Answer
The angle between the two bonds is $109.5^{\circ}$
Work Step by Step
Let $\vec{A}=i+j+k$ and $ \vec{B}=i-j-k$.
We know that $\vec{A} \cdot \vec{B} = AB~cos(\phi)$, where $\phi$ is the angle between the two vectors.
$cos(\phi) = \frac{\vec{A} \cdot \vec{B}}{AB}$
We can find $\vec{A} \cdot \vec{B}$.
$\vec{A} \cdot \vec{B} = A_xB_x + A_yB_y + A_zB_z$
$\vec{A} \cdot \vec{B} = (1)(1)+(1)(-1)+(1)(-1)$
$\vec{A} \cdot \vec{B} = -1$
We can find the magnitude of A.
$A = \sqrt{(A_x)^2+(A_y)^2+(A_z)^2}$
$A = \sqrt{(1)^2+(1)^2+(1)^2}$
$A = \sqrt{3}$
We can find the magnitude of B.
$B = \sqrt{(B_x)^2+(B_y)^2+(B_z)^2}$
$B = \sqrt{(1)^2+(-1)^2+(-1)^2}$
$B = \sqrt{3}$
We can find the angle $\phi$ between the two vectors.
$cos(\phi) = \frac{\vec{A} \cdot \vec{B}}{AB} = \frac{-1}{(\sqrt{3})(\sqrt{3})}$
$\phi = cos^{-1}(\frac{-1}{3})$
$\phi = 109.5^{\circ}$
The angle between the two bonds is $109.5^{\circ}$
