Answer
(a) The angle between the line ab and line ad is $54.7^{\circ}$
(b) The angle between line ac and line ad is $35.3^{\circ}$
Work Step by Step
(a) Let $A$ be (line ab) = $1\hat{z}$
Let $B$ be (line ad) = $1\hat{x}+1\hat{y}+1\hat{z}$
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 $\vec{A} \cdot \vec{B}$.
$\vec{A} \cdot \vec{B} = A_xB_x + A_yB_y + A_zB_z$
$\vec{A} \cdot \vec{B} = 0+0+(1)(1)$
$\vec{A} \cdot \vec{B} = 1$
We can find the magnitude of A.
$A = \sqrt{(A_x)^2+(A_y)^2+(A_z)^2}$
$A = \sqrt{0+0+(1)^2}$
$A = 1$
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.
$cos(\phi) = \frac{\vec{A} \cdot \vec{B}}{AB} = \frac{1}{(\sqrt{3})(1)}$
$\phi = cos^{-1}(\frac{1}{\sqrt{3}})$
$\phi = 54.7^{\circ}$
The angle between the line ab and line ad is $54.7^{\circ}$
(b) Let $A$ be (line ac) = $1\hat{y}+1\hat{z}$
Let $B$ be (line ad) = $1\hat{x}+1\hat{y}+1\hat{z}$
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 $\vec{A} \cdot \vec{B}$.
$\vec{A} \cdot \vec{B} = A_xB_x + A_yB_y + A_zB_z$
$\vec{A} \cdot \vec{B} = 0+(1)(1)+(1)(1)$
$\vec{A} \cdot \vec{B} = 2$
We can find the magnitude of A.
$A = \sqrt{(A_x)^2+(A_y)^2+(A_z)^2}$
$A = \sqrt{0+(1)^2+(1)^2}$
$A = \sqrt{2}$
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.
$cos(\phi) = \frac{\vec{A} \cdot \vec{B}}{AB} = \frac{2}{(\sqrt{3})(\sqrt{2})}$
$\phi = cos^{-1}(\frac{2}{\sqrt{6}})$
$\phi = 35.3^{\circ}$
The angle between line ac and line ad is $35.3^{\circ}$
