Science:Math Exam Resources/Courses/MATH221/April 2013/Question 12 (b)/Solution 1

If we apply the projection of (a) to the vector u, we get a vector w that is orthogonal to the plane. To get v, we simply subtract w from u.

To verify that v is indeed on the plane, we compute that: $P_{1}(v)=P_{1}(u-w)=P_{1}(u)-P_{1}(w)=w-w=0$, which shows that the vector v has no competent that is orthogonal to the plane. Thus v must lie in the plane itself.

Let's perform the necessary calculations to get w and v. We have

$w={\frac {u\cdot \left[{\begin{array}{ccc}1&-1&1\\\end{array}}\right]}{3}}\left[{\begin{array}{ccc}1&-1&1\\\end{array}}\right]={\frac {1}{3}}\left[{\begin{array}{ccc}1&-1&1\\\end{array}}\right]$

And hence $v=u-w=\left[{\begin{array}{ccc}{\frac {2}{3}}&{\frac {4}{3}}&{\frac {2}{3}}\\\end{array}}\right]$.

Note: It is also easy to directly verify that v satisfies the equation x - y + z = 0 of the plane, since $2/3-4/3+2/3=0$.