Exercise 3.1.10. Given the two vectors and find a homogeneous system in three unknowns whose solutions are the linear combinations of the vectors.

Answer: In the previous exercise 3.1.9 we showed that the plane spanned by the vectors and was orthogonal to the line passing through the origin and . This means that for any vector in that plane the inner product of and is zero, or

So we have for all vectors in the plane spanned by and . But the plane spanned by and is simply the set of all vectors that are linear combinations of and . So we have found a homogeneous system in three unknowns whose solutions are the linear combinations of and .

(Note that we can multiply the system by -1 on both sides to obtain an equivalent system .)

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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