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Question Video: Finding the Intersection of a Line and a Plane Given Their Vector Equations Mathematics

Find the point of intersection of the straight line −3𝑥 = 4𝑦 − 2 = 𝑧 + 1 and the plane −3𝑥 + 𝑦 + 𝑧 = 13.

02:29

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.

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