Question

Find u, v , u , v , and d(u, v) for the given inner product defined on Rn. u = (0, −4), v = (5, 3), u, v = 3u1v1 + u2v2

Answer 1

Answer:

$<u,v>=-12$

$d(u,v)=2\sqrt{31}$

Step-by-step explanation:

We are given that inner product defined on $R_n$

$<u,v>=3u_1v_1+u_2v_2$

u=(0,-4),v=(5,3)

We have to find the value of <u,v> and d(u,v)

We have $u_1=0,u_2=-4,v_1=5,v_2=3$

Substitute the value then we get

$<u,v>=3(0\cdot5)+(-4)(3)=-12$

Now, $d(u,v)=\left \|v-u\right \|$

Using this formula we get

$d(u,v)=\left \| (5,7) \right \|=\sqrt{3(5)^2+(7)^2}=\sqrt{75+49}=\sqrt{124}$

$d(u,v)=2\sqrt{31}$