Vector Geometry: Solving for Position Vectors and Equations in 3D

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In summary, according to the article referenced, a rhombus must have four sides which all have the same length (distance). The opposite sides of a rhombus must also be equal in length.
  • #1
chwala
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Homework Statement


The points A,B and C have position vectors , relative to the origin O given by ## OA= i+2j+3k, OB=4j+k , OC=2i+5j-k.## A fourth point D is such that the quadrilateral ABCD is a parallelogram.
i) Find the position vector of D and verify that the parallelogram is a Rhombus.
ii)The plane p is parallel to OA and the line BC lies in p. Find the equation of p,giving your answer in the form ##ax+by+cz=d##

Homework Equations

The Attempt at a Solution


##AB= (4j+k)-(i+2j+3k)=-i+2j-2k, AC= (2i+5j-k)-(i+2j+3k)=i+3j-4k, BC= (2i+5j-k)-(4j+k)=2i+j-2k## i am just groping in the dark here, but i know vectors on same line and parallel vectors should have a scalar or something...relating them
 
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  • #2
chwala said:

Homework Statement


The points A,B and C have position vectors , relative to the origin O given by ## OA= i+2j+3k, OB=4j+k , OC=2i+5j-k.## A fourth point D is such that the quadrilateral ABCD is a parallelogram.
i) Find the position vector of D and verify that the parallelogram is a Rhombus.
ii)The plane p is parallel to OA and the line BC lies in p. Find the equation of p,giving your answer in the form ##ax+by+cz=d##

Homework Equations

The Attempt at a Solution


##AB= (4j+k)-(i+2j+3k)=-i+2j-2k, AC= (2i+5j-k)-(i+2j+3k)=i+3j-4k, BC= (2i+5j-k)-(4j+k)=2i+j-2k## i am just groping in the dark here, but i know vectors on same line and parallel vectors should have a scalar or something...relating them
Let's take things one at a time.

What do you know about the sides of a general quadrilateral figure which would make it a rhombus?

https://en.wikipedia.org/wiki/Rhombus

Hint: it's OK to make a sketch, if that helps.
 
  • #3
i am unable to make a sketch on this application, ok for a quadrilateral the opposite sides are equal in terms of distance ##d##
In Euclidean geometry, a rhombus(◊), plural rhombi or rhombuses, is a simple (non-self-intersecting) quadrilateral whose four sides all have the same length. Another name is equilateral quadrilateral, since equilateral means that all of its sides are equal in length. The rhombus is often called a diamond, after the diamonds suit in playing cards which resembles the projection of an octahedral diamond, or a lozenge, though the former sometimes refers specifically to a rhombus with a 60° angle (see Polyiamond), and the latter sometimes refers specifically to a rhombus with a 45° angle.

Every rhombus is a parallelogram and a kite. A rhombus with right angles is a square.
220px-Rhombus.svg.png
 
  • #4
chwala said:
i am unable to make a sketch on this application, ok for a quadrilateral the opposite sides are equal in terms of distance ##d##
You don't have to make a sketch here in the edit box at PF, but just for yourself.

For a rhombus, is it just the opposite sides which are equal in length, or ...?

In the article referenced above, there is a whole list of geometric properties which must be satisfied for a quadrilateral to be called a rhombus.
 
  • #5
agreed , a square is a rhombus too point noted...all sides equal also qualifies to be a rhombus. In that case the two sides opposite to each other in our problem should be equal in terms of distance i.e parallelogram
 
  • #6
chwala said:
agreed , a square is a rhombus too point noted...all sides equal also qualifies to be a rhombus. In that case the two sides opposite to each other in our problem should be equal in terms of distance i.e parallelogram
Good. This should help you find the position vector for point D. Some of the other properties may also be used to check your result.
 
  • #7
ok, ##AD=BC, BC=2i+j-2k, → ##DA=d-a= 2i+j-2k,
##D=(i+2j+3k)+(2i+j-2k)= 3i+3j+k##
 
  • #8
chwala said:
ok, ##AD=BC, BC=2i+j-2k, → ##DA=d-a= 2i+j-2k,
##D=(i+2j+3k)+(2i+j-2k)= 3i+3j+k##
OK, that looks good for OD.

Now, what about part ii)?
 
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  • #9
now what do they mean by part i) of the question, that verify that the parallelogram is a Rhombus? am i supposed to show using ##L=√(x2-x1)^2+(y2-y1)^2+(z2-z1)^2## that the opposite sides are same in terms of length(distance) ie ##AD=BC, AB=DC?##
 
  • #10
chwala said:
now what do they mean by part i) of the question, that verify that the parallelogram is a Rhombus? am i supposed to show using ##L=√(x2-x1)^2+(y2-y1)^2+(z2-z1)^2## that the opposite sides are same in terms of length(distance) ie ##AD=BC, AB=DC?##
That would be one of the ways to show that the figure is a rhombus, according to the wiki article. There are several other methods which are equally valid.
 
  • #11
like finding mid point?...
 
  • #12
for part ii) ##BC= (2i+5j-k)-(4j+k)=2i+j-2k,→2x+y-2z, OA=i+2j+3k→x+2y+3z## what next?
 

1. What is a vector in 3D?

A vector in 3D is a mathematical representation of a physical quantity that has both magnitude and direction. It is typically represented by an arrow in three-dimensional space, with the length of the arrow representing the magnitude and the direction of the arrow indicating the direction of the vector.

2. How do you find the magnitude of a vector in 3D?

The magnitude of a vector in 3D can be found using the Pythagorean theorem, which states that the square of the length of a vector is equal to the sum of the squares of its components. In other words, the magnitude of a vector can be calculated by taking the square root of the sum of the squares of its x, y, and z components.

3. What is the dot product of two vectors in 3D?

The dot product of two vectors in 3D is a mathematical operation that results in a scalar quantity. It is calculated by multiplying the corresponding components of the two vectors and then summing up the products. The dot product can be used to find the angle between two vectors, as well as to determine if two vectors are perpendicular to each other.

4. How do you find the cross product of two vectors in 3D?

The cross product of two vectors in 3D is a mathematical operation that results in a vector perpendicular to both the original vectors. It is calculated by taking the determinant of a 3x3 matrix formed by the two vectors and the unit vectors in the x, y, and z directions. The resulting vector is perpendicular to both original vectors and its magnitude is equal to the area of the parallelogram formed by the two original vectors.

5. What are some real-world applications of vectors in 3D?

Vectors in 3D have many real-world applications, including in physics, engineering, and computer graphics. Some examples include using vectors to model forces and motion in physics, designing 3D structures in engineering, and creating 3D animations and simulations in computer graphics.

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