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Find the angle between the line (x-2)/(3...

Find the angle between the line `(x-2)/(3)=(y-1)/(-1)=(z-3)/(2)" and the plane "3x+4y+z+5=0.`

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The correct Answer is:
` (3)/(2sqrt91) `
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Column I, Column II Image of the point (3,5,7) in the plane 2x+y+z=-18 is, p. (-1,1,-1) The point of intersection of the line (x-2)/(-3)=(y-1)/(-2)=(z-3)/2 and the plane 2x+y-z=3 is, q. (-21 ,-7,-5) The foot of the perpendicular from the point (1,1,2) to the plane 2x-2y+4z+5=0 is, r. (5/2,2/3,8/3) The intersection point of the lines (x-1)/2=(y-2)/3=(z-3)/4a n d(x-4)/5=(y-1)/2=z is, s. (-1/(12),(25)/(12),(-2)/(12))

Find tha angle between the lines (x-20)/(1)=(y+15)/(2)=(z-3)/(-2) and (x+5)/(6)=(y+3)/(3)=(z-16)/(6)

Find the angle between the following lines (x-1)/(2)=(y+1)/(3)=(z-4)/(6) and x+1= (y+2)/(2)=(z-4)/(2) .

Check whether the line (x-1)/(4)=(y+2)/(5)=(z-7)/(6) lines in the plane 3x+2y+z=6 .

Find the equation of the projection of the line (x-1)/2=(y+1)/(-1)=(z-3)/4 on the plane x+2y+z=9.

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