The vector <em>v</em> (-6, 6, -5) points in the direction of itself, so we start there.
We can capture all points on the line through the origin and the point (-6, 6, -5) by scaling <em>v</em> by an arbitrary real number <em>t</em>.
The line through point <em>P</em> and pointing in the same direction as <em>v</em> is parallel to the other line that passes through the origin. Then the line we want can be obtained by translating the line through the origin by a vector <em>p</em> that points to (-4, 5, 2), so the vector equation for this line is
<em>r </em>(<em>t </em>) = <em>p</em> + <em>t v</em>
<em>r </em>(<em>t </em>) = (-4, 5, 2) + <em>t</em> (-6, 6, -5)
<em>r </em>(<em>t </em>) = (-4 - 6<em>t</em>, 5 + 6<em>t</em>, 2 - 5<em>t</em> )
To get the parametric equations, simply take out the components:
<em>x</em> (<em>t</em> ) = -4 - 6<em>t</em>
<em>y </em>(<em>t</em> ) = 5 + 6<em>t</em>
<em>z</em> (<em>t</em> ) = 2 - 5<em>t</em>
because the answer would be
