Given:
P: (2,0,5)
L: (0,6,4)+t(7,-1,5)
and required plane, Π , passes through P and perpendicular to L.
The direction vector of L is V=<7,-1,5>.
For Π to be perpendicular to V, Π has V as the normal vector.
The equation of a plane with normal vector <7,-1,5> passing through a given point P(xp,yp,zp) is
7(x-xp)-1(y-yp)+5(z-zp)=0
Thus the equation of plane Π passing through P(2,0,5) is
7(x-2)-y+5(z-5)=0
or alternatively,
7x-y+5z = 14+25
7x-y+5z = 39
Answer:
Step-by-step explanation:
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Krystal is correct
<h2>Explanation:</h2>
We have the following equation:
In order to solve this, we need to follow this step:
Step 1. <em>Add </em>11 to both sides:
So, Krystal is correct
<h2>
Learn more:</h2>
Solution to equations: brainly.com/question/6251329
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y=(c+b)/ab. Hope this helped, could I possibly get brainliest?
Parallel lines have equal slopes.
The line y = 5x + 3 has slope 5, so the other line also has slope 5.
Answer: The slope is 5.
