1)find the point ,if any,of intersection of the line (x+1)/(-1) =(y-2)/(2)=(z+2)

Question

1)find the point ,if any,of intersection of the line (x+1)/(-1) =(y-2)/(2)=(z+2)/(3) and the plane P: 2x +3y -z=4<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" />

2) find the anglebetween the line the line (x+1)/(-1) =(y-2)/(2)=(z+2)/(3) and the plane P1 :8x-y+2z=1

3) find an equation of the plane joining the point P(2,1,7); Q(-1,2,-1) and R(3,-2,4)

4) used the cross product of vectors to find the area of the triangle P,Q,R

5) Find an equation of the line of intersection of the line 3x - 2y +5z =1 and x +2y -3z =8

Explanation / Answer

1)x=-1-t

y=2+2t

z=3t-2

So

2(-1-t)+3(2+2t)-(3t-2)=4

So t=6

and so the intersection point is (-7,14,16)

2)angle between line and normal = cos^-1((-8-2+6)/sqrt(14*69))=97.4 degrees

So angle between line and plane=7.4 degrees

3)27(x-2)+17(y-1)-8(z-7)=0

4)A=1/2*|PQxQR|=1/2*sqrt(27^2+17^2+8^2)=16.447

5)(0,43/4,9/2) and (43/14,0,-23/14) lie on both planes. So line will be

x/(-43/14)=(y-43/4)/(43/4)=(z-9/2)/(43/7)

So

-14x=4y-43=7z-63/2

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