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If A = 2i + 3j + 4k, B = -1i + 6j - k and C = 1i + 6j + xk represent the sides of right angle triangle then find value of x?
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[Note: Here i, j and k are unit vectors along X-axis , Y-axis and Z-axis respectively]
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This problem is posed and worded INCORRECTLY and HAS NO solution.
According to the problem, the given objects A, B and C are the vectors, representing the sides of a triangle in 3D space.
It means that the first thing which should be checked is that the three given 3D vectors lie in one 2D plane.
For it, the necessary and sufficient condition is that the determinant of the matrix,
comprised of coordinates of these vectors is zero.
det = 0.
The determinant is
2*(6x + 6) - 3*(-x+1) + 4*(-6-6) = 12x + 12 + 3x - 3 - 48 = 15x - 39
The determinant is zero if and only if
x = 39/15 = 13/5.
Then it is easy to check that NO ONE of the tree scalar products of the vectors A, B, C, (A,B), (A,C), (B,C), is zero.
Therefore, no one of the angles of the triangle is right angle.
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The solution given by the other tutor in the post, preceding to mine, is incorrect.
Alan found his value of x, using the Pythagorean equation only, but missed to check if the three vectors
(the sides of the triangle) lie in one plane.
It is just hundreds of times I saw in this forum that the posted problems are incorrect,
which means that either their composers are mathematically illiterate (or inaccurate) OR they recover
their "problems" from UNTRUSTED sources.