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Question Video: Finding the Parametric Equation of the Line of Intersection of Two Planes Mathematics

Find the parametric equations of the line of intersection between the two planes π‘₯ + 𝑧 = 3 and 2π‘₯ βˆ’ 𝑦 βˆ’ 𝑧 = βˆ’2. [A] π‘₯ = 3 + 𝑑, 𝑦 = 4 + 3𝑑, 𝑧 = βˆ’π‘‘ [B] π‘₯ = 3 + 𝑑, 𝑦 = 8 βˆ’ 3𝑑, 𝑧 = βˆ’π‘‘ [C] π‘₯ = 3 + 𝑑, 𝑦 = 8 + 3𝑑, 𝑧 = βˆ’π‘‘ [D] π‘₯ = 1 + 𝑑, 𝑦 = 3 + 3𝑑, 𝑧 = 2 βˆ’ 𝑑 [E] π‘₯ = 3 + 𝑑, 𝑦 = 4 βˆ’ 3𝑑, 𝑧 = βˆ’π‘‘

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Video Transcript

Find the parametric equations of the line of intersection between the two planes π‘₯ plus 𝑧 equals three and two π‘₯ minus 𝑦 minus 𝑧 equals negative two.

Recall that a straight line in 3D space may be described by the set of parametric equations π‘₯ equals π‘₯ naught plus π‘Žπ‘‘, 𝑦 equals 𝑦 naught plus 𝑏𝑑, and 𝑧 equals 𝑧 naught plus 𝑐𝑑. π‘₯ naught, 𝑦 naught, 𝑧 naught can be any point on the line 𝑝 naught, and π‘Ž, 𝑏, 𝑐 is a direction vector that is parallel to the line. These parametric equations are arbitrary because 𝑝 naught can be any point on the line that we choose, and 𝐝 is just one direction vector that is parallel to the line. Any nonzero scalar multiple of 𝐝, 𝛼𝐝, is also parallel to the line.

To find the set of parametric equations for the line of intersection, we set one of these arbitrary parameterizations for one of the variables then substitute it into the two equations of the planes then rearrange the resulting equations to find expressions for the other two variables, also in terms of the parameter. For our two planes, we have the equations π‘₯ plus 𝑧 equals three and two π‘₯ minus 𝑦 minus 𝑧 equals negative two. If we substitute in an arbitrary parameterization for π‘₯, we get π‘₯ naught plus π‘Žπ‘‘ plus 𝑧 equals three and two times π‘₯ naught plus π‘Žπ‘‘ minus 𝑦 minus 𝑧 equals negative two. We can rearrange the first equation to give 𝑧 purely in terms of the parameter: 𝑧 equals three minus π‘₯ naught minus π‘Žπ‘‘.

Remember that π‘₯ naught and π‘Ž are both constants, so 𝑧 is a function of just the parameter 𝑑. We can now substitute this expression for 𝑧 into the second equation, which gives two times π‘₯ naught plus π‘Žπ‘‘ minus 𝑦 minus three minus π‘₯ naught minus π‘Žπ‘‘ equals negative two. Rearranging for 𝑦 gives 𝑦 purely in terms of the parameter: 𝑦 equals three times π‘₯ naught plus π‘Žπ‘‘ minus one. We now have π‘₯, 𝑦, and 𝑧 expressed purely as functions of the parameter 𝑑. So we have the arbitrary set of parametric equations. We are free to choose any values for π‘₯ naught and π‘Ž that we like with the exception that π‘Ž cannot be equal to zero because then changing the value of 𝑑 would not change the position on the line.

If we look at the list of possible answers, we can see that four of them have 𝑧 equals negative 𝑑. If we choose this as our parametric equation for 𝑧, this implies that three minus π‘₯ naught minus π‘Žπ‘‘ is equal to negative 𝑑. So this must mean that π‘₯ naught is equal to three and π‘Ž is equal to one. We can now substitute these values for π‘₯ naught and π‘Ž into the equations for π‘₯ and 𝑦. For π‘₯, we have π‘₯ equals three plus 𝑑, and for 𝑦 we have 𝑦 equals three times three plus 𝑑 minus one, which simplifies to eight plus three 𝑑. Our set of parametric equations therefore matches with answer (c) π‘₯ equals three plus 𝑑, 𝑦 equals eight plus three 𝑑, and 𝑧 equals negative 𝑑.

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