I have no idea how to solve this Vector Problem! doe anybody?

you can find the angle between the two planes, and then add 90 degree to find the answer..

first find the equation of the plane with the three points given..

use the second plane as horizontal plane, take any three points with y as zero, eg 0,0,0 1,0,0 and 0,0,1

and calculate the eqn.

using the two eqns calculate the angle between the planes.. then add 90%

start by drawing up the X,Y,Z axis, and plot the points to get a general feel for the problem. where the points are etc. Then you need to find the equation of the plane in which all points lie.

let point A = (0,0,0)

let point B = (0,1,3)

let point C = ( 1,2,3)

find vector AB = [ 0, 1 , 3]

vector AC = [1, 2, 3]

now you need to find the vector perpendicular to both of these. do a vector cross product

AB X AC = | 0 1 3 |

| 1 2 3 |

= [ -3, 3, -1] this is the normal direction vector to the plane containing all the points A B and C, this is all we need to calculate the angle, the whole equation of the plane is not necessary.

now, look at the problem geometrically. it said the angle between this plane and the vertical plane, in other words, the Y and Z plane.

so the Y Z plane has direction vector [ 0 , 1 ,1]

to find the angle between both planes we use this formula

cos (-) = (N.n)/(|N|.|n|) where N and n are the direction vectors for both of the planes. so doing a dot product of both ve ctors and finding the modulus of both vectors we get this answer:

cos (-) = 2 / (√(19) . √(2))

and so therefore theta equals 71.07 degrees

the points are now in a new plane which is 71.07 degrees away from the vertical. tadam XD

enable P be the factor P ( 6, a million, -2) and O be the beginning O (0, 0, 0) Vector ( OP ) = 6 i + j - 2 ok , the place i, j, and ok are the unit vectors in the guidelines of the x, y and z axes respectively further, if A is a factor A (a million, 3, - 2) , then Vector ( OA ) = i + 3j - 2 ok The length of the Vector OA = ?(a million² + 3² + 2²) = ?14 Unit Vector in the path OA = (a million / ?14)(i + 3 j - 2 ok) in view that OA is perpendicular to the the airplane, then the unit vector popular to the airplane is ( n ) = (a million / ?14)(i + 3 j - 2 ok) Now the projection of OP on OA = ( OP ) ? ( n ) = (6 i + j - 2 ok) ? {(a million / ?14)(i + 3 j - 2 ok)} = (a million / ?14) ( 6 + 3 + 4) = (13 / ?14) = length of the perpendicular from the beginning on the airplane for this reason the vector equation of the airplane is ( r ) ? ( n) = (13 / ?14) For the cartesian equation : positioned ( r ) = x i + y j + z ok and ( n ) = (a million / ?14)(i + 3 j - 2 ok) and you get x + 3y - 2z = 13

The cross products of the vectors (1,2,0) amd (0,1,3) will give a vector that points in the same direction that the tower points now.

The cross product is: 6 i - 3 j + 1 k

The unit vector in that direction is:

( 6, -3, 1 )

--------------

sqrt(46)

Dot this with the unit vector in the z direction (0,0,1)

Thus cos (theta) = 1 / sqrt(46)

theta = 81.521 degrees

Draw a triangle on the coordinate system (with a y-axis and x-axis that you draw your graphs on). Then find the length of each side. Finally, use trig to find the angle you need. Trig is sine=opposite/hypotenuse, cosine=adjacent/hypotenuse, tangent=opposite/adjacent.