A First Course in Linear Algebra, 2017a

178 R n
From the above scalar equation, we have that ⃗n = ⎣
⎡
2 = d. Then, −→ P 0 P = ⎣
3
2
3
⎤
⎡
⎦ − ⎣
1
0
0
Next, compute −→ QP = proj ⃗n
−→
P0 P.
⎤
⎡
⎦ = ⎣
2
2
3
⎤
⎦.
⎡
2
1
2
⎤
−→ −→
QP = proj⃗n P0 P
( −→ )
P0 P •⃗n
=
‖⃗n‖ 2 ⃗n
⎡ ⎤
= 12 2
⎣ 1 ⎦
9
2
⎡ ⎤
= 4 2
⎣ 1 ⎦
3
2
Then, ‖ −→ QP‖ = 4 so the shortest distance from P to the plane is 4.
Next, to find the point Q on the plane which is closest to P we have
−→
0Q = −→ 0P − −→ QP
⎡ ⎤ ⎡ ⎤
3
= ⎣ 2 ⎦ − 4 2
⎣ 1 ⎦
3
3 2
⎡
= 1 ⎣
3
1
2
1
⎤
⎦
⎦. Now, choose P 0 =(1,0,0) so that ⃗n • −→ 0P =
Therefore, Q =( 1 3 , 2 3 , 1 3 ).
♠
4.9 The Cross Product
Outcomes
A. Compute the cross product and box product of vectors in R 3 .
Recall that the dot product is one of two important products for vectors. The second type of product
for vectors is called the cross product. It is important to note that the cross product is only defined in
R 3 . First we discuss the geometric meaning and then a description in terms of coordinates is given, both
of which are important. The geometric description is essential in order to understand the applications to
physics and geometry while the coordinate description is necessary to compute the cross product.