I am working on my maths homework and encounter the following question which I have no clue to answer:

Let A = (1, 1, 2), B = (-3, 1, 4), C = (-1, -1, 0) be points in space.

Q1: Find all values x such that the vector u = [2, 14, x] is a linear combination of vectors AB and AC.

I got vectors AB [ -4, 0, 2] and AC [-2, -2, -2], and combination of vectors AB and AC [-6, -2, 0]. Then I equated that with vector u in the following manner:

[2, 14, x] = a[-4, 0, 2] + b[-2, -2, -2] where a = 3 and b = -7 which gives x = 20. Is this the only value of x?

Q2: Find a vector form and parametric equations for the line passing through A and perpendicular to the triangle ABC.

This one I surrender. No clue.

Please kindly help cos I have been stuck in this question for hours. Many many thanks!!!

  • $\begingroup$ What have your tried? $\endgroup$ – Paras Khosla Mar 22 at 13:36
  • $\begingroup$ Hint: perpendicular is related with dot product of vectors. $\endgroup$ – Ertxiem - reinstate Monica Mar 22 at 13:53

Q1 Yes, it's correct (unless you miscalculated), $x=10$ is the only value that makes it linearly dependent to $AB$ and $AC$.

Q2 $AB\times AC$ gives a normal vector to $ABC$ plane, so it's the direction vector of the line perpendicular to this plane, and you also know a point on the line, namely $A$.

  • $\begingroup$ Thank you very much Berci! In fact, there is a question before Q2 that I have to answer is the area of the triangle ABC. The answer for that is cross product of AB x AC, take the strength of it and divide it by 2, which I got 2(14)^(1/2). Looking at Q2 I have no clue because I forgot the plane was 3D instead of 2D, and so for a line through A perpendicular to the triangle would be a line forming a right angle with BC. Now I got the cross product of AB x AC = [4, -12, 8] and given A [1, 1, 2], the vector in vector form = [1, 1, 2] + t[4, -12, 8] by treating OA from origin to A. Am I correct? $\endgroup$ – Axis Mar 22 at 14:34
  • $\begingroup$ I don't entirely follow, but the conclusion seems correct, yes. $\endgroup$ – Berci Mar 22 at 16:36

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