Question

Find the point of intersection the plane 3x-y+2z-3=0 and the straight line (x+1)/3=(y+1)/2=(z-1)/-2

Answer 1

Answer on question #68629, Math / Other

Question Find the point of intersection the plane $3x-y+2z-3=0$ and the straight line $(x+1)/3=(y+1)/2=(z-1)/-2$

Solution We find this point by solving system of equations

$3x-y+2z-3=0$

$(x+1)/3=(y+1)/2$

$(y+1)/2=(z-1)/-2$

From first:

$y=3x+2z-3$

Substituting into second and third

$(x+1)/3=(3x+2z-2)/2$

$(3x+2z-2)/2=(1-z)/2$

From last one:

$3x+2z-2=1-z$

$x+z=1$

$z=1-x$

Hence

$(x+1)/3=(3x+2-2x-2)/2$

$(x+1)/3=x/2$

$3x=2x+2$

$x=2$

Then

$z=1-2=-1$

$y=3\cdot 2+2\cdot(-1)-3=1$

Coordinates of point is (2,1,-1).

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