vb.net generate qr code If you multiply x by 1 and do not change the values of y or z, then point P will move parallel in Visual Studio .NET

Creator Code-39 in Visual Studio .NET If you multiply x by 1 and do not change the values of y or z, then point P will move parallel

If you multiply x by 1 and do not change the values of y or z, then point P will move parallel
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to the x axis to the opposite side of the yz plane, but P will end up at the same distance from the yz plane as it was before If you multiply y by 1 and do not change the values of x or z, then point P will move parallel to the y axis to the opposite side of the xz plane, but P will end up at the same distance from the xz plane as it was before If you multiply z by 1 and do not change the values of x or y, then point P will move parallel to the z axis to the opposite side of the xy plane, but P will end up at the same distance from the xy plane as it was before
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In Cartesian three-space, the distance of a point from the origin depends on all three of the coordinates in the ordered triple representing the point The formula for this distance resembles the formula for the distance of a point from the origin in Cartesian two-space
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The general formula It s not difficult to derive a general formula for the distance of a point from the origin in Cartesian three-space, as long as we re willing to use our spatial mind s eye Suppose we name the point P, and assign it the coordinates
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P = (xp,yp,zp) Figure 7-4A shows this situation, along with a point P* = (xp,yp,0), which is the projection of P onto the xy plane We ve moved again back to the perspective of Fig 7-2, looking in toward the origin from somewhere far out in space near the negative y axis To find the distance of P* from the origin, we can work entirely in the xy plane This gives us a two-dimensional distance problem, which we learned how to handle in Chap 1 Let s call the distance of P* from the origin by the name a Using the formula we learned in Chap 1 for the distance of a point from the origin in Cartesian xy plane, we have a = (xp2 + yp2)1/2
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First, we find the distance from the origin to P*, and call it a
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P* (xp, yp, 0) x a
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z P (xp, yp, zp)
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Figure 7-4A Finding the distance of point P from the origin: step 1
Distance of Point from Origin
This completes the first step in a three-phase process Figure 7-4B shows the second step Here, we find the distance between P* and P Let s call that distance b It s the perpendicular distance of P from the xy plane, which is simply the coordinate value zp Therefore, we have b = zp That s the end of the second step In Fig 7-4C, the distance from the origin to P is labeled c Note that we now have a right triangle with sides of lengths a, b, and c The right angle is between the sides whose lengths are a and b The Pythagorean theorem therefore allows us to make the claim that a2 + b2 = c2 Substituting the previously determined values for a and b into this formula gives us [(xp2 + yp2)1/2]2 + zp2 = c2 which simplifies to xp2 + yp2 + zp2 = c2
Second, we find the distance from P* to P, and call it b
P* (xp, yp, 0) x a
z P (xp, yp, zp)
Figure 7-4B Finding the distance of point P from the origin: step 2
Cartesian Three-Space
Third, we call the distance from the origin to P by the name c +z
P* (xp, yp, 0) x Right angle b a c
z P (xp, yp, zp)
and note that c is the length of the hypotenuse of a right triangle!
Figure 7-4C Finding the distance of point P from the origin: step 3
When we switch the right-hand and left-hand sides of this equation and then take the 1/2 power of both sides, we get the formula we ve been looking for, which is c = (xp2 + yp2 + zp2)1/2
An example Let s find the distance from the origin to the point P = ( 5, 4,3) as shown in Fig 7-3 We have xp = 5, yp = 4, and zp = 3 If we call the distance c, then
c = (xp2 + yp2 + zp2)1/2 = [( 5)2 + ( 4)2 + 32]1/2 = (25 + 16 + 9)1/2 = 501/2
Another example Now let s find the distance from the origin to Q = (3,5, 2) as shown in Fig 7-3 This time, the coordinates are xq = 3, yq = 5, and zq = 2 We can again call the distance c, so
c = (xq2 + yq2 + zq2)1/2 = [32 + 52 + ( 2)2]1/2 = (9 + 25 + 4)1/2 = 381/2
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