Show there is no pointof intersection of the line: (x-2)/3 = (y+5)/1 =(z-6)/8 an

Question

Show there is no pointof intersection of the line:
        (x-2)/3 = (y+5)/1 =(z-6)/8 and the plane:
        5x+y-2z+2=0 I thought I could dot the normal (5,-2,2) and the directionvector (2,-5,6) and if i got zero that means they areparellel...but that doesnt work out. If they arent parellelthen wouldnt they have to intersect somewhere? Show there is no pointof intersection of the line:
        (x-2)/3 = (y+5)/1 =(z-6)/8 and the plane:
        5x+y-2z+2=0 I thought I could dot the normal (5,-2,2) and the directionvector (2,-5,6) and if i got zero that means they areparellel...but that doesnt work out. If they arent parellelthen wouldnt they have to intersect somewhere?

Explanation / Answer

A very basic way to do the problem is get x,y, and z in termsof t for the equation of the line. We get that x=3t+2, y=t-5, and z=8t+6. Now substitute into theequation of the plane and get 5(3t+2) +(t-5)-2(8t+6)+2=0. This reduces to -5=0 which meansno solution to the equation which means no point ofintersection.

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