Question

Planes: (a) Find a vector equation for the plane P passing through the points A = (-10, 6, -7) , B = (0, -5, -6) , and C = (-8, 4, 9) (b) Convert the vector equation for P from part (b) (a) to a scalar equation for P.

Answer 1

Step-by-step explanat

To find a vector equation for the plane P passing through the points A = (-10, 6, -7), B = (0, -5, -6), and C = (-8, 4, 9), we can start by finding two vectors that lie in the plane.

Let's take vector AB and vector AC:

Vector AB = B - A = (0, -5, -6) - (-10, 6, -7) = (10, -11, 1)

Vector AC = C - A = (-8, 4, 9) - (-10, 6, -7) = (2, -2, 16)

Now we can find the cross product of these two vectors to obtain a normal vector to the plane P. The cross product will be perpendicular to both AB and AC, and therefore, it will lie in the plane.

Normal vector n = AB × AC

= (10, -11, 1) × (2, -2, 16)

To calculate the cross product, we can use the determinant method:

n = ((-11)(16) - (-2)(1), -(1)(16) - (10)(2), (10)(-2) - (-11)(2))

= (-174 + 2, -16 - 20, -20 + 22)

= (-172, -36, 2)

So, we have a normal vector n = (-172, -36, 2) to the plane P.

Now, let's find the equation of the plane in vector form:

r = r₀ + tn

where r is a position vector, r₀ is a known position vector in the plane, t is a scalar parameter, and n is the normal vector to the plane.

We can choose any of the given points A, B, or C as r₀. Let's choose A:

r = (-10, 6, -7) + t(-172, -36, 2)

Thus, the vector equation for the plane P passing through the points A, B, and C is:

r = (-10, 6, -7) + t(-172, -36, 2) where t ∈ ℝ

Now, to convert the vector equation to a scalar equation for the plane P, we can expand it:

x = -10 - 172t

y = 6 - 36t

z = -7 + 2t

The scalar equation for the plane P is:

-10 - 172t = x

6 - 36t = y

-7 + 2t = z

This is the scalar equation for the plane P passing through the points A, B, and C.

Answer 2

(a) To find the vector equation for the plane P, we first need to find two vectors in the plane. We can do this by taking the cross product of the vectors AB and AC.

AB = B - A = (0, -5, -6) - (-10, 6, -7) = (10, -11, 1)

AC = C - A = (-8, 4, 9) - (-10, 6, -7) = (2, -2, 16)

Taking the cross product of AB and AC gives us a normal vector to the plane:

n = AB x AC = (-178, -152, 22)

Now we can use the point-normal form of the equation of a plane:

n · (r - A) = 0

where r is any point on the plane. Substituting in the values we have, we get:

(-178, -152, 22) · (r - (-10, 6, -7)) = 0

Expanding this out, we get:

-178(x + 10) - 152(y - 6) + 22(z + 7) = 0

This is the vector equation for the plane P.

(b) To convert the vector equation for P to a scalar equation, we can expand out the dot product:

-178x - 1780 - 152y + 912 + 22z + 154 = 0

Simplifying, we get:

-178x - 152y + 22z - 714 = 0

This is the scalar equation for the plane P.

Explanation: