#30. How to find distance from point to plane?
We learn to search and calculate the distance from a point to a plane (without the volume method and the coordinate method). №14 EGE in Mathematics
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A new problem continues the theme of distance. Who has difficulties with the construction of sections: our early videos under the "Egyptian" numbers 3, 4 and 5 - to help!
Condition. On the edges CD and BB1 of the cube ABCDA1B1C1D1 with edge 12, the points P and Q are marked, respectively, with DP = 4 and B1Q = 3. The APQ plane intersects the edge CC1 at point M.
a) Prove that the point M is the midpoint of the edge CC1.
b) Find the distance from point C to the plane APQ.
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In the comments to this video asked about how to formally prove the similarity of the triangles MCP and BQA. The answer is so long that it would be useful to fix it here.
1st method. ∠MCP = ∠QBA = 90 °. CP || BA, MP || QA ⇒ ∠CPM = BAQ. This means that △ MPC ~ △ QAB, pt.d.
In this method, we use the similarity feature at two angles, and the parallelism of the lines MO and QA is explained exactly as you realized it - through the intersection theorem of two parallel planes of the third, formally here it can be written as: ((ABB₁) || CDD₁), (SQA) (ABB₁) = MP, (SQA) (CDD₁) = QA) ⇒ MP || QA. We also use a simple fact: parallel transfer of straight lines does not change the angles, that is, in fact, the CPM angle is combined with the angle BAQ by means of parallel transfer, but you can not talk about it in detail: the main thing is to indicate the parallelism of the corresponding straight lines.
2nd way. The planes ABB₁ and CDD₁ intersect the rays SQ, SB and SA at the points Q, B, A and M, C, P, respectively. Therefore, △ QBA is obtained from △ MCP by a homothety with center S. Therefore, △ MPC ~ △ QAB, rd
The essence of this brief proof will become clear if we understand the third method, when we simply work separately with the planes and the corresponding triangles.
3rd way. ∠MCP = ∠QBA = 90 °. And then through the similarity of △ SCM ~ △ SBQ and △ SCP ~ SBA we come to the fact that MC: CP = QB: BA, which implies △ MPC ~ QAB, rd
But this is a completely detour, and in fact, the purpose of point a) will be achieved before we come to the similarity of interest. The same idea can be completely aggravated and the proportionality of all three sides for the MPC and QAB triangles at all can be explained in terms of flat angles, but I note this for fairness, there is no practical use for this.
That is, if to summarize, there are clear signs of similarity of triangles, and for formal proof it is worthwhile to operate with them,