The portal has been deactivated. Please contact your portal admin.

Question Video: Finding the Parametric Equation of the Main Diagonal Line of a Cube Mathematics

A cube of a side length of 3 sits with a vertex at the origin and three sides along the positive axes. Find the parametric equations of the main diagonal from the origin.

02:42

Video Transcript

A cube of a side length of three sits with a vertex at the origin and three sides along the positive axes. Find the parametric equations of the main diagonal from the origin.

Let’s begin by making a sketch of the cube. The main diagonal of the cube goes from the vertex at the origin, which has coordinates zero, zero, zero, to the vertex furthest from this, that is, the vertex at the point with coordinates three, three, three. Now recalling that the parametric equations of a line in space are π‘₯ is equal to π‘₯ sub zero plus π‘Žπ‘‘, 𝑦 is equal to 𝑦 sub zero plus 𝑏𝑑, and 𝑧 is equal to 𝑧 sub zero plus 𝑐𝑑. And that’s where π‘₯ sub zero, 𝑦 sub zero, 𝑧 sub zero is a point on the line, the vector π‘Ž, 𝑏, 𝑐 is a direction vector, and 𝑑 is a real parameter that varies between negative and positive ∞.

Now, if we take the origin, that’s the point zero, zero, zero, to be the point on the line, that’s π‘₯ sub zero, 𝑦 sub zero, 𝑧 sub zero, then our parametric equations become π‘₯ is equal to zero plus π‘Žπ‘‘, 𝑦 is zero plus 𝑏𝑑, and 𝑧 is zero plus 𝑐𝑑. That is, π‘₯ is π‘Žπ‘‘, 𝑦 is 𝑏𝑑, and 𝑧 is 𝑐𝑑.

So now, we need to find the coordinates of a direction vector, that is, the components of our direction vector π‘Ž, 𝑏, and 𝑐. And if we have two points on a line, that’s π‘₯ sub one, 𝑦 sub one, 𝑧 sub one and π‘₯ sub two, 𝑦 sub two, 𝑧 sub two, then these components are given by π‘Ž is equal to π‘₯ sub two minus π‘₯ sub one, 𝑏 is equal to 𝑦 sub two minus 𝑦 sub one, and 𝑐 is 𝑧 sub two minus 𝑧 sub one. So now, if this time we take the origin as our point π‘₯ sub one, 𝑦 sub one, 𝑧 sub one and the vertex at the end of the main diagonal as the point π‘₯ sub two, 𝑦 sub two, 𝑧 sub two β€” so that has coordinates three, three, three β€” then we have π‘Ž is equal to three minus zero, 𝑏 is three minus zero, and 𝑐 is three minus zero. And so our direction vector has components three, three, three. And substituting these values into our parametric equations, we have π‘₯ is three 𝑑, 𝑦 is three 𝑑, and 𝑧 is three 𝑑. And these are the parametric equations of the main diagonal of the cube.

It’s worth noting that we could’ve used the point three, three, three as the point on the line. And for the direction vector, we could’ve chosen any vector parallel to the line. We could also have restricted the values of the parameter 𝑑. In fact, the line we found extends across the whole of space, whereas if we’d restricted the parameter 𝑑 to take values between zero and one only, our equations would describe only the main diagonal of the cube.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.