Find parametric equations and symmetric equations for the line. The line through (-6, 2, 3) and parallel to the line 12x=13y=z+1

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Find parametric equations and symmetric equations for the line.  The line through (-6, 2, 3) and parallel to the line 12x=13y=z+1

Mike Henson · Accepted Answer

Rationalization. Given a dispacement vector v=&amp;lt;a,b,c&amp;gt; which is the difference of the initial position vector r0 and final position vector r as well as the value of an initial position vector r0, we can acquire the parametric equations of a line: {x=x0+aty=y0+btz=z0+ct From this parametric equation, we can acquire the parametric equations of L as follows: x−x0a=y−y0b=z−z0c Deriving the parametric equations: finding displacement vector. 12x+13y=z+1 This is already a symmetric equation of a line parallel to the given line that passes through (-6,2,3) with position vector r0=&amp;lt;−6,2,3&amp;gt;. We can rewrite this as a parametric equation by equating to t: t=12x=13u=z+1⇒{t=12xt=13yt=z+1⇒{x=2ty=3tz=t−1 Deriving the symmetric equations.The symmetric equations are the three-fold equaliy we acquire between x, y and z by eliminating t. It was already mentioned in the chapter that a shorthand formula for this from the parametric equations is: x−x0a=y−y0b=z−z0c Substituting calues, we get: x+62=y−23=z−31=x+62=y−23=z−3

Find parametric equations and symmetric equations for the line. The line

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