A rectangular tank with width of $8 m$ and length $12 m$ , and height $20 m$ lying flat on its $8 \times 12$ base is filled with water to 30% of its height (i.e. $6 m$ ). The tank is then tilted such that one of the space diagonals (the diagonal between two spatially opposite vertices) is vertical to the ground. What is the new depth of water in the tank ? i.e. how high is the surface of water above the ground ?
My Attempt:
The center of the water surface in the tilted tank is the same distance away from the base, i.e. $6 m$ "above" the base. In a proper coordinate system, the original space diagonal vector is $(12, 8, 20)$. We now have have two vectors: $V_1 = (0, 0, 6)$ is the vector representing the distance between the center of the surface of water and the base, and $V_2 = (6, 4, 0)$ is the vector between the center of the base and the lower-most vertex. We just have to find the projection of these two vectors onto the space diagonal. The unit vector along the space diagonal is
$ u = \dfrac{(12, 8, 20) }{ \sqrt{ 12^2 + 8^2 + 20^2 } } = \dfrac{1}{\sqrt{38}} (3, 2, 5) $
Therefore, the new depth of water is
$ d = \dfrac{1}{\sqrt{38}} (3, 2, 5) \cdot ( (0, 0, 6) + (6, 4, 0) ) = \dfrac{1}{\sqrt{38}} (3, 2, 5) \cdot (6, 4, 6) = \dfrac{56}{\sqrt{38}} = 9.0844 $m
Is my attempt correct ?
Yes, you are correct. A different way. After tilting the volume remains the same, i.e. $8\cdot 12\cdot 6=576$. In the coordinate system of the tank, the surface of the water is represented by the plane $8x+12y+20z=t$ where $t$ is given by the volume invariance: $$\frac{1}{20}\int_{x=0}^8\left(\int_{y=0}^{12}(t-8x-12y) dy\right)dx=576\implies t=224.$$ Hence $d$ is the distance of such plane from the origin $$d=\frac{|8\cdot 0+12\cdot 0+20\cdot 0-224|}{\sqrt{ 12^2 + 8^2 + 20^2 }}=\frac{224}{4\sqrt{38}}=\frac{56}{\sqrt{38}}.$$