1. **Problem statement:** Find the length of diagonal $BE$ in cuboid $ABCDEFGH$ with dimensions $AB=11$ cm, $AD=6.1$ cm, and $AE=4.2$ cm. 2. **Formula:** The length of a space diagonal between two opposite vertices in a cuboid is given by $$d=\sqrt{l^2+w^2+h^2}$$ where $l$, $w$, and $h$ are the length, width, and height respectively. 3. **Apply values:** Here, $BE$ is a diagonal from vertex $B$ to $E$. Since $B$ is at $(11,0,0)$ and $E$ is at $(0,6.1,4.2)$, the length is $$BE=\sqrt{(11-0)^2+(0-6.1)^2+(0-4.2)^2}$$ 4. **Calculate:** $$BE=\sqrt{11^2+6.1^2+4.2^2}=\sqrt{121+37.21+17.64}=\sqrt{175.85}$$ 5. **Simplify:** $$BE=13.3\text{ cm (to 1 d.p.)}$$ 1. **Problem statement:** Calculate the angle that diagonal $BH$ makes with the plane $ABFE$. 2. **Understanding:** The plane $ABFE$ is the base plane of the cuboid. The diagonal $BH$ connects vertex $B(11,0,0)$ to $H(0,6.1,4.2)$. 3. **Method:** The angle between a line and a plane is $$\theta=90^\circ - \alpha$$ where $\alpha$ is the angle between the line and the normal vector to the plane. 4. **Find normal vector:** The plane $ABFE$ lies on the base, so its normal vector is vertical, along $AE$, i.e., $$\vec{n} = (0,0,1)$$ 5. **Find vector $\vec{BH}$:** $$\vec{BH} = H - B = (-11,6.1,4.2)$$ 6. **Calculate angle $\alpha$ between $\vec{BH}$ and $\vec{n}$:** $$\cos \alpha = \frac{\vec{BH} \cdot \vec{n}}{|\vec{BH}||\vec{n}|} = \frac{4.2}{13.3 \times 1} = 0.3158$$ 7. **Calculate $\alpha$:** $$\alpha = \cos^{-1}(0.3158) = 71.6^\circ$$ 8. **Calculate angle with plane:** $$\theta = 90^\circ - 71.6^\circ = 18.4^\circ$$ **Final answers:** - Length of $BE = 13.3$ cm - Angle $BH$ makes with plane $ABFE = 18.4^\circ$