course Mth 275
Section 10.6:This is another section that took some thinking about each concept. I did find figure 10.44, which shows the procedure for finding the distance from a point to a line, somewhat misleading. As drawn, the figure appears to show the point (x0, y0) to have the same y-value as the y-intercept (point Q in the diagram). This is obviously not necessarily the case, but the diagram does distract form the explanation.
I have a question about one problem, #9: ?ind the distance between point P (1, 1, -1) and the plane x ?y + 2z = 4.? Using the formula for distance between a point and a plane, I get a numerator of |-6| = 6 and a denominator of sqrt(6), which simplifies to sqrt(6). The book solution is 1. Am I missing something here, or is the book solution incorrect?
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Think in terms of projecting the vector from some point on the plane onto the normal vector.
If you use a normal vector, this projection will be just the dot product of that unit normal with the other vector.
Some of your symbols rendered as ? marks and I'm not going to try to reconstruct the actual equation, but assuming that the equation is x + y + 2z = 4, a normal vector would be i + j + 2 k so the unit normal would be (i + j + 2k) / (sqrt(6)). One point on the plane would be (0, 0, 2) and the vector from this point to (1, 1, -1) would be i + j - 3k. The dot product of this with the unit normal is -4. I'll bet that means that the ? in front of y means -, which would make the dot product 6 and the distance would be 6 / sqrt(6) = sqrt(6). So making certain assumptions, I tend to agree with your solution.