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

Find the vector equation of the line of intersection between the planes π‘₯ + 3𝑦 + 2𝑧 βˆ’ 6 = 0 and 2π‘₯ βˆ’ 𝑦 + 𝑧 + 2 = 0. [A] 𝐫 = 〈0, 2, βˆ’12βŒͺ + π‘‘βŒ©5, 3, βˆ’7βŒͺ [B] 𝐫 = 〈0, 14, 12βŒͺ + π‘‘βŒ©2, βˆ’3, 2βŒͺ [C] 𝐫 = 〈0, 2, 0βŒͺ + π‘‘βŒ©2, βˆ’3, 2βŒͺ [D] 𝐫 = 〈0, 2, 0βŒͺ + π‘‘βŒ©5, 3, βˆ’7βŒͺ [E] 𝐫 = 〈0, 14, 12βŒͺ + π‘‘βŒ©5, 3, βˆ’7βŒͺ

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Video Transcript

Find the vector equation of the line of intersection between the planes π‘₯ plus three 𝑦 plus two 𝑧 minus six equals zero and two π‘₯ minus 𝑦 plus 𝑧 plus two equals zero.

Recall that the vector equation of a line in 3D space is given by 𝐫 equals 𝐫 naught plus 𝑑𝐝. 𝐫 naught is the position vector of a point on the line π‘₯ naught, 𝑦 naught, 𝑧 naught. 𝑑 is a scalar. And 𝐝 is a direction vector parallel to the line. This equation is not unique since we are free to choose any point that we like on the line for 𝐫 naught, and any vector that is a nonzero scalar multiple of 𝐝 will also be parallel to the line. So to find the vector equation for the line of intersection between the two planes, all we need to do is find a point in both planes with position vector 𝐫 naught and a direction vector 𝐝 that is parallel to the line of intersection.

Let’s begin by finding a point that lies in both planes for 𝐫 naught. Assuming the line of intersection is not parallel to any of the coordinate planes, we can choose any value for any one variable that we like and find corresponding values for the other two variables. This will give us one point that lies in both planes. Since every one of the possible answers has π‘₯ equal to zero in the constant vector, let’s try π‘₯ equals zero. If the line of intersection happens to be parallel to the 𝑦𝑧-plane, the value of π‘₯ will be constant and probably not equal to zero. In which case, setting π‘₯ equal to zero will mean that we will not be able to find solutions to the two equations. If this were the case, we would need to set some other variable equal to some other value.

Fortunately, that is not the case here. Setting π‘₯ equal to zero gives us three 𝑦 plus two 𝑧 minus six equals zero and negative 𝑦 plus 𝑧 plus two equals zero. We can simply rearrange the second equation to give 𝑦 equals 𝑧 plus two. And substituting this expression for 𝑦 into the first equation gives us three times 𝑧 plus two plus two 𝑧 minus six equals zero. And rearranging for 𝑧 gives us 𝑧 equals zero. And since 𝑦 is equal to 𝑧 plus two, 𝑦 is equal to two. So by setting π‘₯ equals zero into the two equations of the planes, we’ve obtained 𝑦 equals two and 𝑧 equals zero. So the point zero, two, zero lies in both planes. We therefore have the position vector 𝐫 naught of a point that lies on the line of intersection zero, two, zero.

We now need to find a direction vector 𝐝 that is parallel to the line of intersection. Since the line of intersection lies on both planes, its direction vector is parallel to both planes. If we look down the axis of the two planes intersecting, their line of intersection comes straight out of the screen. Their normal vectors, however, 𝐧 one and 𝐧 two, both lie in the plane of the screen. The cross product of the two normal vectors, 𝐧 one cross 𝐧 two, is perpendicular to both 𝐧 one and 𝐧 two and therefore also either come straight out of the screen or go straight into the screen. So we can use this as our direction vector for the line of intersection.

We can obtain the normal vectors 𝐧 one and 𝐧 two by simply reading off the coefficients of the variables in each of the equations of the planes. 𝐧 one is therefore equal to one, three, two and 𝐧 two is equal to two, negative one, one. Their cross product, 𝐧 one cross 𝐧 two, is given by the determinant of the three-by-three matrix 𝐒, 𝐣, 𝐀 followed by the components of 𝐧 one one, three, two and then the components of 𝐧 two two, negative one, one, where 𝐒, 𝐣, and 𝐀 are the unit vectors in the π‘₯-, 𝑦-, and 𝑧-directions, respectively. Expanding the determinant along the top row gives us 𝐒 times three minus negative two minus 𝐣 times one minus four plus 𝐀 times negative one minus six. Simplifying gives us five 𝐒 plus three 𝐣 minus seven 𝐀, which as a tuple is five, three, negative seven.

We therefore have our direction vector for the vector equation of the line of intersection: five, three, negative seven. Our complete vector equation for the line of intersection between the two planes is therefore 𝐫 equals zero, two, zero plus 𝑑 times five, three, negative seven, which matches with answer (d).

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