Answer to Question #211621 in Linear Algebra for prince

Question

(4.3) Let ~u =< 0, 1, 1 >, ~v =< 2, 2, 0 > and w~ =< −1, 1, 0 > be three vectors in standard form.

(a) Determine which two vectors form a right angle triangle?

(b) Find θ := ~ucw~ , the angel between the given two vectors. (2)

(4.4) Let x < 0. Find the vector ~n =< x, y, z > that is orthogonal to all three vectors (2) ~u =< 1, 1, −2 >, ~v =< −1, 2, 0 > and w~ =< −1, 0, 1 >.

(4.5) Find a unit vector that is orthogonal to both ~u =< 0, −1, −1 > and ~v =< 1, 0, −1 >.

Answer 1

(4.3)

(a)

$$\vec u\cdot\vec v=\langle 0,1,1 \rangle\langle 2,2,0 \rangle=0+2+0=2\not=0$$

$$\vec u\cdot\vec w=\langle 0,1,1 \rangle\langle -1,1,0 \rangle=0+1+0=1\not=0$$

$$\vec w\cdot\vec v=\langle -1,1,0 \rangle\langle 2,2,0 \rangle=-2+2+0=0$$

$$\vec w \perp \vec v$$

Vectors $\vec w$ and $\vec v$ form a right angle triangle.

(b)

$$\vec u\cdot\vec w=\langle 0,1,1 \rangle\langle -1,1,0 \rangle=0+1+0=1$$

$$||\vec u||=\sqrt{(0)^2+(1)^2+(1)^2}=\sqrt{2}$$

$$||\vec w||=\sqrt{(-1)^2+(1)^2+(0)^2}=\sqrt{2}$$

$$\cos \theta=\dfrac{\vec u\cdot\vec w}{||\vec u||\cdot||\vec w||}=\dfrac{1}{\sqrt{2}\sqrt{2}}=\dfrac{1}{2}$$

$$\theta=60\degree$$

(4.4)

$$\vec u\cdot\vec n=\langle 1,1,-2 \rangle\langle x,y,z \rangle=x+y-2z=0$$

$$\vec v\cdot\vec n=\langle -1,2,0 \rangle\langle x,y,z\rangle=-x+2y=0$$

$$\vec w\cdot\vec n=\langle -1,0,1 \rangle\langle x,y,z\rangle=-x+z=0$$

$$x=2y$$

$$z=x=2y$$

$$2y+y-2y=0$$

$$x=0, y=0, z=0$$

Since $x<0,$ then such vector $\vec n$ does not exist.

(4.5)

$$\vec u\times \vec v=\begin{vmatrix} \vec i & \vec j & \vec k \\ 0 & -1 & -1 \\ 1 & 0 & -1 \\ \end{vmatrix}$$

$$=\vec i\begin{vmatrix} -1 & -1 \\ 0 & -1 \end{vmatrix}-\vec j\begin{vmatrix} 0 & -1 \\ 1 & -1 \end{vmatrix}+\vec k\begin{vmatrix} 0 & -1 \\ 1 & 0 \end{vmatrix}$$

$$=\vec i-\vec j+\vec k$$

$$\sqrt{(1)^2+(-1)^2+(1)^2}=\sqrt{3}$$

$$\vec e=\langle \dfrac{\sqrt{3}}{3}, -\dfrac{\sqrt{3}}{3}, \dfrac{\sqrt{3}}{3}\rangle$$