PDF SOLUTIONS TO HOMEWORK ASSIGNMENT #2, Math 253 ~ Top Stories Info

1. Find an equation for a plane which contains the line given by (x, y, z) = t(1, 2, 0) and which forms an angle of π/3 with the xy-plane. This we recognize as the equation of a plane, and since any point on the intersection of the above surfaces must satisfy this equation also, we have shown that the...geometry homework intersection plane + Manage Tags. and also plot a graph in maple showing the two planes and the line of intersection? 5016 views.Direction vector of line of intersection of two planes is the cross product of the normal vectors of the planes, namelyp1: x+y+z=2p2: x+7y+7z=2and the correspon… Thus the parametric equation of the line is L: (2,0,0)+t(0,-1,1) or L: x=2, y=-t, z=t.First, the line of intersection lies on both planes. Therefore, it shall be normal to each of the normals of the planes. Thus, find the cross product. first, you need normal vector of the planes.you can get these by taking coefficients of the equation of planes. second, you need a vector that perpendicular...Parametric Equation of a Plane. Fun Facts. Any Point. Finding the Normal Vector. We will do it by creating a parametric equation of the ray and intersecting that with the implicit equation of the plane. If I is the intersection point of the line and the plane, we have

parametric equations for the line of intersection of the planes

Now I need to find the point where the line intersects the plane, so taking the norm vector of $x - y = 2$ gives $(1,-1,0)$ so this can serve as my point of Since we want the point of intersection between $l$ and the given plane $x-y=2$, simply substitute the parametric equations of $l$ in and solve for...Use T As The Parameter Along The Line And Assign The Equations Of The Line To The List EquationsOfLine In The Form [x=.... , Y =.. , Z =.] . Use t as the parameter along the line and assign the equations of the line to the list EquationsOfLine in the form [x=.... , y =.. , z =.] ., To find the intersection of the line through a point. perpendicular to plane. . This set of parametric equations represents the line through the origin that is perpendicular to.The line of intersection will be parallel to both planes. We can write the equations of the two planes in 'normal form' as r.(2,1,-1)=4 and r.(3,5,2)=13 respectively. To find a point on the line we could substitute z=0 (for example, any value would do) into the equations of the planes and solve the...

(a) find parametric equations for the line of intersection of the...

b. If the planes are neither parallel nor orthogonal, then find the measure of the angle between the planes. Express the answer in degrees rounded to the nearest integer. c. If the planes intersect, find the line of intersection of the planes, providing the parametric equations of this line.Partial Derivatives. Section 6. Tangent Planes and Differentials. Discussion. You must be signed in to discuss.Calculus Parametric Functions Introduction to Parametric Equations. In the drawing below, we are looking right down the line of intersection, and we get an idea as to why the cross product of the normals of the red and blue planes generates a third vector, perpendicular to the normal vectors, that...To find the intersection point P(x,y,z), substitute line parametric values of x, y and z into the plane equation The angle between the line and the plane can be calculated by the cross product of the line vector with the vector representation of the plane which is perpendicular to the plane: v = 4i + k.Solution: (a) The ±rst equation can be rewritten as x + y − z = 0. The direction vector of the line of intersection can be constructed from the normals to the so the acute angle between the two planes is ( π − θ ) = arccos 2 √ 90 . Problem 3. (10 points) Determine whether the following four points A (1...

Find the parametric equations for the line of intersection of the planes:

z = x + y

2x - 5y - z = 1

Let's recast the equations of the planes.

x + y - z = 0

2x - 5y - z = 1

The go product of the commonplace vectors of the two planes will be the directional vector v, of the line of intersection.

v = <1, 1, -1> X <2, -5, -1> = <-6, -1, -7>

Any non-zero more than one of v can be a directional vector of the line. Multiply via -1.

v = <6, 1, 7>

Now find some extent on the line. It can be some extent in both planes. Let y = Zero and clear up for x and z.

x - z = 0

2x - z = 1

Subtract the first equation from the 2d.

x = 1

Plug again into the first equation and solve for z.

1 - z = 0

z = 1

So our point on the line is P(1, 0, 1).

The equation of the line of intersection is:

L(t) = P + television

L(t) = <1, 0, 1> + t<6, 1, 7>

the place t is a scalar ranging over the real numbers

Now put the equation of the line in parametric shape.

L(t):

x = 1 + 6t

y = t

z = 1 + 7t

__________

Both points (0, -1/6, -1/6) and (1/7, -1/7, 0) are on the line as is my point of P(1, 0, 1).

In order for your parametric equation to be correct you want:

1) Any non-zero more than one of the directional vector of the line.

2) Any point on the line.

So any of the 3 issues, and an unlimited quantity of other points on the line, along side a directional vector of the line offers you correct parametric equations. Your point is ok.