Question Video: Finding the Parametric Equation of a Line through the Origin and the Centre of a Sphere Given the Endpoints of a Diagonal of the Sphere Mathematics • Third Year of Secondary School

Video Transcript

Which of the following is the parametric equation of the straight line that passes through the origin point and the center of a sphere in which the points 𝐴 six, five, seven and 𝐵 two, one, three are endpoints of one of the sphere’s diameters? Is it (A) 𝑥 is equal to two 𝑡, 𝑦 is equal to three 𝑡, and 𝑧 is equal to two 𝑡? (B) 𝑥 is four 𝑡, 𝑦 is three 𝑡, and 𝑧 is five 𝑡. Or (C) 𝑥 is six 𝑡, 𝑦 is three 𝑡, and 𝑧 is five 𝑡. (D) 𝑥 is six 𝑡, 𝑦 is three minus three 𝑡, and 𝑧 is five 𝑡. Or (E) 𝑥 is equal to four 𝑡, 𝑦 is equal to three minus three 𝑡, and 𝑧 is equal to two 𝑡.

Let’s begin by making a sketch using the information we’ve been given. We’re told that the endpoints of one of our sphere’s diameters are the points with coordinates 𝐴 six, five, seven and 𝐵 two, one, three. And we want to find a parametric equation of a line that goes through the origin of our coordinate system, that is, the point with coordinates zero, zero, zero, and the center of the sphere. We can use the two diameter endpoints, that’s 𝐴 and 𝐵, to find the coordinates of the center of the sphere. We’ll then have the coordinates of two points on our line, that’s the center of the sphere and the origin of the coordinate system, which we can use to find the parametric equation of the line.

So we have our sphere, and since the center is the midpoint of any diameter, we can use the midpoint formula as shown. This tells us that the midpoint of a line between two points 𝑥 one, 𝑦 one, 𝑧 one and 𝑥 two, 𝑦 two, 𝑧 two has coordinates 𝑥 one plus 𝑥 two over two, 𝑦 one plus 𝑦 two over two, and 𝑧 one plus 𝑧 two over two. So if we take point 𝐴 is 𝑥 one, 𝑦 one, 𝑧 one and point 𝐵 is 𝑥 two, 𝑦 two, 𝑧 two, the midpoint of the line joining 𝐴 and 𝐵 has coordinates six plus two over two, five plus one over two, and seven plus three over two, that is, eight over two, six over two, 10 over two. This gives us the coordinates for the center of our circle 𝐶 as four, three, five.

So now we have two points on the line of interest; the center with coordinates four, three, five and the origin with coordinates zero, zero, zero. So now making some space so that we can concentrate on our line, we recall that the parametric equation of a line in space is given by 𝑥 is equal to 𝑥 zero plus 𝑡𝑎, 𝑦 is 𝑦 zero plus 𝑡𝑏, and 𝑧 is 𝑧 zero plus 𝑡𝑐 for real parameter 𝑡 and where the line has the direction vector 𝐝 is equal to 𝑎 multiplied by unit vector 𝐢 plus 𝑏 multiplied by the unit vector 𝐣 plus 𝑐 multiplied by the unit vector 𝐤. And the line passes through the point with coordinates 𝑥 zero, 𝑦 zero, and 𝑧 zero.

Now we have two points on our line. That’s the center with coordinates four, three, five and the origin with coordinates zero, zero, zero. And if we choose the origin to be the point 𝑥 zero, 𝑦 zero, 𝑧 zero through which the line passes, then our parametric equations are 𝑥 is equal to zero plus 𝑡𝑎, that’s 𝑥 zero plus 𝑡𝑎, 𝑦 is equal to zero plus 𝑡𝑏, and 𝑧 is equal to zero plus 𝑡𝑐. And remember, 𝑎, 𝑏, and 𝑐 are the coefficients of the direction vector. And these are also called the direction ratios.

Now we also know that if a line passes through two points 𝑥 one, 𝑦 one, 𝑧 one and 𝑥 two, 𝑦 two, 𝑧 two, then a direction vector of the line is given by 𝐝 is equal to 𝑥 two minus 𝑥 one times the unit vector 𝐢 plus 𝑦 two minus 𝑦 one times the unit vector 𝐣 plus 𝑧 two minus 𝑧 one multiplied by the unit vector 𝐤. The direction ratios 𝑎, 𝑏, and 𝑐 are therefore 𝑥 two minus 𝑥 one, 𝑦 two minus 𝑦 one, and 𝑧 two minus 𝑧 one, respectively.

So now again, taking our two points zero, zero, zero and four, three, five so that zero, zero, zero now corresponds to 𝑥 one, 𝑦 one, 𝑧 one and four, three, five corresponds to 𝑥 two, 𝑦 two, 𝑧 two, we have 𝑎 is equal to four minus zero, 𝑏 is three minus zero, and 𝑐 is five minus zero. That is, 𝑎 is four, 𝑏 is three, and 𝑐 is equal to five. Our direction vector is therefore 𝐝 is equal to four multiplied by 𝐢 plus three 𝐣 plus five 𝐤. Now substituting these values for 𝑎, 𝑏, and 𝑐 into our parametric equations, we have 𝑥 is equal to four 𝑡, 𝑦 is three 𝑡, and 𝑧 is five 𝑡. And that’s where each real value of the parameter 𝑡 gives a unique point on the line.

The parametric equation of the straight line passing through the origin and the center of a sphere in which the points 𝐴 six, five, seven and 𝐵 two, one, three are endpoints points of one of the sphere’s diameters is therefore 𝑥 is equal to four 𝑡, 𝑦 is three 𝑡, and 𝑧 is five 𝑡. And this corresponds to option (B).