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.
There is more than one way to
approach this. We are going to use the equation of
the line to eliminate the variables 𝑥 and 𝑦 from the equation of the plane. The description that we have of the
straight line in three-dimensional space equates three expressions: negative three
𝑥, four 𝑦 minus two, and 𝑧 plus one.
First, we separate this into two
equations: negative three 𝑥 equals 𝑧 plus one and four 𝑦 minus two equals 𝑧 plus
one. We rearrange these two equations to
the equations 𝑥 equals negative 𝑧 plus one over three and 𝑦 equals 𝑧 plus three
over four, expressing 𝑥 and 𝑦 in terms of 𝑧.
We now take the equation of the
plane, negative three 𝑥 plus 𝑦 plus 𝑧 equals 13, and substitute in our
expressions for 𝑥 and 𝑦 in terms of 𝑧. We now have the equation negative
three times negative 𝑧 plus one over three plus 𝑧 plus three over four plus 𝑧
equals 13, which contains only the variable 𝑧. We now simplify this equation by
first multiplying out the first term and then multiplying everything by four to
clear the denominator in the second term.
The result is a linear equation in
𝑧, which we can solve. We find that 𝑧 equals five. Recall that the first thing we did
was use our straight line equation to express 𝑥 and 𝑦 in terms of 𝑧. We can now substitute our value of
𝑧 equals five into these expressions to find that 𝑥 equals negative two and that
𝑦 equals two.
We have calculated that the point
of intersection of the line and the plane has coordinates negative two, two,
five.
