Question

In the $CH_4$ molecule, there are four C-H bonds. If two adjacent bonds are in $\hat i + \hat j + \hat k$ and $\hat i - \hat j - \hat k$ direction, then find the angle between these bonds.
a. $\left( {\text{A}} \right){\text{ }}{\sin ^{ - 1}}\left( {\dfrac{{ - 1}}{3}} \right)$
b. $\left( {\text{B}} \right){\text{ }}{\cos ^{ - 1}}\left( {\dfrac{1}{3}} \right)$
c. $\left( {\text{C}} \right){\text{ }}{\sin ^{ - 1}}\left( {\dfrac{1}{3}} \right)$
d. $\left( {\text{D}} \right){\text{ }}{\cos ^{ - 1}}\left( {\dfrac{{ - 1}}{3}} \right)$

Answer 1

Here we have to find out the angle between two bands. To simplify the given data as per we use the formula to find the angle. After doing some simplification we get the required answer.

Formula used:
$\vec A.\vec B = \left| A \right|\left| B \right|\cos \theta$
Where, $\left| A \right| =$modulus or value of $\vec A$
$\left| B \right| =$modulus or value of $\vec B$
$\theta =$The angle between the two vectors.

Complete step by step answer:
The directions of two bonds of CH4 molecule out of four, are given by $\hat i + \hat j + \hat k$and $\hat i - \hat j - \hat k$. Since they are vector quantity, let us put take vector A and vector B as,
$\vec A = \hat i + \hat j + \hat k$
$\vec B = \hat i - \hat j - \hat k$
The dot product of two vectors is, $\vec A.\vec B = \left| A \right|\left| B \right|\cos \theta ..........(1)$
Where, $\left| A \right| =$modulus or value of $\vec A$
$\left| B \right| =$modulus or value of $\vec B$
$\theta =$The angle between two vectors
So, we can find the angle between the given vectors from the formula of dot products, i.e. from equation$\left( 1 \right)$,
$\cos \theta = \dfrac{{\vec A.\vec B}}{{\left| A \right|\left| B \right|}}$
On rewriting we get,
$\Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{{\vec A.\vec B}}{{\left| A \right|\left| B \right|}}} \right).............(2)$
Now, we have to find the value
$\vec A.\vec B = (\hat i + \hat j + \hat k).(\hat i - \hat j - \hat k)$
$\Rightarrow \vec A.\vec B = \hat i.\hat i - \hat j.\hat j - \hat k.\hat k$
$\vec A.\vec B = 1 - 1 - 1$
$\Rightarrow \vec A.\vec B = - 1$
Alsi we can find,
$\Rightarrow \left| A \right| = \sqrt {{1^2} + {1^2} + {1^2}} = \sqrt 3$
$\Rightarrow \left| B \right| = \sqrt {{1^2} + {1^2} + {1^2}} = \sqrt 3$
Therefore, putting the value of $\vec A.\vec B$and$\left| A \right|,\left| B \right|$we get from equation$\left( 2 \right)$,
$\Rightarrow \cos \theta = \dfrac{{ - 1}}{{\sqrt {3.} \sqrt 3 }}$
On simplify we get,
$\Rightarrow \cos \theta = - \dfrac{1}{3}$
Thus we get,
$\Rightarrow \theta = {\cos ^{ - 1}}\left( { - \dfrac{1}{3}} \right)$
So, the angle between the two adjacent bonds that are in $\hat i + \hat j + \hat k$and$\hat i - \hat j - \hat k$ direction, ${\cos ^{ - 1}}\left( {\dfrac{{ - 1}}{3}} \right)$.

Hence, the correct answer is option (D).

Additional information
We get the scalar quantity from the dot multiplication of two vectors and get the vector quantity from the cross multiplication of two vectors.
The scalar product increases If you increase the magnitude of the vectors or the length of the vectors.
At angle 0, the scalar product is at its highest value. As the product is equal to both of them times cosθ. The value of Cosθ is maximum when θ = 0.

Note: The dot product between a unit vector is computed through the following steps,
Suppose $\hat i$is the unit vector of a vector $\vec A$ and also a vector$\vec B$.
If we want to, $\hat i.\hat i$the result will be 1.
Because $\hat i.\hat i = 1.1\cos 0$ and $\hat i.\hat i = 1$
Since the value of the unit vector is 1 and the angle between them is 0 as they are in the same direction.