Quiz 3: Lines and Planes

A vector parallel to the line is the vector Hence the line can be represented by the vector equation Note that there are many other ways of giving a vector 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 equations of a line are given in parametric form we can identify the coordinates of two points on the line by letting the parameter take two different values, such as t = 0 and t = 1. Hence (2,1,-6) is on the line, and so is (5,3,-7). The vector from the first to the second point, namely , is therefore parallel to the line. The vector is a scalar multiple of and is also parallel to the line.

The point (2,1,-6) is on the line (corresponding to t = 0), and so is (5,3,-7) (corresponding to t = 1). The vector from the first to the second point, namely , is therefore parallel to the line.

In vector form, the equation of a line is written x = p + td, where p is the position vector (relative to the origin) of a point P on the line and d is a direction vector for the line. The components of the direction vector are simply the coefficients of t in the parametric equations.

Line ℓ must be parallel to . The line given in this option is parallel to , which is not a direction vector for ℓ.

A vector parallel to the line is (from the coefficients of t in the parametric equations). So the line is also parallel to . Therefore since the line passes through the point (1,2,1), it has vector equation x = + t.

This line is parallel to the vector , but does not contain the point (1,2,1). If it did, there would be a value of the parameter t satisfying all three of the following equations simultaneously: It is easy to see that no such t exists!

The line given in this option is parallel to the vector , which is not parallel to the line given in the question.

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

The general equation of the plane through the point (p,q,r) perpendicular to the vector is a(x-p) + b(y -q) + c(z -r) = 0. In this particular case, the equation becomes 1(x - 3) - 1(y - 1) + 2(z - 0) = 0, that is, x - y + 2z = 2.

First find a vector perpendicular (normal) to the plane by calculating ×. Then remember that the formula ax + by + cz = d for the equation of a plane gives us the information that the vector [a,b,c] is normal to the plane. So the only unknown constant left to find is the constant d. This can be evaluated by substituting the coordinates of any of the points A, B, C into the equation.

First find a vector perpendicular (normal) to the plane by calculating ×. Then remember that the formula ax + by + cz = d for the equation of a plane gives us the information that the vector [a,b,c] is normal to the plane. So the only unknown constant left to find is the constant d. This can be evaluated by substituting the coordinates of any of the points A, B, C into the equation.

First find a vector perpendicular (normal) to the plane by calculating ×. Then remember that the formula ax + by + cz = d for the equation of a plane gives us the information that the vector [a,b,c] is normal to the plane. So the only unknown constant left to find is the constant d. This can be evaluated by substituting the coordinates of any of the points A, B, C into the equation.

The vector equals [-2,3,4] and the vector equals [-3,3,-5]. These two vectors are parallel to the plane and so their cross product is perpendicular to the plane. We find that × = [-27,-22,3]. The equation of the plane has the form -27x- 22y + 3z = d where we can find d by substituting the coordinates of any of the three original points. This gives d = -34 and the answer follows.

The first line is parallel to [1,0,3] and the second is parallel to [4,-2,7]. A vector perpendicular to both lines is therefore the cross product of these two vectors.

The first line is parallel to [1,0,3] and the second is parallel to [4,-2,7]. A vector perpendicular to both lines is therefore the cross product of these two vectors.

The first line is parallel to [1,0,3] and the second is parallel to [4,-2,7]. A vector perpendicular to both lines is therefore the cross product of these two vectors.

The first line is parallel to [1,0,3] and the second is parallel to [4,-2,7]. A vector perpendicular to both lines is therefore the cross product of these two vectors.

This is just a scalar multiple of u×v and is therefore a vector in the same or opposite direction, not perpendicular to u×v.

This is the negative of u × v and is therefore a vector in the opposite direction to u × v.

The properties of cross product give Therefore this option is certainly not perpendicular to u × v – it is equal to it.

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

The two planes are not parallel, since the first plane is perpendicular to and the second plane is perpendicular to , and 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 the (acute) angle θ = cos^-1 ≈ 1.32 radians.

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

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