Let v1, v2 and v3 be the vectors in the plane from the points P1, P2 and P3 respectively to a point P. Express in terms of the dot product and these three vectors the condition that P is on the altitude of the triangle P1P2P3 from the vertex P1. (By altitude we mean the entire line through a vertex perpendicular to the opposite side, not just the segment from the vertex to the side.) How to go about it?
I have a solution but I am not sure if it is correct. Can someone ascertain? |dw:1436768645510:dw| From point P3 to P2, the vector will be (v3 - v2). For P to be on the altitute, v1 which passes through P should be perpendicular to P3P2, i.e (v3 - v2). For it to be perperndicular, dot product should be 0. Thereby , v1.(v3 - v2) = 0. v1.v3 = v1.v2.
Yes, your reasoning is good. Why are you unsure? by definition of "altitude", the line (or vector) from P1 extended to the opposite side P2 P3 is perpendicular to the side P2 P3 as you note, the direction of the altitude is defined by the vector P1 to P i.e. P-P1= v1 also, vector v3 takes us from P3 to P, then going "backwards" along vector v2 (i.e. -v2) takes us to P2. in other words, v3+ - v2 or v3-v2 represents the direction from P3 to P2 and as you stated, the dot product of v1 with (v3-v2) must be 0 for these vectors to be orthogonal.
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