, v, and w:
V = |u . (v x w)|
We already found u . (v x w) in part (a), so we just need to take the absolute value:
V = |10| = 10 cubic units
To find the distance from point P to the plane, we can use the formula:
d = |ax + by + cz + d| / sqrt(a^2 + b^2 + c^2)
where the equation of the plane is given by ax + by + cz + d = 0. In this case, we have:
a = 4, b = -1, c = 3, and d = -9
So the equation of the plane is 4x - y + 3z - 9 = 0. To find the distance from point P(2, 8, -7), we plug in these values:
d = |(4)(2) + (-1)(8) + (3)(-7) - 9| / sqrt(4^2 + (-1)^2 + 3^2) ≈ 5.61 units
Therefore, the distance from point P to the plane is approximately 5.61 units.
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