Video Transcript
Find the point of intersection of
the straight line negative three 𝑥 equals four 𝑦 minus two equals 𝑧 plus one and
the plane negative three 𝑥 plus 𝑦 plus 𝑧 equals 13.
If they are not parallel or
coplanar, a line and a plane in 3D space will intersect at a single point 𝑥 naught,
𝑦 naught, 𝑧 naught. This point is the unique solution
to the equation of the line and the equation of the plane. We have three unknowns 𝑥, 𝑦, and
𝑧. We have one equation, the equation
of the plane. And the equation of the straight
line is effectively two equations since there are two equalities. We can rewrite the equation of the
line as two distinct equations: negative three 𝑥 equals 𝑧 plus one, and four 𝑦
minus two equals 𝑧 plus one. We therefore effectively have three
distinct equations for three unknowns. So as long as the line and the
plane are not parallel or coplanar, there should be one unique solution to these
three equations, which is the point where they intersect.
We can rewrite these two equations
to give 𝑥 and 𝑦 both purely in terms of 𝑧. 𝑥 equals negative one-third times
𝑧 plus one, and 𝑦 equals one-quarter times 𝑧 plus three. We can then substitute these
expressions for 𝑥 and 𝑦 into the equation of the plane, which will give us one
equation for one variable 𝑧, which we can solve to find the value of 𝑧 and
therefore the corresponding values of 𝑥 and 𝑦. So substituting in these
expressions into the equation of the plane gives us negative three times negative
one-third times 𝑧 plus one plus one-quarter times 𝑧 plus three plus 𝑧 equals
13.
Distributing the parentheses gives
us 𝑧 plus one plus 𝑧 over four plus three-quarters plus 𝑧 equals 13. Rearranging by collecting terms in
𝑧 on the left-hand side gives us nine 𝑧 over four equals 45 over four. Multiplying by four and dividing by
nine gives us 𝑧 equals five. We already have expressions for
both 𝑥 and 𝑦 in terms of 𝑧, so we can substitute in this value of 𝑧 to give the
values of 𝑥 and 𝑦. 𝑥 is equal to negative one-third
times five plus one, which equals negative two, and 𝑦 is equal to one-quarter times
five plus three, which is equal to two. The point of intersection between
the line and the plane is therefore negative two, two, five.
