Answer to Question #211621 in Linear Algebra for prince

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)


"\\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}\n \\vec i & \\vec j & \\vec k \\\\\n 0 & -1 & -1 \\\\\n 1 & 0 & -1 \\\\\n\\end{vmatrix}"

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


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