**A methane molecule CH**

_{4}may be fitted in a cube of side 2a such that the C atom lies at the body centre and four H-atoms at non-adjacent corners of the cube as shown in figure. The angle between any two C-H bonds is**(a) 120º**

**(b) cos**

**–**

^{–1}(**1/3)**

**(c) 150º**

**(d) sin**

^{–1}(1/3)

**(e) cos**

^{–1}(1/3)

This MCQ appeared in Kerala Engineering Entrance 2006 question paper. You are required to find the angle between the lines CH

_{1}and CH_{2}. Treat these lines as vectors. The coordinate axes are marked (by the question setter) in the figure with H_{3}as the origin.
The vectors

**CH**and_{1}**CH**have X, Y and Z components of the same magnitude ‘a’ since the carbon atom C is at the centre of the cube of side 2a. These vectors can be written as_{2}**CH**–a

_{1}=**i**+a

**j**+a

**k**

**CH**+a

_{2 }=**i**–a

**j**+a

**k**

[Note that bold face letters represent vectors and

**i**,**j**,**k**are unit vectors in the X, Y and Z directions respectively].
The magnitudes

*CH*_{1}_{ }and_{ }*CH*of these vectors are the same, equal to a√3._{2}
The angle between the vectors

**CH**and_{1}**CH**is given by_{2}
Cos(

**CH**,_{1}**CH**) = (_{2}**CH**_{1}.**CH**)/_{2}*(**CH*)(_{1}*CH*) = (–a_{2}^{2}–a^{2}+a^{2})/ (a√3)( a√3)
= –(1/3)

The angle between the vectors, which is the bond angle required, is therefore

**cos****–**^{–1}(**1/3)**.
Now, consider the following question. This is a simple one, but similar questions are often seen in entrance test papers.

Angle (in rad) made by the vector

**j**+**√3****k**with the Y-axis is
(a) π/8

(b) π/6

(c) π/4

(d) π/3

(e) π/2

You should remember that the direction cosines (the cosines of the angles between coordinate axes and the vector itself) are given by

*Cos(***A**,x) =*A*/_{x}*A*, Cos(**A**,y) =*A*/_{y}*A*and*Cos(***A**,z) =*A*/_{z}*A.*
Since the Y-component has magnitude 1 and the vector has magnitude 2, the cosine of the angle between the Y-axis and the given vector is ½. The angle therefore is π/3.

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