Question #211621

(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 >.


1
Expert's answer
2021-06-29T11:06:47-0400

(4.3)

(a)


uv=0,1,12,2,0=0+2+0=20\vec u\cdot\vec v=\langle 0,1,1 \rangle\langle 2,2,0 \rangle=0+2+0=2\not=0

uw=0,1,11,1,0=0+1+0=10\vec u\cdot\vec w=\langle 0,1,1 \rangle\langle -1,1,0 \rangle=0+1+0=1\not=0

wv=1,1,02,2,0=2+2+0=0\vec w\cdot\vec v=\langle -1,1,0 \rangle\langle 2,2,0 \rangle=-2+2+0=0

wv\vec w \perp \vec v

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


(b)


uw=0,1,11,1,0=0+1+0=1\vec u\cdot\vec w=\langle 0,1,1 \rangle\langle -1,1,0 \rangle=0+1+0=1

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

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

cosθ=uwuw=122=12\cos \theta=\dfrac{\vec u\cdot\vec w}{||\vec u||\cdot||\vec w||}=\dfrac{1}{\sqrt{2}\sqrt{2}}=\dfrac{1}{2}

θ=60°\theta=60\degree

(4.4)


un=1,1,2x,y,z=x+y2z=0\vec u\cdot\vec n=\langle 1,1,-2 \rangle\langle x,y,z \rangle=x+y-2z=0

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

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


x=2yx=2y

z=x=2yz=x=2y

2y+y2y=02y+y-2y=0

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

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


(4.5)


u×v=ijk011101\vec u\times \vec v=\begin{vmatrix} \vec i & \vec j & \vec k \\ 0 & -1 & -1 \\ 1 & 0 & -1 \\ \end{vmatrix}

=i1101j0111+k0110=\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}

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

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


e=33,33,33\vec e=\langle \dfrac{\sqrt{3}}{3}, -\dfrac{\sqrt{3}}{3}, \dfrac{\sqrt{3}}{3}\rangle


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