Linear Algebra and Its Applications, Review Exercise 1.24

Review exercise 1.24. The equation u + 2v - w = 6 defines a plane in 3-space. Find equations that define the following:

(a) a plane parallel to the first plane but going through the origin

(b) a second plane that (like the original plane) contains the points (6, 0, 0) and (2, 2, 0)

(c) a third plane that intersects the original plane and the one from (b) in the point (4, 1, 0)

Answer: (a) A plane passing through the origin must correspond to an equation that holds true for u = v = w = 0. The equation u + 2v - w = 0 satisfies this condition, and produces a plane parallel to the original plane.

(b) The points (6, 0, 0) and (2, 2, 0) satisfy the original equation u + 2v - w = 6 and are in the plane defined by that equation. For both these points we have w = 0. Therefore if we take the equation for the original plane and change the term involving w we can produce a new and different equation corresponding to a new plane containing these points. One such equation is u + 2v + w = 6.

(c) For the point (4, 1, 0) we have w = 0, and so as in (b) the equation we produce may have any term involving w. We simply need to ensure that the equation is satisfied when u = 4 and v = 1. One such equation is 2u - 2v + w = 6. Since the point (4, 1, 0) satisfies both this equation, the original equation, and the equation from (b) above, it is at the intersection of all three planes.

NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.

If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.

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