TO find a unit vector perpendicular to a 3-D plane formed by points (1,01),(0,2,2) and (3,3,0)?
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Since you already know what vectors are, take two vectors from this plane as a basis for all vectors in the plane. The easiest basis is just to take two sides from the triangle formed by the 3 given points. \nFrom your study of linear algebra, you may remember that one way of checking whether a set of n vectors was linearly independent was to take the n-volume of their parallelotope (their determinant). If the n-volume was 0, then obviously the set of vectors makes less than an n-dimensional parallelotope, so they can't span an n-dimensional space. We also define a vector v to be orthogonal to a vector w if their inner product is 0.\nThus, we are looking for the vector w = (w1, w2, w3) such that the determinant (u, v, w) is non-zero and u*w=0 and v*w=0. If we expand the determinant above along the unknown vector w, choosing the n-volume to be the same as the area of the parallelegram spanned by u and v and expand the dot products, we find we have system of 3 linear equations in 3 variables. The solution vector of this system is called cross product of u and v, the formula for which you can find in the source below. It is a vector perpendicular to u and v that has as magnitude the area of the parallelegram spanned by u and v.
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The strategy is as follows:\n\n1.  Find two vectors in the plane.  One can be found using the line.  Another can be found using the point A and any point you like on the line.  \n\n(If (x,y,z) and (u,v,w) are two points in the plane, then the vector (x-u)i + (y-v)j + (z-w)k lies in the plane; or more accurately, this vector is parallel to the plane.)\n\n2. Find the cross product of the two vectors in part 1.  Call this new vector N (for normal).\n\n3. The equation of the plane is m(x-a) + n(y-b) + p(z-c) = 0, where N = mi + nj + pk and where A = (a,b,c).\n\n(Why does step 3 work?  A point B = (x,y,z) lies in the plane if and only if the vector AB is orthogonal to N.  In other words, their dot product is zero.  If you write this out in coordinates, you get exactly the condition that m(x-a) + n(y-b) + p(z-c) = 0.)
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Perpendicular lines have slopes with a product of -1. Your line (y=3x-5) has a slope of 3, so any line with a slope of -1/3 would be perpendicular to it. \n\nThus, any line with a slope *not equal to* -1/3 would not be perpendicular to it.
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To find a point on the plane, you simply need to find any values of x, y, and z which make the equation true. The easiest way to find a point is to choose two of the variables and set them equal to zero. So in your example, I might choose to set x and z to zero. Then -y must equal 4, so y must be -4 when x and z are zero. This point is written (0,-4,0). Other points include (2,0,0), (0,0,-2). In this plane, we can pick any values for two of the variables and find the third. This wouldn't be true for planes which are parallel to one of the axes.
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1. For two given points P(p1, p2, p3) and Q(q1, q2, q3), the vector v between them is v = (x, y, z) = (q1 - p1, q2 - p2, q3 - p3).\n\nThe magnitude or length of v, denoted by ||v|| is\n||v|| = sqrt[ x^2 + y^2 + z^2 ]\n\nJust do that for every line that you have.\n\n2. This one's a little harder. Let's define the first line as the vector u, and your second line that you're using as the vector v.\n\nYou first need to write each one as a unit vector, which means dividing all of the values through by a value so that the magnitude will be 1. For each of these vectors, this means dividing by the length of the vector.\n\nAs an example, if I wanted to turn w = (1, 1, 1) into a unit vector, I would notice that the magnitude is ||w|| = sqrt[3] and divide all the values through by that to get (1 / sqrt[3], 1 / sqrt[3], 1 / sqrt[3]).\n\nSo, let us say that the new unit vectors are u / ||u|| = (a, b, c) and v / ||v|| = (x, y, z)\n\nWhat these values give you now are called the direction cosines of each vector. If A is the angle between u and the x axis, then a = cos A, and similarly for b with the y-axis and c with the z-axis.\n\nSo, to find the angles between each vector and the x, y, and z axes (also called the direction angles), you take the inverse cosine (cos -1) of the components (values) of the vector.\n\nLet us say that after taking the inverse cosine of u / ||u|| = (a, b, c) = (cos A, cos B, cos C), you get (A, B, C), and the inverse cosine of v / ||v|| gives you (X, Y, Z).\n\nThe angles between the two vectors in the x, y, and z directions are (X - A, Y - B, Z - C).\n\nSo the main steps are finding the magnitude of each vector, normalizing each vector (making them into unit vectors), taking the inverse cosine of each vector, and subtracting the two vectors.\n\nHope that helps.
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1) Find the line that is perpendicular to the line 3x -4y = 18 and passes through the point (3,4)\n2) Find the point where the two lines intersect.\n3) Find the distance between that point and the point (3,4).
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find the slope, m, of your line (-2/3 in this case). \n\nThe slope of any perpendicular line will be -1/m (3/2 in this case).\n\nSo for any real number b,\n. . . . . . . y = (-1/m)x + b = (3/2)x + b\nis the equation of a line perpendicular to the original line.
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Use this formula(it uses the third-order determinant)\n\nIf the 3 points are\n(x1,y1)\n(x2,y2)\n(x3,y3)\nThen the area A is:\n\nA = \n|x1 y1 1|\n|x2 y2 1|\n|x3 y3 1|\n\n\n\n^_^
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if these are vectors then,\nvector c has a direction of (i+j+k)/√3\nand since vector b is along c vector b is 3(i+j+k)/√3 =     √3(i+j+k).\n Now finding a dot b  and a cross b will be easy.\na.b=4√3   and    a*b=√3(i-k)  by normal vector multiplication.
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Perpendicular lines have slopes whose product is -1.\n\nIf the original line has slope 3, the line that is perpendicular has slope -(1/3)\n\nNow use the point slope form of the line.\n\ny - y1 = m(x - x1)\n\nIf (x1,y1) = (2,2) - your point, then:\n\ny - 2 = (-1/3)(x - 2)\n\nWe can convert this into slope-intercept form.\n\ny - 2 = (-1/3)x + (2/3)\ny = (-1/3)x +(8/3)\n\nBoth equations represent the same line.
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They're vectors, so take into account the direction. Thankfully, they're all in the same dimension, so you can make north positive and south negative (or vice versa, if you prefer), then add. -1 + 3 - 6 + 2 = -2, that is, 2 km south.\nIf they were in different directions, you'd do the same thing, adding horizontal and vertical separately, then using the Pythagorean theorem to find the resultant vector.
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There is several way to solve the problem, but I think it is easies, to find the parametric representation, and then use, that the scalar-product of the orientations-vectors is zero, if the line are perpendicular, and the orientations-vectors is linear depended if parallel.\n\nL1:  y = -½x-3 => (x,y)= (1,-½)*t+(0,-3)\n4x-2y+5 = 0 <=> y=2x+5/2=>\nL2: (x,y) = (1,2)*t + (0+5/2)\n\nbecause (1,-½)*(1,2)=1-1=0 the lines are perpendicular
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Best to plot this on a piece of graph paper; "+i" is  to the right' , "+j" is upwards, "-i" is to the left, "-j" is downwards, all with respect to the origin.\n\n1. Draw the vectors on the graph paper. "u" is in the second quadrant (i is neg and j is pos). "v" is in the fourth quadrant, with (i is pos, j neg) . There are two answers to this question, since you could go from u to v or from u to v in the opposite direction. I always take the angle from zero to "u" and add it to the angle from 0 to "v".  For "u", the tangent of the angle to the vertical axis is 1/4, then angle = 14%. So the angle from zero degrees to the vert axis = 90, and "u"s angle from zero degrees is 90 + 14 = 104 degrees. For "v", the tangent of the angle to the horizontal axis is 4/3 = 1.333.. and the angle = 53 degrees. Therefore the angle from "u" to "v" is 104 + 53 = 157 degrees. (the other answer is 360 degrees - 157 degrees = 203 degrees if you rotate in the other direction).\n\n2. Again use graph paper. 240 degrees is 30 degrees less that 270 degrees (the vertical axis, down) draw a line 10 graph units long. It will be in the third quadrant where i and j will both be negative. At this point you can use a polar to rectangular converter, or do it by hand. By hand: the i side of the triangle formed with the down axis can be found by saying sin(angle) = i/10, and cos(angle) = j/10. The angle is 30 degrees (from vert downwards). so sin(30) = 0.5, and, cos(30) = 0.866; thus i= 0.5 and j = 0.866 (ignoring pos/neg). Since the vector is in the third quadrant, both i and j are negative.\nAnswer = -0.5i -0.866j\n\nHope this helps!
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Remeber that perpendicular lines have negative reciprocal slopes, so the slope of the line you are trying to find is 1/3 (that is the negative reciprocal of -3).  You know that the y intercept is 5, remember that slope intercept form is y = mx + b where m is slope and b is y intercept.  You know both of these things now, the m = 1/3 and the b = 5.  Your equation is y = 1/3x + 5.
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The slope of y = (1/3)x - 5 is m = 1/3\nA line that is perpendicular to the above line will have slope m = -3.\nIf two lines are perpendicular, then the product of their slopes is -1.  For this case, we have (1/3)(-3) = -1, so -3 must be the slope.\nSince you're not given a point, any equation of the form \ny = -3x+b will work.\nJust pick whatever value of b that you want.\nSome examples, \ny = -3x\ny = -3x + 5\ny = -3x - 7
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The direction (angle) of |u| and |v|?
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Hi I assume you mean an unsymmetrical item. I also assume you want a best fit. I have done similar things on a computer but not by hand. The computer finds three point most distant from the centroid and uses these points as pivots for three orthogonal planes. Not sure of the rest.