1. Determine the scalar equation of the plane with vector equation Vector r= (3,-1,4) +s(2,-1,5) + t(-3,2,-2).

2. Determine the value of k so that the line with parametric equations x = 2 + 3t, y = -2 + 5t, z = kt is parallel to the plane with equation 4x + 3y – 3z -12 = 0.

3. The planes A1x + B1y + C1z + D1 = 0 and A2x + B2y + C2z + D2 = 0 are perpendicular. Find the value of A1A2 + B1B2 + C1C2.

1. Well, the scalar equation of a plane is not the most exciting thing to talk about at parties, but I'll give it a shot. To determine the scalar equation, we need to find the normal vector of the plane. The coefficients of x, y, and z in the vector equation will give us the components of the normal vector. So, the normal vector will be (2,-1,5) in this case.

Now, let's take a point on the plane, which is given by (3,-1,4). We can substitute this point and the normal vector into the scalar equation of a plane, which is Ax + By + Cz = D. Plugging in the values, we get 2x - y + 5z = 16. So, the scalar equation of the plane is 2x - y + 5z = 16. Now, isn't that plane-tastic?

2. Parallel lines and planes, oh my! To determine the value of k, we need to find the direction vector of the line and compare it to the normal vector of the plane.

The direction vector of the line is (3,5,k), and the normal vector of the plane is (4,3,-3). To ensure that the line is parallel to the plane, the direction vector and the normal vector must be orthogonal (perpendicular), which means their dot product should be zero.

For the vectors to be perpendicular, we have 4(3) + 3(5) - 3(k) = 0. Simplifying this, we get 12 + 15 - 3k = 0. Solving for k, we find k = (12+15)/3 = 9.

So, the value of k that makes the line parallel to the plane is k = 9. Now, go and show off your parallel lines and planes knowledge!

3. Ah, the joy of perpendicular planes. To find the value of A1A2 + B1B2 + C1C2, we'll use the fact that perpendicular planes have normal vectors that are orthogonal (perpendicular).

Let's call the normal vectors of the planes (A1,B1,C1) and (A2,B2,C2). Since the planes are perpendicular, their dot product must be zero. So, A1A2 + B1B2 + C1C2 = 0.

Voila! We have our answer. The value of A1A2 + B1B2 + C1C2 is zero. Now, wasn't that plane-perfect? It's geometry, but with a touch of humor!

1. To determine the scalar equation of the plane, we can start by finding two vectors in the plane.

Given the vector equation of the plane: Vector r = (3,-1,4) + s(2,-1,5) + t(-3,2,-2).

Let's take two arbitrary points on the plane, say (3,-1,4) and (5,0,2). Subtracting the first point from the second, we get the vector (5,0,2) - (3,-1,4) = (2,1,-2).

This vector lies in the plane, so it is one of the required vectors.

Now, take the cross product of the two direction vectors: (2,-1,5) × (-3,2,-2). The cross product gives us a vector perpendicular to both of them and hence, perpendicular to the plane.

The cross product is calculated as:
(2,-1,5) × (-3,2,-2) = (3,16,8).

This vector (3,16,8) is perpendicular to the plane, so it can be utilized to determine the scalar equation of the plane.

The equation of the plane is of the form Ax + By + Cz + D = 0, where A, B, C, and D are constants.

We know that the direction vector (3,16,8) is perpendicular to the plane, so it satisfies the scalar equation of the plane.

Substituting the coordinates of a point on the plane (say (3,-1,4)) into the equation, we get:
3A - 16B + 8C = 0.

Thus, we now have two equations:
3A - 16B + 8C = 0, and
A(2) + B(-1) + C(5) = 0.

Solving these two equations simultaneously will give us the values of A, B, and C, thus providing the scalar equation of the plane.

2. To determine the value of k such that the line with parametric equations x = 2 + 3t, y = -2 + 5t, z = kt is parallel to the plane with equation 4x + 3y – 3z - 12 = 0, we need to find the direction vector of the line and check if it is parallel to the normal vector of the plane.

The direction vector of the line is given by coefficients of t in the parametric equations, which are 3, 5, and k.

The normal vector of the plane is the coefficients of x, y, and z in the equation of the plane, which are 4, 3, and -3.

For the line to be parallel to the plane, the direction vector of the line must be perpendicular to the normal vector of the plane.

Hence, the dot product of the direction vector of the line and the normal vector of the plane should be zero.

Calculating the dot product: (3, 5, k) · (4, 3, -3) = 3(4) + 5(3) + k(-3) = 12 + 15 - 3k = 27 - 3k.

For the dot product to be zero, we set 27 - 3k = 0 and solve for k:
27 - 3k = 0
3k = 27
k = 9.

Therefore, the value of k that makes the line parallel to the given plane is k = 9.

3. To find the value of A1A2 + B1B2 + C1C2 when the planes A1x + B1x + C1z + D1 = 0 and A2x + B2y + C2z + D2 = 0 are perpendicular, we need to consider that two planes are perpendicular if their normal vectors are perpendicular.

The normal vectors of the two planes are (A1, B1, C1) and (A2, B2, C2) respectively.

The scalar product (dot product) of two vectors is defined as: A1A2 + B1B2 + C1C2. So, we are required to find this scalar product.

Since the planes are perpendicular, their normal vectors are perpendicular, and hence, their scalar product is zero.

Therefore, A1A2 + B1B2 + C1C2 = 0.

1. notice that you have two direction vectors , find a vector [a,b,c] which is perpendicular to both of those. Such a vector is called the normal and is obtained by taking the cross-product of the 2 direction vectors.

So your equation of the plane will be ax + by + cz = k
plug in the point (3,-1,4) to find k

2. As I said in #1, the normal to 4x + 3y – 3z -12 = 0 is [4,3,-3]
to be parallel with the line, the dot-product of [4,3,-3] with [3,5,k] mus
be zero. go for it!

3. Well duhh!

Have you done any of the work??