# Math Help - analytic geometry helpp !! plz it is urgent

1. ## analytic geometry helpp !! plz it is urgent

well i have some problems i have tried hard to solve them but wasnt succeed

1- find the equation of the plane throw (2,-1,3) , (-3,4,3) and parallel to the vector <-5,-3,2>

2- find the distance between the two planes 4x-3y+2z-9=0 .
8x-6y+4z=0

3-find the equation of the plane through (1,2,3) and perpendicular to
3x-y+2z+5=0 , 4x+2y-3z+8=0

help plz my exams is next week

2. Originally Posted by mohamedsafy
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1- find the equation of the plane throw (2,-1,3) , (-3,4,3) and parallel to the vector <-5,-3,2>

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to #1: You know one point of the plane and one vector (parallel to the plane). You need a second vector such that the 2 known vectors span the plane. The general equation is:

$\vec r = \vec a + s\cdot \vec u + t\cdot \vec v$

$\vec a = (2, -1, 3)$

$\vec u = (-5, -3, 2)$

$\vec v = (2, -1, 3) - (-3, 4, 3) = (5, -5, 0) = k\cdot (1, -1, 0)$

Therefore the equation of the plane is:

$\vec r = (2, -1, 3) + s\cdot (-5, -3, 2) + t\cdot (1, -1, 0)$

3. Originally Posted by mohamedsafy
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2- find the distance between the two planes 4x-3y+2z-9=0 .
8x-6y+4z=0

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The equations of both planes are in normal form. They are parallel because $(8, -6, 4) = 2\cdot (4,-3,2)$

The second plane passes through the origin. You only have to calculate the distance of the origin to the first plane:

Re-write the first equation in Hesse normal form, that means divide the equation by $|\vec n| = \sqrt{29}$

Then the distance of the origin to the first plane is $d = \dfrac{9}{\sqrt{29}}\approx 1.67$

4. Originally Posted by mohamedsafy
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3-find the equation of the plane through (1,2,3) and perpendicular to
3x-y+2z+5=0 , 4x+2y-3z+8=0

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You need a vector which is perpendicular to both planes:

$\vec u = (3, -1, 2) \times (4, 2, -3) = (-1, 17, 10)$

Write the equation of the plane in normal form:

$(-1, 17, 10)\cdot ((x, y, z) - (1,2,3))=0~\implies~-(x-1)+17(y-2) + 10(z-3)=0$ $~\implies~-x+17y+10z-63=0$