Find parametric equations and symmetric equations for the line through P0 and pe

Question

Find parametric equations and symmetric equations for the line through P0 and perpendicular to both given vectors. (P0 corresponds to t=0.)
p0= (5,3,0)
i+j and j+k

x=_______
y=_______
z=t
________=_________=z

Explanation / Answer

x + 5 = y/2 = z + 5 The general form of the parametric equation of a line is: (x-ax)/(m1)=(y-ay)/(m2)=(z-az)/(m3) Parametric equations look like: x = x0 + at, y = y0 + bt, z = z0 + ct. So t = (x-x0)/a, t = (y-y0)/b, t = (z-z0)/c. To form this equation, treat each expression in the parametric equation of a line as an equality (for example, in the parametric equation, you have x: ax + (t)(m1)It should be treated as if it was x=ax + (t)(m1) Solve each expression for the parameter, t. Since each of the resultant expressions is equal to the constant t, they are also equal to each other. Express them this way. => symmetric form : (x- x0)/a= (y- y0)/b= (z- z0)/c parametric form: (x- x0)/a= (y- y0)/b= (z- z0)/c= t => (x- x0)/a= t, (y- y0)/b= t, and (z-z0)/c= t. Solve those for x, y, and z to get parametric equations for the same line. From parametric equations, x= At+ x0, y= Bt+ y0, z= Ct+ z0, a vector parallel to the line is . This tutorial may help you - you can click on the graphs too (interactive): http://www.ehow.com/how_4742009_equation… Solution for your probem: P0 = (3,-1,2) and the cross product of the 2 vectors will be a vector perpendicular to both, so thakes the form: (i + j)x(j + k) = i + j + k L: (3, -1, 2) + t(1, 1, 1) x = 3 + t y = t - 1 z = 2 + t Solve for t: x - 3 = y+1 = z - 2

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