Find the points of intersection of the line x=3+2t,y=5+7t,z=−3+t, that is, I(t)=3+2t,5+7t,−3+t, with the coordinate planes.

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yunitsiL · Accepted Answer

The coordinate planes&#039; equations are provided below:yz:x=0,zx:y=0,xy:z=0Consider in xy plane z=0.Substitute 0 for z in the equation of line as given below,z=0⇒−3+t=0t=3, now&amp;nbsp;x=3+2tAt&amp;nbsp;t=3,x=3+2∗3=3+6=9At t=3,&amp;nbsp;y=5+7∗3=5+21=26Point is (9,26,0)Consider in yz plane x=0.Substitute 0 for x in the equation of line as given below,x=0⇒3+2t=02t=-3. nowAt&amp;nbsp;t=−32y=5+7⋅(−32)=10−212=−112At&amp;nbsp;t=−32z=−3−32=−6−32=−92Point is&amp;nbsp;(0,−112,−92)Consider in zx plane y=0.Substitute 0 for y in the equation of line as given below,y=0⇒5+7t=07t=-5, nowAt&amp;nbsp;t=−57x=3+2⋅(−57)=21−107=117At&amp;nbsp;t=−57z=−3−57=−21−57=−267Point is&amp;nbsp;(117,0,−267)Thus, the point of intersections on the coordinate planes are as given below:xy:(9,26,0)yz:(0,112,92)zx:(117,0,−267)

Find the points of intersection of the line x = 3+2t, y =5+7t, z = -3+t, that is, I(t) = 3+2t, 5+7t, -3+t, with the coordinate planes.

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