Rationalize. An equation in R3 will intersect the xy - plane when z=0, intersect the xz - plane when y=0, and intersect the yz-plane when x=0. We acqured from the previous item:

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pattererX · Accepted Answer

{x=2+ty=4−tz=6+3t Letting x=0. We find the point that intersects the yz- plane by letting x=0 as follows and then solving for t such as x=0. Afterwards, we substitute this value of t to find y and z: ⇒ Starting equation, x=0 {x=2+ty=4−tz=6+3t 0=2+t t=−2 z=6+3(−2) 0=2+t t=−2 ⇒ Solve for y and z. y=4−(−2) =6 z=6+3(−2) =0 Thus the point that intersects the yz-plane is (0,6,0). Also observe that the z-coordinate is also 0 so this must also be the point that intersects the xy-plane. In short, this is the point where the line passes through the y-axis. Letting y=0. We find the point that intersects the xz-plane by letting y=0 as follows: ⇒ Starting equation, y=0 {x=2+ty=4−tz=6+3t 0=4−t t=4 ⇒ Solve for x and z. x=2+4 =6 z=6+3(4) =18 Thus, the point that intersects the xz-plane is (6,0,18) Conclusion. We skip letting z=0 since we have solved (0,6,0) as a point. Thus, the two points that passes through the coordinate planes are: (0,6,0) and (6,0,18)

Rationalize. An equation in R^3 will intersect the xy - plane when z=0, in

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