Question #274142

Find an orthonormal basis of R^3, of which (0,3√13, 2√13) is one element

1
Expert's answer
2021-12-02T07:22:53-0500

The length of the vector v1\vec v_1 is 11 as


v1=(0)2+(3/13)2+(2/13)2=1||\vec v_1 ||=\sqrt{(0)^2+(3/\sqrt{13})^2+(2/\sqrt{13})^2}=1

We want to find two vectors v2,v3\vec v_2,\vec v_3 such that {v1,v2,v3}\{\vec v_1,\vec v_2,\vec v_3\}is an orthonormal basis for R3.\R^3.

Let u=x,y,z\vec u=\langle x, y, z \rangle be a vector that is perpendicular to v1.\vec v_1.

Then we have uv1=0,\vec u\cdot \vec v_1=0, and hence we have the relation


0(x)+(3/13)(y)+(2/13)(z)=00(x)+(3/\sqrt{13})(y)+(2/\sqrt{13})(z)=0

3y+2z=03y+2z=0

For example, the vector u=1,2,3\vec u=\langle 1, 2, -3 \rangle satisfies the relation, anf hence uv1=0.\vec u\cdot \vec v_1=0.

Let us define the third vector w\vec w to be the cross product of u\vec u and v1\vec v_1


w=u×v1=ijk12303/132/13\vec w=\vec u\times\vec v_1=\begin{vmatrix} \vec i & \vec j & \vec k \\ 1 & 2 & -3 \\ 0 & 3/\sqrt{13} & 2/\sqrt{13} \end{vmatrix}

=i233/132/13j1302/13=\vec i\begin{vmatrix} 2 & -3 \\ 3/\sqrt{13} & 2/\sqrt{13} \end{vmatrix}-\vec j\begin{vmatrix} 1 & -3 \\ 0 & 2/\sqrt{13} \end{vmatrix}

+k1203/13+\vec k\begin{vmatrix} 1 & 2 \\ 0 & 3/\sqrt{13} \end{vmatrix}


=(13/13)i(2/13)j+(3/13)k=(13/\sqrt{13})\vec i-(2/\sqrt{13})\vec j+(3/\sqrt{13})\vec k

By the property of the cross product, the vector w\vec w is perpendicular to both u,v1.\vec u, \vec v_1.

u=(1)2+(2)2+(3)2=14||\vec u||=\sqrt{(1)^2+(2)^2+(-3)^2}=\sqrt{14}

w=(13/13)2+(2/13)2+(3/13)2||\vec w||=\sqrt{(13/\sqrt{13})^2+(-2/\sqrt{13})^2+(3/\sqrt{13})^2}

=14=\sqrt{14}

Therefore, the set


{v1,v2,v3}\{\vec v_1,\vec v_2,\vec v_3\}


={[03/132/13],[1/142/143/14],[13/1823/1822/182]}=\bigg\{\begin{bmatrix} 0 \\ 3/\sqrt{13}\\ 2/\sqrt{13} \end{bmatrix},\begin{bmatrix} 1/\sqrt{14} \\ 2/\sqrt{14}\\ -3/\sqrt{14} \end{bmatrix},\begin{bmatrix} 13/\sqrt{182} \\ 3/\sqrt{182}\\ 2/\sqrt{182} \end{bmatrix}\bigg\}

is an orthonormal basis for R3.\R^3.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS