Hi, my exam is tomorrow and I'm here with my girlfriend who's also in the same class trying to figure out how to do the exam review but we're stuck.. please be kind and help us out, we'd really appreciate it. Thank you in advance.
QUESTION:
Write a vector equation and parametric equations for a plane that satisfies each set of conditions.
a) contains the point P̥(2,-4,6); has direction vectors a = [6,-7,0] and b = [4,1,3]
b) contains the point P̥(-5,9,-3); is parallel to the lines [x,y,z] = [1,-2,7] + s[4,-1,-3] and [x,y,z] = [7,-2,15] + t[1,6,-8]
c) contains the points A(4,-2,5), B(3,-3,1), and C(5,2,8)
d) has x-intercept 12, y-intercept -4, and z-intercept 7
please and thank you!
To write the vector equation of a plane you need two different direction vectors on the plane, and a point on the plane
a) you are given exactly what I described above
so
r = (2,-4,6) + s(6,-7,0) + t(4,1,3)
It is easy to write these in parametric form
x = 2 + 6s + 4t
y = -4 -7t
z = 6 = 3t , it should be obvious what I did.
b) each of the lines gives you a direction vector, and you are given a point, so ....... (same question)
c) You can find vector AB and vector AC or vector BC , and you have the choice of 3 different points,
d) "has x-intercept 12, y-intercept -4, and z-intercept 7" translates to ...
you are given the points (12,0,0), (0,-4,0) and (0.0.7). Now it becomes the same kind of problem as c)
Of course, I'd be happy to help you with your exam review. Let's go through each question one by one and find the vector equation and parametric equations for the planes.
a) To find the vector equation of a plane, you need a point on the plane and two direction vectors. In this case, the point P̥(2, -4, 6) is given, and the direction vectors are a = [6, -7, 0] and b = [4, 1, 3].
The vector equation of a plane is given by r = r0 + s*a + t*b, where r = [x, y, z] is a point on the plane, r0 is the given point, and s and t are scalar parameters.
Substituting the values, the vector equation for this plane is:
r = [2, -4, 6] + s*[6, -7, 0] + t*[4, 1, 3]
or, in component form:
x = 2 + 6s + 4t
y = -4 - 7s + t
z = 6 + 3t
These equations represent the vector equation and parametric equations for the plane that satisfies the given conditions.
b) For this question, the plane contains the point P̥(-5, 9, -3) and is parallel to the given lines: [x, y, z] = [1, -2, 7] + s[4, -1, -3] and [x, y, z] = [7, -2, 15] + t[1, 6, -8].
A plane parallel to two lines is perpendicular to their direction vectors. So, to find the direction vectors of the plane, we take the cross product of the direction vectors of the lines.
Let's call the direction vectors of the lines u = [4, -1, -3] and v = [1, 6, -8]. Their cross product will give the direction vector of the plane.
u x v = [(-1 * -8) - (6 * -3), (-3 * 1) - (-8 * 4), (4 * 6) - (1 * -3)]
= [5, -29, 27]
Now we have the direction vector of the plane. Using the given point P̥(-5, 9, -3), and the direction vector a = [5, -29, 27], the vector equation of the plane can be written as:
r = [-5, 9, -3] + s * [5, -29, 27]
In parametric form, the equations would be:
x = -5 + 5s
y = 9 - 29s
z = -3 + 27s
These equations represent the vector equation and parametric equations for the plane that satisfies the given conditions.
c) For this question, the plane contains the points A(4, -2, 5), B(3, -3, 1), and C(5, 2, 8).
To find the vector equation of a plane passing through these points, we need to find two vectors lying on the plane. One way to do this is by taking the vectors formed by subtracting a point from another point.
Let's take vector AB = B - A = [3 - 4, -3 - (-2), 1 - 5] = [-1, -1, -4]
And vector AC = C - A = [5 - 4, 2 - (-2), 8 - 5] = [1, 4, 3]
Now, we have two vectors, and we can take their cross product as the direction vector of the plane.
AB x AC = [-1 * 4 - 1 * 3, -4 * 1 - 1 * 3, -1 * 1 - (-1 * 4)]
= [-7, -1, 3]
Using the given point A(4, -2, 5) and the direction vector b = [-7, -1, 3], the vector equation of the plane can be written as:
r = [4, -2, 5] + s * [-7, -1, 3]
In parametric form, the equations would be:
x = 4 - 7s
y = -2 - s
z = 5 + 3s
These equations represent the vector equation and parametric equations for the plane that satisfies the given conditions.
d) For this question, you have the x-intercept (where the plane crosses the x-axis) as 12, the y-intercept as -4, and the z-intercept as 7.
To find the vector equation of a plane from intercepts, we use the reciprocal of each intercept as the direction vector components.
The direction vector for the x-intercept is i = [1/12, 0, 0]
The direction vector for the y-intercept is j = [0, 1/-4, 0]
The direction vector for the z-intercept is k = [0, 0, 1/7]
Now, using the given intercepts, the vector equation of the plane can be written as:
r = [0, 0, 0] + s * [1/12, 0, 0] + t * [0, 1/-4, 0] + u * [0, 0, 1/7]
In parametric form, the equations would be:
x = s/12
y = t/-4
z = u/7
These equations represent the vector equation and parametric equations for the plane that satisfies the given conditions.
I hope this helps you understand how to find the vector equation and parametric equations for planes in different scenarios. Best of luck with your exam tomorrow!