The projection of a point on the coordinate axis of the space coordinate system How to find. Can give an example to illustrate, as long as I can understand. Success can be + points How to find it when drawing

The projection of a point on the coordinate axis of the space coordinate system How to find. Can give an example to illustrate, as long as I can understand. Success can be + points How to find it when drawing


(1.2.3)
Projection on X-axis (1.0.0)
It's equivalent to keeping the number of that axis
The rest is zero



In the space rectangular coordinate system, the distance from a certain point to the three coordinate axes is 1?
I made root 3, but the answer is root 6 / 2. Why?


√3
It's obvious that we use the Pythagorean theorem twice
Distance = √ {[√ (1 ^ 2 + 1 ^ 2)] ^ 2 + 1 ^ 2}
for the first time
The second time



If the square difference of the distance between the summation point O (0,0), a (C, 0) is constant, the trajectory equation of the point C should be solved in detail. Should we consider who minus whom?


Let such a point be p (x, y), then Po ^ 2 = (x-0) ^ 2 + (y-0) ^ 2 = x ^ 2 + y ^ 2PA ^ 2 = (x-C) ^ 2 + (y-0) ^ 2 = x ^ 2-2cx + C ^ 2 + y ^ 2, so | Po ^ 2-pa ^ 2 | = C | 2cx-c ^ 2 | = square 4C ^ 2x ^ 2-4c ^ 3x + C ^ 4 = C ^ 2. If C = 0, then OA coincides, which is obviously not right, because P is the origin, so C is not equal to 0



Through the ellipse C x ^ 2 / 8 + y ^ 2 / 4 = 1 point P (x0, Y0) to the circle ox ^ 2 + y ^ 2 = 4 lead two tangents PA Pb AB as the tangent point AB and X axis Y axis intersection Mn
1. If the vector PA * vector Pb = 0, find the P coordinate
2. Solving AB equation (expressed by x0, Y0)
3. Find the minimum area of triangle mon


1. Because the meaning vector PA and the vector ob are not zero vectors, so PA ⊥ Pb, so OA ⊥ ob, and because OA = ob, so the quadrilateral OAPB is a square, so Po & sup2; = x0 & sup2; + Y0 & sup2; = 8 ① and the point P is on the ellipse. So x0 & sup2; + 2y0 & sup2; = 8 ② from ① and ②, we get the P (± 2 root) of x0 = ± 2 root sign 2 Y0 = 0