masses at A, B, and C, the center of mass of that system is given by the intersections of the medians of the triangle. 9. (a) Let u, v e R2. Describe the vectors x su + tv, where s +t = 1. Pay particular attention to the location of x when s > 0 and whent > 0. (b) Let u, v, w e Ri. Describe the vectors x ru +sV + tw, wherer+s +t = 1. Pay particular attention to the location of x when each of r, s, and t is positive. 10. Suppose x, y e R" are nonparallel vectors. (Recall the definition on p. 3.)

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I need help for problem 9. Thanks!

masses at A, B, and C, the center of mass of that system is given by the intersections of the medians
of the triangle.
9. (a) Let u, v e R2. Describe the vectors x su + tv, where s +t = 1. Pay particular attention
to the location of x when s > 0 and whent > 0.
(b) Let u, v, w e Ri. Describe the vectors x ru +sV + tw, wherer+s +t = 1. Pay particular
attention to the location of x when each of r, s, and t is positive.
10. Suppose x, y e R" are nonparallel vectors. (Recall the definition on p. 3.)
Transcribed Image Text:masses at A, B, and C, the center of mass of that system is given by the intersections of the medians of the triangle. 9. (a) Let u, v e R2. Describe the vectors x su + tv, where s +t = 1. Pay particular attention to the location of x when s > 0 and whent > 0. (b) Let u, v, w e Ri. Describe the vectors x ru +sV + tw, wherer+s +t = 1. Pay particular attention to the location of x when each of r, s, and t is positive. 10. Suppose x, y e R" are nonparallel vectors. (Recall the definition on p. 3.)
Expert Solution
Step 1

For (a) 

We are given that u, v e R2. And x = su + tv, where s t
t 1- s
х%3D su + (1 —s)v
х %3Ds (и — v) + v
1
So,
which implies that x is a vector which is a linear
v passing through v.
(which can be observed by putting the value of s 0)
combination of the vector u
Step 2

Now we will check the behaviour of x for values of s and t.

Now we will check the behaviour of x for values of s and t.
When s > 0 then the position of x on the line u - v passing through v
will move towards an increasing direction. Similarly, when s < 0, the
position of x on the line u - v passing through v will move towards a
decreasing direction
Again, to visualize for t 0,we can rearrange the vector
x ut(vu). Similar explanation can be made for t as above.
Step 3

If both s ≥ 0, t ≥ 0 and satisfying s + t = 1, then the position of x will lie on the u-v vector as shown in fig.

u-v
V
Step 4

Now for (b),

We are given that u, v e R3. And x
where rs+t = 1.
ru + sv + tw,
t 1-r-s
So,
x rusv + (1 -r-s)w
x r(u w) s(v -w)w
Which implies that x is a vector which is a linear combination of the
w passing through w. (which can be observed
0 and r 0)
vector u w and v
by putting the value of s
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