Let l be the line that passes through the point P=(−5,−7,−1) and is perpendicular to the plane 8x−6y−1z=15, how do you find the parametric equations?

Question

Cameron Alexander · Accepted Answer

# { (x=-5+8lamda), (y=-7-6lamda), (z=-1-lamda) :} #

Consider the plane equation:

# 8x-6y-1z=15 #

The Normal vector to this plane is given by the coefficient of #x#, #y# and #z#, and so we have;

# vec(n) = ( (8), (-6), (-1) ) #

And so the equation of the line #L# that passes through #P(-5,-7,-1) and has the same directional as the above normal is give by;

# L: \ \ \ vec(r) = ( (-5), (-7), (-1) ) + lamda ( (8), (-6), (-1) ) #

If we assign #vec(r)# a generic coordinate #(x,y,z)# the we have;

# ( (x), (y), (z) ) = ( (-5), (-7), (-1) ) + lamda ( (8), (-6), (-1) ) #
# \ \ \ \ \ \ \ \ \ \ = ( (-5+8lamda), (-7-6lamda), (-1-lamda) ) #

Leading to the parametric equations:

# { (x=-5+8lamda), (y=-7-6lamda), (z=-1-lamda) :} #

Which we can see graphically;

Here the plane is drawn in red and the line #L# in green

Charles Anctil · Answer

undefined To find the parametric equations of the line $l$ passing through point $P(-5, -7, -1)$ and perpendicular to the plane $8x - 6y - z = 15$, we first find the direction vector of the line. Since the line is perpendicular to the plane, its direction vector must be orthogonal to the normal vector of the plane.

The normal vector of the plane $8x - 6y - z = 15$ is $\langle 8, -6, -1 \rangle$.

So, the direction vector of the line is parallel to $\langle 8, -6, -1 \rangle$, which is also perpendicular to the plane.

Now, we have a point on the line ($-5, -7, -1$) and its direction vector $\langle 8, -6, -1 \rangle$.

The parametric equations of the line can be written as:
$$x = -5 + 8t$$
$$y = -7 - 6t$$
$$z = -1 - t$$

where $t$ is any real number.