Find the Equation of the Normal to the Center of PQ | Linear Algebra Question

[sonya]

Let P=(9,-7,-1) and Q=(-5,-5,-5). The set of all points that are equidistant from P and Q are given by the equation __x + __y + __z + __ = 0.

^^The question!

All I want to know is how to start this question. I thought I was supposed to find the equation of the normal to the center of PQ but that didnt work out. Thanks!

 

[quantumdude]

You need to set up the distance formula twice. Once for the distance between point P and the point (x,y,z) and once for the distance between point Q and the point (x,y,z). And since (x,y,z) is equidistant from P and Q, you do...what?

 

[wais]

To find the equation of the normal to the center of PQ, we first need to find the center of PQ. This can be done by finding the midpoint of the line segment connecting P and Q. The midpoint formula is given by:

Midpoint = ( (x1 + x2)/2, (y1 + y2)/2, (z1 + z2)/2)

Substituting the coordinates of P and Q, we get the midpoint as:

Midpoint = ( (9 + (-5))/2, (-7 + (-5))/2, (-1 + (-5))/2) = (2, -6, -3)

This is the center of PQ. Now, to find the equation of the normal to this point, we need to find the vector perpendicular to the line segment PQ. This can be done by taking the cross product of the vectors PQ and the vector (1, 1, 1).

PQ = Q - P = (-5, -5, -5) - (9, -7, -1) = (-14, 2, -4)

Cross product = (-14, 2, -4) x (1, 1, 1) = (-6, 10, -16)

So, the equation of the normal to the center of PQ is given by:

-6x + 10y - 16z = d

To find the value of d, we can substitute the coordinates of the midpoint (2, -6, -3) into this equation.

-6(2) + 10(-6) - 16(-3) = d

d = 12 - 60 + 48 = 0

Therefore, the equation of the normal to the center of PQ is:

-6x + 10y - 16z = 0

This is the equation of the plane that is perpendicular to the line segment PQ and passes through the center of PQ.

Additionally, the set of all points that are equidistant from P and Q can be found by taking the average of the coordinates of P and Q.

Average = ( (9 + (-5))/2, (-7 + (-5))/2, (-1 + (-5))/2) = (2, -6, -3)

This is the same point we found earlier, which confirms that it is the center of PQ.

Therefore