Question

10

mathstina
In a methane(CH4). The position of the hydrogen atoms are (0,0,0),(1,1,0),(1,0,1)and (0,1,1);the positionof the carbon atom is (0.5,0.5,0.5). Compute the angle HCH or the angle formed by the hydrogen-carbon-hydrogen bond.

Answer 1

calculus?

Answer 2

yes,vectors.

Answer 3

nope 3d

Answer 4

yeah 3d and vectors do the same

Answer 5

yes

Answer 6

[-0.5,-0.5,-0.5] . [0.5,-0.5,0.5] = 2*sqrt(0.75) cos(A)
=>arccos(-.25/(2*sqrt(0.75)) = A

Answer 7

the angle is 98.3 degrees?

Answer 8

atchyut is the ans right?

Answer 9

actually i donno but i found this once take a look of this

Answer 10

Since we want the angle between ANY H-C-H bond, choose any two vertices of the tetrahedron to be your two hydrogen atoms (the angles are all the same, so it doesn't matter which two you choose). Here, I chose (0, 0, 1) and (0, 1, 0) to be the two hydrogen atoms (they seem to be the easiest to pick).

We want the angle between the following two vectors:
(a) one with endpoints <1/2, 1/2, 1/2> and <0, 0, 1>
(b) one with endpoints <1/2, 1/2, 1/2> and <0, 1, 0>.

Subtracting head from tail, the two vectors are:
<1/2, 1/2, 1/2 - 1> and <1/2, 1/2 - 1, 1/2>
==> <1/2, 1/2, -1/2> and <1/2, -1/2, 1/2>.

The two vectors above both have magnitudes √(1/4 + 1/4 + 1/4) = √3/2. Using:
a • b = |a||b|cosθ,

and:
a • b = <1/2, 1/2, -1/2> • <1/2, -1/2, 1/2>
= (1/2)(1/2) + (1/2)(-1/2) + (-1/2)(1/2)
= 1/4 - 1/4 - 1/4
= -1/4,

we have:
θ = arccos[(a • b)/(|a||b|)]
= arccos{(-1/4)/[(√3/2)(√3/2)]}
≈ 109.5°,

which is to be expected.

I hope this helps!

Answer 11

ok. Thanks. is this the simpler method?

Answer 12

yeah best method but the points here varies and not the same points given by you

Answer 13

(1,0,0), (0,1,0), (0,0,1), and (1,1,1). the centroid is at (1/2,1/2,1/2).

Answer 14

using dot product is always best and easier way than to remember some formulae

Answer 15

ok. i'll try using using my points .

Answer 16

Thats good goood luck and do post the answer

Answer 17

I will check with mine and we can have the correct answer then

Answer 18

ok.

Answer 19

Answer will be the same

Answer 20

atchyut i got the same ans , using my points. thanks a lot!!!!

Answer 21

109 degrees itself

Answer 22

no prob

Answer 23

yes,
109.47 degrees.

Answer 24

atchyut cani ask u one more qn on vectors?

Answer 25

yeah sure

Answer 26

two planes with eqns 3x+4y=1 & x+y-z=3
a) find the angle btwnthe planes.
b)determine the symmetric eqns fot the line of intersection L of these two planes.

Answer 27

i have done part a), i got 36.07 degrees. am i right?
hw abt part b)?

Answer 28

hmmm i am not so sure about it... But symmetric equationis something like
(x-x1)/a=(y-y1)/b=(z-z1)/c isnt it???

Answer 29

yes!

Answer 30

hmmmm then can u find the point of intersection?????

Answer 31

no, i'm nt sure hw to find?

Answer 32

Yeah listen in order to find a point in this situation just put z=0 and solve the other two equations then it will be helpful not that this is the exact procedure but this is taught to me by some prof and so it will be helpful sometimes as in this case

Answer 33

Keeping z=0 and solving other two u will get (11,-8,0) which lies on both the planes

Answer 34

ok, then?

Answer 35

It will be helpful because they are planes and u will definitely find a pint where it meet x,y plane and thats z=0 understood

Answer 36

and now find any two points on each planes get the directional ratios and substitute them in a,b,c

Answer 37

is the ans: x=11-4t
y=-8+3t
z=-t

Answer 38

Yeah right if d.r's are correct

Answer 39

i used cross product to get the ans. was that ur method?

Answer 40

not exactly mine but yeah answer matched

Answer 41

if possible can teach me ur method?

Answer 42

First tell me about ur cross product method dude coz i am confused how we can do that

Answer 43

cross product using 2 eqns, get -4i+3j-k. then using (11,-8,0). so,x=11-4t,y=-8+3t,z=-t.

Answer 44

Yeah u can do it and it is the good procedure...
Mine is somewhat trialing type..
For theoretical proofs ur's is correct and mine is for to choose answer from given options.
yeah u can follow that and that's right

Answer 45

ok,thanks for ur time n help.

Answer 46

when we get that point (11,-8,0) it is (x1,y1,z1) and a b c which are called d.r's are obtained by ur cross product then symmetric equations are
x-11/-4 = y+8/3 =z-0/-1 = t
and each of the equations u have written for x y z are called parametric equations in t.
ie, for each value of t u will get each point on the line which is intersection of the two given planes

Answer 47

is parametric eqn & symmetric eqn has same meaning? am i right?

Answer 48

thanks n bye!

Answer 49

sorry dude sys restart

Answer 50

Actually no symmetric equation is what i told u in the first
(x-x1)/a=(y-y1)/b=(z-z1)/c
while parametric eqns are individually for each of x y and z
x=11-4t,y=-8+3t,z=-t.

Answer 51

ok.understood, i learnt a lot

Answer 52

fine good luck

Answer 53

which country r u frm?

Answer 54

BYE

Answer 55

i am from India

Answer 56

i'm frm singapore. nice to meet u

Answer 57

there should be abt 1pm?? here 3:27 pm

Answer 58

yeah thats right

Answer 59

and what are u studying in btw?

Answer 60

in college.r u a student too?

Answer 61

yeah

Answer 62

ok i'veto go out .bye

Answer 63

k k tc bubye