$15.00 all questions answered with full work
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- Posted on Jan. 14, 2009 at 10:16:58AM
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Preview: ... ane passing through the intersection of two planes is<br><br> 4x-3y-z-1+k(2x+4y+z-5) =0 <br><br> x(4+2k) +y(4k-3)+z(k-1) +(-1-5k) =0 (eqn1)<br><br> d.rs of the normal to the plane are < 4+2k,4k-3, k-1><br><br>since the above plane is parallel to x axis normal will be perpendicular to x axis <br><br> so (4+2k)*1 +(4k-3)*0 +(k-1)*0 =0<br><br> so 2k+4 =0<br><br> so 2k =-4<br><br> k=-2<br><br> so eqn of the required plane is<br><br> 4x-3y-z-1-2( 2x+4y+z-5) =0<br><br> so -11y -3z +9 =0<br><br> so 11y +3z -9 =0 (answer)<br><br>4) eqn ofany plane passing through intersection of two given planes is<br><br> x-3y-2z-1 +k(2x+4y+z-5)=0<br><br> so x(1+2k) +y( 4k-3) +z(k-2) +(-5k-1) =0<br><br> D.rs of the normal to the plane are < 1+2k,4k-3,k-2><br><br> plane is parallel to x axis so <br><br> 1*(1+2k)=0<br><br> so 2k=-1<br> so k=-1/2<br><br> so eqn of the reqd plane is<br> x-3y-2z-1 -(1/2)(2x+4y+z-5) =0<br><br>so 2x-6y-4z-2-2x-4y-z+5=0<br><br> so -10y -5z +3 =0<br><br> so 10y +5z -3 =0 (answer)<br><br> 1) the given planes intersect each othe ...
The full tutorial is about 1491 words long .
$10.00 Analytic geometry
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- Posted on Jan. 14, 2009 at 10:40:53AM
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Preview: ... - z - 1 = 0 <br>is (4,-3,-1), that is, it has components equal to the coefficients of x,y,z in the equation of the plane.<br><br>Similarly, a vector that is normal to the second plane is <br>(2,4,1).<br><br>Now, the two planes are parallel if the two vectors <br>are parallel, that is, one is a multiple of the other, which is not true because you cannot find number t so that<br>(2,4,1)=t*(4,-3,-1).<br><br>In conclusion, the two planes intersect and the cartesian equation of the line of intersection is<br>given by system of the two equations put together.<br><br>4x - 3y - z - 1 = 0 <br>2x + 4y + z - 5 = 0<br><br><br>That is, all points (x,y,z) on the line satisfy the above system of equations.<br><br>This is the cartezian equation.<br><br>You can go further and solve the system. Multiply the second eq by 2 annd subtract from the first, we get<br><br>-11y-3z+9=0<br><br>Add the two equations together and get<br>6x+y-6=0.<br><br>You can write the general solution as<br><br>((6-y)/6,y,(-11y+9)/3)<br><br>This is also called the parametric equation of the line.<br><br><br><br><br>2.A plane ...
The full tutorial is about 901 words long .
$10.00 Equations of planes
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- Posted on Jan 25, 2009 at 12:58:19PM
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Preview: ... n part of such a plane in the drawing below. This time, we have to be able to get to the plane first from the origin, so we must know the position vector of some particular point in the plane from the origin. In my drawing, this point is M with position vector m. <br> <br>If P is any general point in the plane, so that the vector MP = p, then N and p are perpendicular to each other.<br>Therefore N.p = 0 but p = r - m. <br>N.(r - m) = 0 or N.r = N.m = C <br>where C is the number we get from working out the dot product of the two known vectors N and m. <br> <br>Now suppose we are starting with the plane below, with the 2 known vectors s and t lying in this plane.<br> <br>How could we use s and t to find a vector N which would be perpendicular to this plane? <br>It's worth thinking about this before finding out. <br><br>Finding a normal vector<br>Here's the plane again with s and t being known vectors which lie in it. <br> <br>Working out the cross product of s and t will automatically give us a vector perpendicular to the plane in which s and t lie. This is the speediest method for finding a normal vector to this plane. <br>As an example, suppose s = 2i + j + 3k and t = 3i + 2j + 4k.<br>Then s x t = - 2i + j + k = N, a normal vector to the plane. (The working out of s x t is described at the end of the cross product section where I used the same vectors in an example there.) <br>It's also possible to find a normal vector by using the dot product. If we call the normal vector N, then N must be perpendicular to both s and t.<br>Therefore N.s = 0 and N.t = 0.<br>Now, using algebra it is possible to find an N which fits these two equations.<br><br>Equations of planes in the form Ax + By + Cz = D<br><br>This is another useful way to describe planes. It is known as the cartesian form of the equation of a plane because it is in terms of the cartesian coordinates x, y and z.<br><br>The form Ax + By + Cz = D is particularly useful because we can arrange things so that D gives the perpendicular distance from the origin to the plane.<br>To get this nice result, we need to work with the unit normal vector. This is the vector of unit length which is normal to the surface of the plane. (There are two choices here, depending on which direction you choose, but one is just minus the other). <br>I'll call this unit normal vector n. <br>Next we see how using n will give us D, the perpendicular distance from the origin to the plane.<br>In the picture below, P is any point in the plane. It has position vector r from the origin O.<br> <br>Now we work out the dot product of r and n. <br>This gives us r.n = |r||n|cos A.<br>But |n| = 1 so we have r.n = |r|cos A = D.<br>This will be true wherever P lies in the plane. <br>Next, we split both r and n into their components.<br>We write r = xi + yj + zk and n = n1i + n2j + ...
The full tutorial is about 2794 words long plus attachments.
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Equations of planes in the form Ax.doc (189K) (Preview)
