The equation for the plane that contains the line and is perpendicular to the plane 2 · x + y - 3 · z = - 4 is - 10 · x + 17 · y - z = 76.
From the equation of the plane 2 · x + y - 3 · z = - 4 we know that its normal vector is (2, 1, - 3). Hence, we have found two direction vectors and already know a point, which are part of this parametric equation:
(x, y, z) = (- 3, 4, 5) + t · (3, 2, 4) + u · (2, 1, - 3) (1)
The normal vector of the plane is equal to the cross product of the direction vectors of (1):
[tex]\vec n = (3, 2, 4) \,\times\,(2, 1, -3)[/tex]
[tex]\vec n = \left|\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\3&2&4\\2&1&-3\end{array}\right|[/tex]
[tex]\vec n = (-6-4,8+9, 3-4)[/tex]
[tex]\vec n = (-10, 17, -1)[/tex]
Since the plane contains the point (x, y, z) = (- 3, 4, 5), then we find that the independent constant of the equation of the plane is:
- 10 · x + 17 · y - z = k
- 10 · (- 3) + 17 · 4 - 5 = k
30 + 51 - 5 = k
k = 76
The equation for the plane that contains the line and is perpendicular to the plane 2 · x + y - 3 · z = - 4 is - 10 · x + 17 · y - z = 76.
To learn more on planes: https://brainly.com/question/1962726
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