Question

In Exercises 53 and 54 , sketch a graph of the line and find the points (if any) where the line intersects the \(x y-, x z-,\) and \(y z\) -planes. $$ x=1-2 t, \quad y=-2+3 t, \quad z=-4+t $$

Short answer

The intersection points of the line with the three planes are \((-7, 10, 0)\) with the \(xy\)-plane, \((5/3, 0, -2/3)\) with the \(xz\)-plane, and \((0, 0.5, -3.5)\) with the \(yz\)-plane.

Step-by-step solution

Find the intersection of the line with the \(xy\)-plane

The \(xy\)-plane is given by \(z=0\). Substituting \(z=0\) in the equation of the line, we get \(0=-4+t\). Solving for \(t\), we get \(t=4\). Substituting \(t=4\) in the equations \(x=1-2t\) and \(y=-2+3t\), we get the intersection point as \((-7, 10, 0)\).

Find the intersection of the line with the \(xz\)-plane

The \(xz\)-plane is given by \(y=0\). Substituting \(y=0\) in the equation of the line, we get \(0=-2+3t\). Solving for \(t\), we get \(t=2/3\). Substituting \(t=2/3\) in the equations \(x=1-2t\) and \(z=-4+t\), we get the intersection point as \((5/3, 0, -2/3)\).

Find the intersection of the line with the \(yz\)-plane

The \(yz\)-plane is given by \(x=0\). Substituting \(x=0\) in the equation of the line, we get \(0=1-2t\). Solving for \(t\), we get \(t=1/2\).Substituting \(t=1/2\) in the equations \(y=-2+3t\) and \(z=-4+t\), we get the intersection point as \((0, 0.5, -3.5)\).

Sketch the graph

Draw the three-dimensional coordinate system and plot the three points \((-7, 10, 0)\), \((5/3, 0, -2/3)\), and \((0, 0.5, -3.5)\). Join these points to form the line. This step could be better shown using a graphing tool or a software.