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Gre4nikov [31]
3 years ago
10

Please help me I have been stuck

Mathematics
1 answer:
Mnenie [13.5K]3 years ago
8 0
I think you have to pay attention in class for this
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Find the vector and parametric equations for the line through the point P(0, 0, 5) and orthogonal to the plane −1x+3y−3z=1. Vect
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Answer:

Note that orthogonal to the plane means perpendicular to the plane.

Step-by-step explanation:

-1x+3y-3z=1 can also be written as -1x+3y-3z=0

The direction vector of the plane -1x+3y-3z-1=0 is (-1,3,-3).

Let us find a point on this  line for which the vector from this point to (0,0,5) is perpendicular to the given line. The point is x-0,y-0 and z-0 respectively

Therefore, the vector equation is given as:

-1(x-0) + 3(y-0) + -3(z-5) = 0

-x + 3y + (-3z+15) = 0

-x + 3y -3z + 15 = 0

Multiply through by - to get a positive x coordinate to give

x - 3y + 3z - 15 = 0

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Which graph can be used to find the solution to x^2-1=x+1
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