Parametric equations, polar coordinates, and vector-valued functions
How do you find a vector parametric equation r(t) for the line through points
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Answered question
2022-05-28
How do you find a vector parametric equation r(t) for the line through points and
Answer & Explanation
Scarlet Reid
Beginner2022-05-29Added 8 answers
Step 1 The general form of the 3D vector equation of a line is a point plus a vector multiplied by a scalar, t: r(t)=(xp,yp,zp)+t(xv,yv,zv) The parametric equations are: x=txv+xp y=tyv+yp z=tzv+zp Equations (1) and (2) are the x parametric equation evaluated at 6 and 10 respectively: -3=6xv+xp [1] -8=10xv+xp [2] Eliminate xp by subtracting equation (1) from equation (2) -5=4xv Solve for xv : xv=-54 Substitute -54 for xv in the general equation: r(t)=(xp,yp,zp)+t(-54,yv,zv) Substitute for xv in equation (1): 6(-54)+xp=-3 Solve for xp : xp=92 Substitute 92 for xp in the general equation: r(t)=(92,yp,zp)+t(-54,yv,zv) Equations (3) and (4) are the y parametric equation evaluated at 6 and 10 respectively: -1=6yv+yp [3] -4=10yv+yp [4] Subtract (3) from (4): -3=4yv Solve for yv yv=-34 Substitute into the general equation: r(t)=(92,yp,zp)+t(-54,-34,zv) Substitute -34 for yv in equation (3): -1=6(-34)+yp Solve for yp : yp=72 Substitute into the general equation: r(t)=(92,72,zp)+t(-54,-34,zv) Equations (5) and (6) are the z parametric equation evaluated at 6 and 10 respectively: 1=6zv+zp [5] 5=10zv+zp [6] Subtract equation (5) from equation (6): 4=4zv Solve for zv zv=1 Substitute into the general equation: r(t)=(92,72,zp)+t(-54,-34,1) Substitute 1 for zv in equation (5): 1=6+zp zp=-5 Substitute into the general equation: r(t)=(92,72,-5)+t(-54,-34,1) The above is the vector equation. The parametric equations are: x=-54t+92 y=-34t+72 z=t-5 The symmetric equations are: x-92-54=y-72-34=z-51