Cartesian equation of plane using two lines?

In summary: It is impossible for me to say anything about this conversation because your summary is not clear or complete.
  • #1
alakd
1
0
Can someone help with this multi-part question. i did the first three but it doesn't seem right!
--------------------------------------…
(a) Find the equation of the line l through P(1, 1, 2) and Q(1, 0, 4) in vector, parametric and Cartesian forms.
(b) Find the vector form of the line k through R(0, 3, 1) which is perpendicular to l.
(c) Find the Cartesian form of the plane containing the lines l and k.
(d) Find the vector form of the line through T(0, 0, 3) which is normal to this plane.
--------------------------------------…
so i guess i did the first bit alright : cartesian : z=6+4x , y=6
for the second one i got this line : (-5,4,2)+t(28,-37,7) , i timed it by 17 to get rid of the denominator
third part : the lines seem to have no intersection so i am stuck!

appreciate it if anyone can help out
 
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  • #2
alakd said:
Can someone help with this multi-part question. i did the first three but it doesn't seem right!
--------------------------------------…
(a) Find the equation of the line l through P(1, 1, 2) and Q(1, 0, 4) in vector, parametric and Cartesian forms.
(b) Find the vector form of the line k through R(0, 3, 1) which is perpendicular to l.
(c) Find the Cartesian form of the plane containing the lines l and k.
(d) Find the vector form of the line through T(0, 0, 3) which is normal to this plane.
--------------------------------------…
so i guess i did the first bit alright : cartesian : z=6+4x , y=6
To which is this an answer? Clearly not (a) because x is always equal to 1 but y is NOT always equal to 6 and z is not always equal to 6+ 4(1)= 10.

for the second one i got this line : (-5,4,2)+t(28,-37,7) , i timed it by 17 to get rid of the denominator
third part : the lines seem to have no intersection so i am stuck!

appreciate it if anyone can help out
It's impossible to tell what you are saying because you talk about ""first", "second", and "third" when the problems are labeled "a", "b", "c", and "d" and "a" has three parts!
 

1. What is the Cartesian equation of a plane using two intersecting lines?

The Cartesian equation of a plane using two intersecting lines is determined by taking the cross product of the direction vectors of the two lines. This will give the normal vector of the plane, which can then be used in the general form of the equation: Ax + By + Cz + D = 0, where A, B, and C are the components of the normal vector and D is a constant.

2. Can the Cartesian equation of a plane be determined using two parallel lines?

No, the Cartesian equation of a plane cannot be determined using two parallel lines. This is because parallel lines do not intersect, and therefore do not determine a unique plane. In this case, an infinite number of planes can be drawn through the parallel lines.

3. How do you find the Cartesian equation of a plane if the two lines are not given in the same form?

If the two lines are not given in the same form, they can be converted to the same form by manipulating the equations. For example, a line in point-slope form (y - y1 = m(x - x1)) can be converted to slope-intercept form (y = mx + b) by solving for y. Once both lines are in the same form, the Cartesian equation of the plane can be determined using the cross product method mentioned in question 1.

4. Can the Cartesian equation of a plane be used to find the distance between a point and the plane?

Yes, the Cartesian equation of a plane can be used to find the shortest distance between a point and the plane. This distance can be calculated by taking the absolute value of the dot product between the normal vector of the plane and a vector from the point to any point on the plane. The resulting value will be the perpendicular distance between the point and the plane.

5. Is the Cartesian equation of a plane unique for a given set of two lines?

Yes, the Cartesian equation of a plane is unique for a given set of two lines. This is because the cross product of the direction vectors of the lines will always result in the same normal vector, and the general form of the equation (Ax + By + Cz + D = 0) is unique for a given normal vector.

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