H2 Maths 3.3 - Planes: Equations, Angles, and Distances

Step 1 - State the problem

Plane pi passes through three points: A at one, zero, two; B at three, negative one, four; and C at two, two, zero.

Step 1 - State the problem

Line l passes through the point negative one, two, three with direction vector two, negative one, four.

Step 1 - State the problem

Three tasks: find the Cartesian equation of pi, find the acute angle between l and pi, and find the distance from point P at four, one, negative one to pi.

Step 2 - Find two direction vectors in $\pi$

Any two non-parallel vectors lying in the plane will give us the normal when crossed.

Step 2 - Find two direction vectors in $\pi$

Subtract A from B: two minus one, negative one minus zero, four minus two - giving vector AB as two, negative one, two.

Step 2 - Find two direction vectors in $\pi$

Subtract A from C: two minus one, two minus zero, zero minus two - giving vector AC as one, two, negative two.

Step 3 - Normal vector by cross product

The normal n equals vector AB cross vector AC, set up as a three-by-three determinant.

Step 3 - Normal vector by cross product

i component: negative one times negative two, minus two times two - that is two minus four, giving negative two. j component: negative of two times negative two, minus two times one - that is negative of negative six, giving positive six. k component: two times two, minus negative one times one - that is four plus one, giving five.

Step 3 - Normal vector by cross product

So the normal vector n is negative two, six, five.

Step 4 - Cartesian equation of $\pi$

The plane satisfies n dot r minus OA equals zero.

Step 4 - Cartesian equation of $\pi$

Substituting n and point A at one, zero, two: negative two times x minus one, plus six y, plus five times z minus two, equals zero.

Step 4 - Cartesian equation of $\pi$

Expanding and collecting constants gives the Cartesian equation: negative two x plus six y plus five z equals eight.

Step 5 - Acute angle between $l$ and $\pi$

The sine of the acute angle between a line and a plane equals the absolute dot product of the direction and normal vectors, divided by their magnitudes.

Step 5 - Acute angle between $l$ and $\pi$

d dot n is two times negative two, plus negative one times six, plus four times five - that is negative four minus six plus twenty, giving ten. The magnitude of d is root twenty-one, and the magnitude of n is root sixty-five.

Step 5 - Acute angle between $l$ and $\pi$

So $\sin\theta = \tfrac\{10\}\{\sqrt\{1365\}\}$, giving $\theta \approx 15.7^\circ$.

Step 6 - Distance from $P$ to plane $\pi$

For a plane n dot r equals k, the distance from any point P to the plane is the absolute value of n dot OP minus k, all divided by the magnitude of n.

Step 6 - Distance from $P$ to plane $\pi$

n dot OP is negative two times four, plus six times one, plus five times negative one - that is negative eight plus six minus five, giving negative seven. The plane constant k is eight.

Step 6 - Distance from $P$ to plane $\pi$

Distance $= \tfrac\{15\}\{\sqrt\{65\}\} = \tfrac\{3\sqrt\{65\}\}\{13\}$.