Finding the vector equation for a line that intersects two planes – Linear Algebra –
Finding the vector equation for a line that intersects two planes – Linear Algebra –

### Video Transcript

Which of the following is the
direction vector of the line of intersection between the two planes 𝑥 plus three 𝑦
plus 𝑧 minus two is equal to zero and 𝑥 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 𝐢, 𝐣, and 𝐤 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 𝐢 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 𝐤.

And recalling that the determinant
of a two-by-two matrix whose elements are 𝑎, 𝑏, 𝑐, 𝑑 is 𝑎𝑑 minus 𝑏𝑐, this
gives us three times negative three minus one times three times 𝐢 minus one times
negative three minus one times one times 𝐣 plus one times three minus three times
one multiplied by 𝐤. That is negative nine minus three
𝐢 minus negative three minus one 𝐣 plus three minus three 𝐤. This evaluates to negative 12𝐢
plus four 𝐣. And since three minus three is
equal to zero, the coefficient of 𝐤 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.

You are watching: Question Video: Determining the Direction Vector of the Line of Intersection between Two Planes. Info created by Bút Chì Xanh selection and synthesis along with other related topics.