Quiz 4: Planes

A vector parallel to the line is the vector . Hence the line can be represented by the parametric vector equation Note that there are many other ways of giving a vector parametric equation for this line. For example, instead of using the position vector of the known point P in the equation, we could have used the position vector of Q, to give another version of the equation,

When the cartesian equations of a line are given in the form we can identify the coordinates of one point on the line (namely (α,β,γ)) and a vector parallel to the line, namely ai + bj + ck. In this case, a = 3, b = 2, c = -1.

In the three parametric scalar equations, the coefficients 5, - 1, 2 of t are the components of a vector parallel to the line. (Therefore the line has direction parallel to 5i - j + 2k.) The numbers 5, - 1, 2 become the denominators in the cartesian equations of the line. The constants 3, 2, 3 in the scalar parametric equations are (in that order) the coordinates of a particular point (3,2,3) on the line. This information allows us to form the numerators in the cartesian form, x - 3, y - 2, z - 3.

A vector parallel to the line is 3i + j + 4k (from the denominators in the cartesian form). So the line is also parallel to 6i + 2j + 8k. Therefore since the line passes through the point (1,2,1), it has vector parametric equation r = i + 2j + k + t(6i + 2j + 8k).

Line (1) is parallel to k and line (2) is parallel to 4i + j + 3k. Hence the required line, which is perpendicular to both, must be parallel to the cross product of these two vectors, k × (4i + j + 3k) = -i + 4j. Its parametric vector equation is therefore r = i + j + k + t(-i + 4j).

There are infinitely many planes containing any two given points. To see this, visualise the line joining the points as the spine of a book, and the infinitely many planes as pages of the book.

The equation of the plane through the point (p,q,r) perpendicular to ai + bj + ck is a(x - p) + b(y - q) + c(z - r) = 0. Hence in this case, the cartesian equation is 1(x - 3) - 1(y - 1) + 2(z - 0) = 0, that is, x - y + 2z = 2.

In order to know the equation of a plane, we must know a point on the plane and a vector perpendicular to the plane. We know three points on the plane, so we’re OK there. We can find a vector perpendicular to the plane by using the vector cross product. Notice that the vectors and are both parallel to the plane, so their cross product is normal to the plane. Since = -2j + 6k and = -3i - 2j + k, we calculate × = 10i - 18j - 6k. This is a normal vector to the plane. The cartesian equation is therefore 10(x - 1) - 18(y - 3) - 6(z + 1) = 0, or 10x - 18y - 6z = -38.

A vector perpendicular to the plane is So the cartesian equation of the plane is -5(x - 1) + 3(y - 1) - 2(z - 1) = 0 i.e 5x - 3y + 2z = 4.

Hint: Find the angles between the normals to the two planes.

The two planes are not parallel, since the first plane has normal vector and the second plane has normal ; so n is not parallel to m. Denote the two planes by ABCD and EBCF, respectively, so that BC is the line of intersection. The diagram above shows that the angle between two planes is the same as the angle between the normals to the planes, so we need to find the angle between the vectors n and m. This angle θ is given by the now well-known formula So, the angle between the two planes is θ ≈ 1.32 radians.

Hint: Find the angles between the normals to the two planes.

Hint: Find the angles between the normals to the two planes.