Question Video: Determining the Direction Vector of the Line of Intersection between Two Planes Mathematics

Which of the following is the direction vector of the line of intersection between the two planes π‘₯ + 3𝑦 + 𝑧 βˆ’ 2 = 0 and π‘₯ + 3𝑦 βˆ’ 3𝑧 + 2 = 0? [A] [βˆ’3 and βˆ’1 and 0] [B] [βˆ’3 and βˆ’1 and 3] [C] [βˆ’3 and 1 and 3] [D] [1 and 9 and βˆ’3] [E] [βˆ’3 and 1 and 0]

03:27

Video Transcript

Which of the following is the direction vector of the line of intersection between the two planes π‘₯ plus three 𝑦 plus 𝑧 minus two is equal to zero and π‘₯ plus three 𝑦 minus three 𝑧 plus two is equal to zero? Is it (A) negative three, negative one, zero? Or (B) negative three, negative one, three. (C) Negative three, one, three. Or (D) one, nine, negative three. Or (E) negative three, one, zero.

We’re given the equations for two planes in general form. Let’s call them P one and P two. We know that if two planes in our three have nonparallel normal vectors, they intersect over a straight line and that this line is the set of solutions to the set of equations P one and P two. Now we’re asked which of the given options (A), (B), (C), (D), or (E) is the direction vector of the line of intersection, where the direction vector is a vector that’s parallel to and in the direction of the line of intersection. We find this direction vector by taking the cross product of the normal vectors for each of the two planes. And we can find the normal vectors 𝑛 sub one and 𝑛 sub two by reading off the coefficients of the variables in the two plane equations. 𝑛 one then has components one, three, and one. And 𝑛 two has components one, three, and negative three.

We can work out their cross product by evaluating the determinant of the three-by-three matrix whose first row is the unit vectors 𝐒, 𝐣, and 𝐀 and whose second and third rows are the two normal vectors. And we can work this out by expanding along the first row to give the determinant of the two-by-two matrix with elements three, one, three, negative three multiplied by 𝐒 minus the determinant of the two-by-two matrix with elements one, one, one, negative three multiplied by 𝐣 plus the determinant of the matrix with elements one, three, one, three multiplied by the unit vector 𝐀.

And recalling that the determinant of a two-by-two matrix whose elements are π‘Ž, 𝑏, 𝑐, 𝑑 is π‘Žπ‘‘ minus 𝑏𝑐, this gives us three times negative three minus one times three times 𝐒 minus one times negative three minus one times one times 𝐣 plus one times three minus three times one multiplied by 𝐀. That is negative nine minus three 𝐒 minus negative three minus one 𝐣 plus three minus three 𝐀. This evaluates to negative 12𝐒 plus four 𝐣. And since three minus three is equal to zero, the coefficient of 𝐀 is zero. So we found our direction vector 𝐝 in component form. And making some space, we can write this in vector form negative 12, four, zero, and similarly in column vector form.

And we see that this doesn’t actually match with any of the five options (A), (B), (C), (D), or (E). But noticing that there’s a common factor of four in the first two components, we can take this scalar factor outside so that we have 𝐝 is equal to four times the column vector with components negative three, one, and zero. Now since four is a scalar multiple and any nonzero scalar multiple of this vector is a direction vector of the line, we see that our direction vector actually matches with option (E). And hence, the direction vector of the line of intersection between the two planes is option (E). That’s column vector negative three, one, zero.

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