Question

Find the equation of the plane passing through the points A=(1,1,1), B=(1,4,5), C=(−3,-2,0).

Find the area of the triangle the 3 points from the first equation.

Find the angle between the 2 vectors; (1) from A to B and (2) from C to B.

Answer 1

• Equation of the plane

Take any two pairs of the given points and make vectors out of them. For example, the vector from A to B is

`\vec v_1 = \langle 1,4,5\rangle - \langle 1,1,1\rangle = \langle0,3,4\rangle`

and the vector from A to C is

`\vec v_2 = \langle -3,-2,0\rangle - \langle1,1,1\rangle = \langle-4,-3,-1\rangle`

These vectors lie in the same plane (the one we want). We can get a third vector that is normal to the plane by taking their cross product (details omitted).

`\vec n = \vec v_1 * \vec v_2 = \langle 9,-16,12 \rangle`

If
`\vec u = \langle x,y,z\rangle` is an arbitrary vector, then the vector from any of the points A, B, or C to
`\vec u` will lie in our plane. That is, if we start from A,

`(\vec u - \langle1,1,1\rangle) \cdot \vec n = 0`

and this reduces to the equation of the plane,

`\langle x - 1, y - 1, z - 1 \rangle \cdot \langle 9, -16, 12 \rangle = 0`

`9 (x - 1) - 16 (y - 1) + 12 (z - 1) = 0`

`\boxed{9x - 16y + 12z = 5}`

• Area of triangle ABC

This follows immediately from the cross product identity

`\|\vec x * \vec y\| = \|\vec x\| \|\vec y\| \sin(\theta)`

where
`\theta` is the angle between
`\vec x` and
`\vec y`. The left side corresponds to the area of the parallelogram spanned by
`\vec x` and
`\vec y`; half of this area would be that of a triangle. (see attached)

In our case, we have

`\|\vec n\| = \|\vec v_1 * \vec v_2\| = √(481)`

so the area of ABC is
`\boxed{\frac{√(481)}2}`.

• Angle between A to B and between C to B

We already know the first vector,
`v_1`.

The vector from C to B is

`\vec v_3 = \langle 1,4,5 \rangle - \langle -3,-2,0 \rangle = \langle 4, 6, 5 \rangle`

Recall the dot product identity,

`\vec x \cdot \vec y = \|\vec x\| \|\vec y\| \cos(\theta)`

Then

`\vec v_1 \cdot \vec v_3 = \|\vec v_1\| \|\vec v_3\| \cos(\theta) \implies \cos(\theta) = (38)/(5√(77)) \implies \theta \approx \boxed{29.9914^\circ}`