The figure shows a cube of side length six. The point 𝑀 is the midpoint of line segment 𝐴𝐵. Which of the following are the parametric equations of the line 𝑂𝑀? Is it (A) 𝑥 is equal to six 𝑡, 𝑦 is equal to three 𝑡, and 𝑧 is equal to three 𝑡. (B) 𝑥 is equal to six 𝑡, 𝑦 is equal to six minus six 𝑡, and 𝑧 is equal to three minus three 𝑡. Option (C) 𝑥 is equal to six 𝑡, 𝑦 is equal to six 𝑡, and 𝑧 is equal to three 𝑡. (D) 𝑥 is equal to three 𝑡, 𝑦 is equal to three 𝑡, and 𝑧 is equal to six 𝑡. Or (E) 𝑥 is equal to six minus six 𝑡, 𝑦 is equal to six minus six 𝑡, and 𝑧 is equal to three 𝑡.
We are told in the question that we have a cube of side length six. We are told that 𝑀 is the midpoint of the line segment 𝐴𝐵. And we’re asked to find the parametric equations of the line 𝑂𝑀.
We recall that the parametric equations of a line are a nonunique set of three equations of the form 𝑥 is equal to 𝑥 sub zero plus 𝑡𝑙, 𝑦 is equal to 𝑦 sub zero plus 𝑡𝑚, and 𝑧 is equal to 𝑧 sub zero plus 𝑡𝑛, where the point with coordinates 𝑥 sub zero, 𝑦 sub zero, 𝑧 sub zero lies on the line. 𝑙, 𝑚, 𝑛 is a direction vector of the line, and 𝑡 is a real number known as the parameter that varies from negative ∞ to ∞.
We know that the origin has coordinates zero, zero, zero. And since this point lies on line 𝑂𝑀, we can let 𝑥 sub zero, 𝑦 sub zero, and 𝑧 sub zero all equal zero. In order to work out a direction vector of this line, we firstly need to work out the coordinates of point 𝑀. Point 𝐴 has coordinates six, six, six. And point 𝐵 has coordinates six, six, zero.
Since 𝑀 is the midpoint of 𝐴𝐵, we can calculate its coordinates by finding average of the corresponding coordinates of 𝐴 and 𝐵. The average of six and six is six, and the average of six and zero is three. Therefore, 𝑀 has coordinates six, six, three. We are now in a position where we can find a direction vector of the line 𝑂𝑀.
One way of doing this is to subtract the position vectors of the two points. We can do this in either order. In this question, we’ll subtract the vector zero, zero, zero from the vector six, six, three. One direction vector of line 𝑂𝑀 is therefore equal to six, six, three. These are the values of 𝑙, 𝑚, and 𝑛 that we will substitute into the general form.
We could use either point 𝑂 or point 𝑀 for 𝑥 sub zero, 𝑦 sub zero, 𝑧 sub zero, as both of these lie on the line. As we are looking for a specific solution that matches one of our options, we will use the origin. Substituting our values of 𝑥 sub zero and 𝑙 into the general form gives us 𝑥 is equal to zero plus six 𝑡. This simplifies to just six 𝑡. Likewise, we have 𝑦 is equal to zero plus six 𝑡 and 𝑧 is equal to zero plus three 𝑡.
Simplifying our equations, we notice that they match those in option (C). From the options given, the parametric equations of line 𝑂𝑀 are 𝑥 equals six 𝑡, 𝑦 equals six 𝑡, and 𝑧 equals three 𝑡. As already mentioned, there are many other sets of parametric equations we could have chosen based on the information in this question.