course Mth 275
Section 10.5:This section took a lot of thought about each concept to understand them. The only question I have is about the standard form of the equation of a plane. On page 710, the book states: ?the direction numbers of the normal vector are proportional to A, B, C.? Assuming that any common factor is factored out of A,B, and C and similarly for the direction numbers of the normal vector, then the direction numbers are in fact A,B, and C, correct? This is what I believe is the meaning of the discussion on page 711, which states that the normal vector N = Ai + Bj + Ck must have the Cartesian equation?Ax + By + Cz +D = 0.
Any multiple of a normal vector will be a normal vector.
However the unit normal vector is often the most useful; it's obtained by dividing a normal vector by its magnitude.
Problem #63: ?how that a plane with x-intercept a, y-intercept b, and z-intercept c has the equation
x/a + y/b + z/c = 1
My solution seems so simple, that I want to confirm that it is correct:
At the x-interecept, the point (x, 0, 0) satisfies the equation above. Substituting for y and z gives x/a + 0 + 0 =1, thus x = a. Similarly for y and z. Therefore, since this equation results in the producing the three intercepts a, b, and c, it must be correct.
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That will do it.
Note also, related to your previous question, that the vectors a i - b j and a i - c k are in the plane; their cross product is b c i + ac j + a b k; this is a normal vector. Dividing it by abc gives i/a + j / b + k / c and if you use this as the normal vector with any of the above points you quickly get the same result.