It was the best thing I could have done at short notice and with not having pots of money. We were matched with a young man who was studying Biomedics at university and the initial conversation with my son, who was slightly reluctant, went so well that we began booking lessons. I saw an add for MyTutor and after a very friendly and helpful telephone conversation I booked a chat with a tutor. I felt I needed to help him as he was getting stressed and anxious so I looked into tutoring. Nevertheless, here it is! However, this is not a ‘proper’ proof since the problem uses a specific example rather than generalizing it with variables, but this is good enough.Julie Plant Published My son was falling behind in A level Biology before his end of year tests. My teacher was also a bit hesitant in proving this apparently random number. The reasoning in my Chemistry textbook was that it was simply the ‘optimum’ angle for a tetrahedral arrangement. Many of you would have learned that for a tetrahedral shaped molecule, or a molecule with four electron bonding pairs, the bonding angle between the hydrogen(s) and the central carbon is 109.4 degrees. In really general terms, VSEPR considers the repulsion between electron bonding and non-bonding pairs in order to deduce the theoretical bond angles between atoms. Using the dot product formula, we can find the bond angle, for the methane molecule:Īnyone doing chemistry would have learned VSEPR theory. Now that point G is known, all that needs to be done is to determine the direction vectors for and. Since, we can solve for and so the central point G is at. The line of intersection between the two planes must pass through G as well due to the symmetry of the molecule. In order to determine the coordinates of the central carbon, or point G, we follow the same steps as we did in part (e). Substituting in in equation, we get the point. Since these atoms are equally spaced in a perfect tetrahedral arrangement, the line of intersection between and must pass through D (and also G, which is the next problem). however is perpendicular to the line connecting hydrogen A and B and goes through the middle of that joining line. This plane passes through hydrogen D! Similarly, also passes through hydrogen D as well.
is the plane connecting the ‘bottom’ of the molecule and is the basin. First, we can label the hydrogen atoms on the ‘bottom’ as A,B and C in an anti-clockwise direction.
This part takes a bit visual and spatial recognition. We know this is an ideal tetrahedral arrangement since the distance between each hydrogen atom is the same and the distance from the central carbon atom to each hydrogen is also the same. Now, we are given a tetrahedral arrangement with points A,B,C and D arranged as hydrogen atoms on a central carbon. You can do this since I simply multiplied the vector by a value of. I did this to make things look nicer (?) and generally easier to work with. Notice I switched the signs on the direction vector. Taking the cross product of the normals of and :Īnd the two planes intersect at point since by substituting into the two plane equations we get one coordinate at which the planes intersect. Since the normal vector of a plane is perpendicular to the plane, the perpendicular of the two normals of two planes is simply the line of intersection that lies on both planes. This is because the cross product of two vectors result in a different vector that is perpendicular to the original two vectors. The direction of the line of intersection of two planes is simply the cross product between the two normals. By following the same procedure as done in part (b), we can obtain:
Now, we are also given the normals of planes and, which are just the direction of and respectively. The midpoint of the lines and can be determined by halving the respective direction vector and then adding this direction to point. Simply take the cross product of and in order to determine the normal vector for. And although the question doesn’t ask, as well.
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