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Let v = (v1, v2) be a vector in r2. show that (v2, −v1) is orthogonal to v, and use this fact to find two unit vectors orthogonal to the given vector. v = (9, 40)

Answer :

crisforp
Because v1 x v2 + v2 x (-v1) = v1 x v2 - v2 x v1 = v1 x v2 - v1 x v2 = 0, vector (v1,v2) is orthogonal to (v2,-v1);
Let (a,b) be an unit vector orthogonal to (9,40);
So, a^2 + b^2 = 1 and a x 9 + b x 40 = 0;
Then, a = - 40 x b / 9;
Finally, 160 x b^2 / 81 + b^2 = 1;
160 x b^2 + 81 x b^2 = 81;
241 x b^2 = 81;
b^2 = 81 / 241;
b = + or - 9/[tex] \sqrt{241} [/tex]
a = + or - 40/[tex] \sqrt{241} [/tex]
We have (+40/[tex] \sqrt{241} [/tex];9/[tex] \sqrt{241} [/tex]) and (-40/[tex] \sqrt{241} [/tex];-9/[tex] \sqrt{241} [/tex]);
lublana

Answer:

[tex]\hat{u_1}=(\frac{40}{41},-\frac{9}{40})[/tex] and [tex]\hat{u_2}=(\frac{-40}{41},\frac{9}{41})[/tex]

Step-by-step explanation:

We are given that [tex]v=(v_1,v_2)[/tex] be a vector in [tex]R^2[/tex].

We have to show that [tex](v_2,-v_1)[/tex] is orthogonal to v.We have to find the two unit vector orthogonal to the given vector v=(9,40).

Orthogonal vectors:If two vectors u and v are orthogonal then

[tex]u\cdot v=u_1v_2+u_2v_1=0[/tex]

Using this condition

[tex](v_1,v_2)\cdot (v_2,-v_1)[/tex]

[tex]=v_1v_2-v_2v_1=0[/tex]

Hence, [tex](v_2,-v_1)[/tex] is orthogonal to v.

By using this fact then, we get

[tex]u_1=[/tex](40,-9) is orthogonal to v=(9,40)

Unit vector=[tex]\frac{\vec{a}}{\mid a\mid}[/tex]

[tex]\mid u_1\mid=\sqrt{(40)^2+(-9)^2}=41[/tex]

[tex]\hat{u_1}=(\frac{40}{41},-\frac{9}{40})[/tex]

[tex]u_2=(-40,9)[/tex] is orthogonal to v=(9,40)

[tex]\mid u_2\mid=\sqrt{(-40)^2+9^2}=41[/tex]

[tex]\hat{u_2}=(\frac{-40}{41},\frac{9}{41})[/tex]

The two unit vectors which are orthogonal to given vector v=(9,40) are given by

[tex]\hat{u_1}=(\frac{40}{41},-\frac{9}{40})[/tex] and [tex]\hat{u_2}=(\frac{-40}{41},\frac{9}{41})[/tex]

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