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beccastark1733
14.02.2020 •
Mathematics
Given the planes P1 : x + 4y − z = 10, P2 : 3x − y + 2z = 4, and the point A (−2, 1, 0).
(a) Determine an equation for the plane that contains A and is parallel to P1.
(b) Determine whether A lies on P2. If it doesn’t, find the distance from A to P2.
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Ответ:
a) x+4y-z=2
b) -7/![\sqrt{15}](/tpl/images/0511/9936/11f23.png)
Step-by-step explanation:
a)
Equation of a plane containing point A is given by;
[x-(-2)]+[y-1]+[z-0]=0 ----(1)
If this plane is parallel to the plane P1 whose normal vector is {1.4.-1] - then this will also be normal to plane (1)
so,
(1)[x-(-2)]+(4)[y-1]+(-1)[z-0]=0
==> x+2+4y-4-z=0
==> x+4y-z=2
will be the plane parallel to P1![\sqrt{15}](/tpl/images/0511/9936/11f23.png)
b) If A lies on P2 then it should satisfy the equation of P2
Putting x= -2, y=1 and z= 0
==> 3(-2)-(1)+2(0)=4
==>-6-1+0=4
==> 5=4
which is not true i-e equation is not satisfied ! So the point lies out of plane. Lets find distance from P2 of point A
We know distance of point (x1,y1,z1) from a plane Ax+By+CZ+D=0 is
d= Ax1+By1+Cz1+D /![\sqrt{A^{2}+B^{2}+C^{2} }](/tpl/images/0511/9936/6bf79.png)
So distance of point A from P2 is;
d= 3×(-2)-1(1)+2(0)/![\sqrt{9+1+4}](/tpl/images/0511/9936/7391f.png)
==> d= -7/![\sqrt{15}](/tpl/images/0511/9936/11f23.png)
Ответ:
Jim’s statement is correct for all numbers greater than or equal to 0, and negative integers. It does not work for negative non-integer numbers. For example, the floor of –3.5 is –4, where –4 is not equal to –3, the number before the decimal point.
Step-by-step explanation: Copy- Paste