Answer:
a) x+4y-z=2
b) -7/[tex]\sqrt{15}[/tex]
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[tex]\sqrt{15}[/tex]
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 /[tex]\sqrt{A^{2}+B^{2}+C^{2} }[/tex]
So distance of point A from P2 is;
d= 3×(-2)-1(1)+2(0)/ [tex]\sqrt{9+1+4}[/tex]
==> d= -7/[tex]\sqrt{15}[/tex]
