Example 1 Given u = (9,3,−4,0,1) and v
= (0,−3,
2,−1,7) compute.
(a)
u − 4v
(b)
v u
(c)
u u
(d)
u
(e)
d (u, v)
Solution
There really
isn’t much to do here other than use the appropriate definition.
(a)
u – 4v =
(9, 3, – 4, 0, 1) – 4 (0, – 3, 2, – 1, 7)
= (9, 3, – 4, 0, 1) – 4 (0, – 12, – 4, 28)
= (9, 15, – 12, 4, – 27)
(b)
v ▪ u = (0)(9) + (−3)(3) + (2)(−4) + (−1)(0) + (7)(1) = −10
(c)
u▪u = 92 + 32 + (−4)2 + 02 +12 = 107
(d)
(e)
Example 2 Given u
= (−2, 3,
1, −1) and v
= (7, 1,
−4, −2) verify
the Cauchy-Schwarz inequality and the Triangle Inequality.
Solution
Let’s first verify the Cauchy-Schwarz inequality. To do this we
need to following quantities.
u▪v = – 14 + 3 – 4 + 2 = – 13
Now, verify the Cauchy-Schwarz inequality.
Sure enough the Cauchy-Schwarz inequality holds true.
To verify the Triangle inequality all we need is,
Now verify the Triangle Inequality.
So, the Triangle Inequality is also verified for this problem.
Example 3 Show that u
= (3,0,1,0,
4,−1) and v
= (−2,5,0,
2,−3,−18) are
orthogonal and verify that the Pythagorean Theorem holds.
Solution
Showing that these two vectors is easy enough.
u▪v = (3)(−2) + (0)(5) + (1)(0) + (0)(2) + (4)(−3) + (−1)(−18) = 0
So, the Pythagorean Theorem should hold, but let’s verify that.
Here’s the sum
u + v = (1,5,1, 2,1,−19)
and here’s the square of the norms.
‖u + v‖2 = 12
+ 52 +12 +22 +12 +(– 19)2
= 393
‖u‖2 = 32 + 02
+12 +02 +42 +(– 1)2 = 27
‖v‖2 = ( – 2)2 + 52
+02 +22 +( – 3)2 +(– 18)2 = 366
A quick computation then confirms that ‖u
+ v‖2 = ‖u‖2 + ‖v‖2.
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