Exercise 3.2.3. Find the multiple of the vector that is closest to the point . Also find the point on the line through that is closest to .

Answer: The first problem amounts to finding the projection of onto . From formula (5) on page 147 (theorem 3H) we have .

Given the definitions of and we then have

We thus have with being the multiple of we were asked to find.

The second problem amounts to finding the projection of onto . Adapting formula (5) we have .

Given the definitions of and we then have

We thus have or

so that is the point on the line through we were asked to find.

UPDATE: Fixed the computation of and thus of . Thanks to Ann and universitypika for the correction.

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.

Hey, for your calculation of bTb I think you miswrote your vector…i got 2*2 + 4*4 + 4*4 = 38.

i got 38 instead of 24 for bTb… 2*2 + 4*4 + 4*4

36 instead of 38, sorry!

Ann and universitypika: Thanks to both of you for catching this error. I have corrected the post to reflect your fix.