I am tackling a question about vectors as shown below:
The coordinates of A,B,C, are given as (5,-2,5),(5,4,-1),and
(-1,-2,-1) respectively.(A)show that AB=AC and that angleBAC=60
degree.(B)find the Cartesian equation of Π, the plane passing through A,B, and C.
I proved the first question and found Π:x-y-z=2
(C) Find the Cartesian equation of Π1, the plane perpendicular to (AB)
Passing through the midpoint of [AB], and Π2, the plane perpendicular
to (AC)passing through the midpoint of [AC].
For this question, I got Π1:y-z=-1, Π2:x+z=4
(D)find the vector equation of L, the line of intersection of Π1 and
Π2, and show that it is perpendicular to Π
The problem is , the direction vector I found for L is (1,-1,-1), which is equal to the one for Π! Where did I made mistakes? Can anyone help me ??
(E)Suppose there is a methane molecule in three dimension, the
position of the centers of three of the hydrogen atoms are A,B,and C
as given. The position of the center of the fourth hydrogen atom is D.
using the fact that AB=AD, show that the coordinates of the fourth
hydrogen is (-1,4,5)
For this question, Can I just suppose D is (-1,4,5) and shows that AB=AD, and thus prove the question?
(F)Leting D be (-1,4,5),who’s that the coordinates of G, the position
of the center of the carbon atom are (2,1,2), hence calculate the
angle DGA, the bonding angle of carbon
My thoughts for this question is that maybe I can found the plane where G exists….I don’t know, how I can solve this problem??
Best Answer
(A),(B),(C): Looks right.
The equation for $\Pi$ is, as you found, $x-y-z=2$, or equivalently $(x,y,z)\cdot (1,-1,-1)=2$. That vector $(1,-1,-1)$? It's perpendicular to the plane, not any of the directions in it. You haven't made any calculation mistakes here; it's just a question of how to interpret the plane's equation.
(E): The hydrogen atoms in a methane molecule form a regular tetrahedron, centered at the carbon atom. If we know the positions of three of the points $A,B,C$, the fourth point $D$ lies on the line through the center of triangle $ABC$ perpendicular to the plane those points are on, and also satisfies $AD=AB$. There are of course two points on the line that satisfy this, and both of them work in that they're possible fourth vertices of the tetrahedron.
So, then, there are two things we need to verify about that point $(-1,4,5)$: the distance equality, and that it's on the line we found in part (D).
(F): We need to find the center of that regular tetrahedron. Building on what we already know, it will be a point equidistant from all four vertices. We found planes in part (C) that were the locus of points equidistant from $A$ and $B$ and equidistant from $A$ and $C$ respectively. We found a line in part (D) that was the intersection of those planes - the locus of all points equidistant from $A,B,C$. To find the point equidistant from all four points $A,B,C,D$, we should intersect this line with one more plane, perpendicular to segment $AD$ through its midpoint. (Or, since the statement gave you a set of coordinates, we could just test those for equal distances.)