Find an equation of the plane that passes through the points (4, 1, 4), (5, -8, 6), and (-4, -5, 1)? 

Solution:

Given three points (4, 1, 4), (5, -8, 6), and (-4, -5, 1)

Equation of a plane passing through three points is

(\(\overrightarrow r - \overrightarrow a).{(\overrightarrow b - \overrightarrow a) × (\overrightarrow c - \overrightarrow a)} =0\)

 A(4, 1, 4) = 4i + j + 4k = a

 B(5, -8, 6) = 5i - 8j + 6k = b

 C(-4, -5, 1) = -4i - 5j + k = c

(b - a) = (5i - 8j + 6k) - (4i + j + 4k)

= i - 9j + 2k

(c - a) = (-4i - 5j + k) - (4i + j + 4k)

= -8i - 6j - 3k

Finding the cross product of the vectors, we get

{(b - a) × (c - a)} = 39i -13j -78k

(r - a) = (xi + yj + zk) - (4i + j + 4k)

= (x - 4)i + (y - 1)j + (z - 4)k

Finding the dot product of the vectors, we get

(r - a). {(b - a) × (c - a)}= {(x - 4)i + (y - 1)j + (z- 4)k}.{39i -13j - 78k}

= 39(x - 4) - 13(y - 1) - 78(z - 4)

= 39x - 156 -13y + 13 - 78z + 312

= 39x -13y - 78z + 169

The required equation 39x - 13y - 78z + 169 = 0

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Find an equation of the plane that passes through the points (4, 1, 4), (5, -8, 6), and (-4, -5, 1)?

Summary:

The equation of the plane that passes through the points (4, 1, 4), (5, -8, 6), and (-4, -5, 1) is 39x - 13y - 78z + 169 = 0.