Video Transcript
Find the vector equation of the line of intersection between the two planes three 𝑥 plus 𝑦 minus five 𝑧 is equal to zero and 𝑥 plus two 𝑦 plus 𝑧 plus four is equal to zero.
To find the vector equation of the line of intersection between two planes, we need to find the position vector 𝐫 naught of a point that lies in both planes and therefore on the line of intersection. We then need to find a nonzero direction vector 𝐝, which is parallel to the line. Now, to find a point that lies in both planes, we begin by choosing a value of one of the variables. Let’s choose 𝑦 is equal to zero. And we note that here we could choose any of the three variables and assign any value because there are an infinite number of points on any line in space.
And by choosing a particular variable value, let’s say 𝑦 is equal to zero, the corresponding point on the line of intersection has specific values for the other two variables. So, with our specific value for 𝑦, that’s 𝑦 zero is equal to zero, we can solve the two plane equations for 𝑥 sub zero and 𝑧 sub zero with this value of 𝑦 is equal to 𝑦 sub zero.
Substituting 𝑦 is equal to zero into our plane equations then, in our first equation, we find three 𝑥 minus five 𝑧 is equal to zero. And in our second, we have 𝑥 plus 𝑧 plus four is equal to zero. Adding five 𝑧 and dividing through by three in our first equation, we then have 𝑥 is equal to five over three 𝑧, which we then substitute into our second equation for 𝑥. Multiplying through by three and then collecting like terms gives us eight 𝑧 plus 12 is equal to zero. And solving for 𝑧 gives us negative three over two; that is negative 1.5.
So now substituting this back into our equation obtained from 𝑃 one, we find 𝑥 is equal to five over three multiplied by negative three over two. Canceling the threes, that gives us 𝑥 is negative five over two, which is negative 2.5. And so, the position vector 𝐫 naught of a point on the line has components 𝑥 sub zero is negative 2.5, 𝑦 sub zero is zero, and 𝑧 sub zero is negative 1.5.
Now making some space, our next step is to find a direction vector parallel to the line of intersection. And to do this, we take the cross product of two normal vectors of the two planes. We can easily find normal vectors from the equations of the planes by reading off the coefficients of the variables for each plane. So, a normal vector for plane one has components three, one, and negative five. And similarly, for plane two, a normal vector has components one, two, and one. Now recalling that the cross product of two vectors in three dimensions is the determinant of the matrix whose first row is the unit vectors 𝐢, 𝐣, and 𝐤 and whose second and third rows are the components of the two vectors.
Expanding along the first row, this then gives us the two-by-two determinant with elements one, negative five, two, one multiplied by 𝐢 minus the two-by-two determinant three, negative five, one, one multiplied by 𝐣 plus the determinant three, one, one, two multiplied by 𝐤. Now recalling that the determinant of a two-by-two matrix with elements 𝑎, 𝑏, 𝑐, 𝑑 is 𝑎𝑑 minus 𝑏𝑐, our direction vector 𝐝 is then one minus negative 10𝐢 minus three minus negative five 𝐣 plus six minus one 𝐤. That is 11𝐢 minus eight 𝐣 plus five 𝐤. The direction vector for the line of intersection of the two planes then has components 11, negative eight, and five.
And making some space, we now have the position vector 𝐫 naught of a point on the line of intersection and the direction vector 𝐝. Hence, the vector equation of the line of intersection of the two given planes is 𝐫 is equal to the vector with components negative 2.5, zero, and negative 1.5 plus 𝑡 multiplied by the vector with components 11, negative eight, and five. It’s worth noting that the point 𝐫 sub zero resulted from our choice of 𝑦 is equal to zero. In fact, we could have alternatively chosen a specific value for 𝑥 or a specific value for 𝑧. And this would have given us a different point 𝐫 sub zero but still on the line of intersection.
