IGNOU BMTE-141 Solved Assignment 2024 for Academic Excellence

IGNOU BMTE-141 Solved Assignment 2024 | B.Sc (G) CBCS

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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IGNOU BMTE-141 Assignment Question Paper 2024

IGNOU BMTE-141 Assignment Question Paper 2024

PART – A (30 Marks)

  1. i) Find the angle between the vectors 2 i + 2 j + 2 k 2 i + 2 j + 2 k sqrt2i+2j+2k\sqrt{2} \mathbf{i}+2 \mathbf{j}+2 \mathbf{k}2i+2j+2k and i + 2 j + 2 k i + 2 j + 2 k i+sqrt2j+sqrt2k\mathbf{i}+\sqrt{2} \mathbf{j}+\sqrt{2} \mathbf{k}i+2j+2k.
ii) Find the vector equation of the plane determined by the points ( 1 , 0 , 1 ) , ( 0 , 1 , 1 ) ( 1 , 0 , 1 ) , ( 0 , 1 , 1 ) (1,0,-1),(0,1,1)(1,0,-1),(0,1,1)(1,0,1),(0,1,1) and ( 1 , 1 , 0 ) ( 1 , 1 , 0 ) (-1,1,0)(-1,1,0)(1,1,0).
iii) Check whether W = { ( x , y , z ) R 3 x + y z = 0 } W = ( x , y , z ) R 3 x + y z = 0 W={(x,y,z)inR^(3)∣x+y-z=0}W=\left\{(x, y, z) \in \mathbb{R}^3 \mid x+y-z=0\right\}W={(x,y,z)R3x+yz=0} is a subspace of R 3 R 3 R^(3)\mathbb{R}^3R3.
iv) Check whether the set of vectors { 1 + x , x + x 2 , 1 + x 3 } 1 + x , x + x 2 , 1 + x 3 {1+x,x+x^(2),1+x^(3)}\left\{1+x, x+x^2, 1+x^3\right\}{1+x,x+x2,1+x3} is a linearly independent set of vectors in P 3 P 3 P_(3)\mathbf{P}_3P3, the vector space of polynomials of degree 3 3 <= 3\leq 33.
v) Check whether T : R 2 R 2 T : R 2 R 2 T:R^(2)rarrR^(2)T: \mathbb{R}^2 \rightarrow \mathbb{R}^2T:R2R2, defined by T ( x , y ) = ( y , x ) T ( x , y ) = ( y , x ) T(x,y)=(-y,x)T(x, y)=(-y, x)T(x,y)=(y,x) is a linear transformation.
vi) If { v 1 , v 2 } v 1 , v 2 {v_(1),v_(2)}\left\{v_1, v_2\right\}{v1,v2} is an ordered basis of R 2 R 2 R^(2)\mathbb{R}^2R2 and { f 1 ( v ) , f 2 ( v ) } f 1 ( v ) , f 2 ( v ) {f_(1)(v),f_(2)(v)}\left\{f_1(v), f_2(v)\right\}{f1(v),f2(v)} is the corresponding dual basis find f 1 ( 2 v 1 + v 2 ) f 1 2 v 1 + v 2 f_(1)(2v_(1)+v_(2))f_1\left(2 v_1+v_2\right)f1(2v1+v2) and f 2 ( v 1 2 v 2 ) f 2 v 1 2 v 2 f_(2)(v_(1)-2v_(2))f_2\left(v_1-2 v_2\right)f2(v12v2).
vii) Find the kernel of the linear transformation T : R 2 R 2 T : R 2 R 2 T:R^(2)rarrR^(2)T: \mathbb{R}^2 \rightarrow \mathbb{R}^2T:R2R2 defined by T ( x , y ) = ( 2 x + 3 y , 2 x 3 y ) T ( x , y ) = ( 2 x + 3 y , 2 x 3 y ) T(x,y)=(2x+3y,2x-3y)T(x, y)=(2 x+3 y, 2 x-3 y)T(x,y)=(2x+3y,2x3y).
viii) Describe the linear transformation T : R 2 R 2 T : R 2 R 2 T:R^(2)rarrR^(2)T: \mathbb{R}^2 \rightarrow \mathbb{R}^2T:R2R2 such that
[ T ] B = [ 1 2 2 0 ] [ T ] B = 1      2 2      0 [T]_(B)=[[1,2],[2,0]][T]_B=\left[\begin{array}{ll} 1 & 2 \\ 2 & 0 \end{array}\right][T]B=[1220]
where B B BBB is the standard basis of R 2 R 2 R^(2)\mathbb{R}^2R2.
ix) Find the matrix of the linear transformation T : R 2 R 2 T : R 2 R 2 T:R^(2)rarrR^(2)T: \mathbb{R}^2 \rightarrow \mathbb{R}^2T:R2R2 defined by T ( x , y ) = ( 2 y , x y ) T ( x , y ) = ( 2 y , x y ) T(x,y)=(2y,x-y)T(x, y)=(2 y, x-y)T(x,y)=(2y,xy) with respect to the ordered basis { ( 0 , 1 ) , ( 1 , 0 ) } { ( 0 , 1 ) , ( 1 , 0 ) } {(0,-1),(-1,0)}\{(0,-1),(-1,0)\}{(0,1),(1,0)}.
x) Let A A AAA be a 2 × 3 2 × 3 2xx32 \times 32×3 matrix, B B BBB be a 3 × 4 3 × 4 3xx43 \times 43×4 matrix and C C CCC be a 3 × 2 3 × 2 3xx23 \times 23×2 matrix and D D DDD be a 3 × 4 3 × 4 3xx43 \times 43×4 matrix. Is A B + C t D A B + C t D AB+C^(t)DA B+C^t DAB+CtD defined? Justify your answer.
xi) Verify Cayley-Hamilton theorem for the matrix A = [ 1 1 0 2 ] A = 1 1 0 2 A=[[1,-1],[0,2]]A=\left[\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right]A=[1102].
xii) Check whether [ 1 1 1 1 ] 1 1 1 1 [[1],[1],[1],[1]]\left[\begin{array}{l}1 \\ 1 \\ 1 \\ 1\end{array}\right][1111] is an eigenvector for the matrix [ 1 0 0 1 0 1 1 0 0 0 1 1 0 1 0 1 ] 1      0      0      1 0      1      1      0 0      0      1      1 0      1      0      1 [[1,0,0,1],[0,1,1,0],[0,0,1,1],[0,1,0,1]]\left[\begin{array}{llll}1 & 0 & 0 & 1 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 0 & 1\end{array}\right][1001011000110101]. What is the corresponding eigenvalue?
xiii) Let C [ 0 , 1 ] C [ 0 , 1 ] C[0,1]C[0,1]C[0,1] be the inner product space of continous real valued functions on the interval [ 0 , 1 ] [ 0 , 1 ] [0,1][0,1][0,1] with the inner product
f , g = 0 1 f ( t ) g ( t ) d t f , g = 0 1 f ( t ) g ( t ) d t (:f,g:)=int_(0)^(1)f(t)g(t)dt\langle f, g\rangle=\int_0^1 f(t) g(t) d tf,g=01f(t)g(t)dt
Find the inner product of the functions f ( t ) = 2 t , g ( t ) = 1 t 2 + 5 f ( t ) = 2 t , g ( t ) = 1 t 2 + 5 f(t)=2t,g(t)=(1)/(t^(2)+5)f(t)=2 t, g(t)=\frac{1}{t^2+5}f(t)=2t,g(t)=1t2+5.
xiv) Find adjoint of the linear operator T : C 2 C 2 T : C 2 C 2 T:C^(2)rarrC^(2)T: \mathbb{C}^2 \rightarrow \mathbb{C}^2T:C2C2 defined by T ( z 1 , z 2 ) = ( z 2 , z 1 + i z 2 ) T z 1 , z 2 = z 2 , z 1 + i z 2 T(z_(1),z_(2))=(z_(2),z_(1)+iz_(2))T\left(z_1, z_2\right)=\left(z_2, z_1+i z_2\right)T(z1,z2)=(z2,z1+iz2) with respect to the standard inner product on C 2 C 2 C^(2)C^2C2.
xv) Find the signature of the quadratic form x 1 2 2 x 2 2 + 3 x 3 2 x 1 2 2 x 2 2 + 3 x 3 2 x_(1)^(2)-2x_(2)^(2)+3x_(3)^(2)x_1^2-2 x_2^2+3 x_3^2x122x22+3x32
Part-B (40 Marks)
  1. a) Let S S SSS be any non-empty set and let V ( S ) V ( S ) V(S)V(S)V(S) be the set of all real valued functions on R R R\mathbb{R}R. Define addition on V ( s ) V ( s ) V(s)V(s)V(s) by ( f + g ) ( x ) = f ( x ) + g ( x ) ( f + g ) ( x ) = f ( x ) + g ( x ) (f+g)(x)=f(x)+g(x)(f+g)(x)=f(x)+g(x)(f+g)(x)=f(x)+g(x) and scalar multiplication by ( α f ) ( x ) = α f ( x ) ( α f ) ( x ) = α f ( x ) (alpha*f)(x)=alpha f(x)(\alpha \cdot f)(x)=\alpha f(x)(αf)(x)=αf(x). Check that ( V ( S ) , + , ) ( V ( S ) , + , ) (V(S),+,*)(V(S),+, \cdot)(V(S),+,) is a vector space.
b) Check that B = { 1 , 2 x + 1 , ( x 1 ) 2 } B = 1 , 2 x + 1 , ( x 1 ) 2 B={1,2x+1,(x-1)^(2)}\boldsymbol{B}=\left\{1,2 x+1,(x-1)^2\right\}B={1,2x+1,(x1)2} is a basis for P 2 P 2 P_(2)\mathbf{P}_2P2, the vector space of polynomials with real coefficients of degree 2 2 <= 2\leq 22.
  1. a) Let T : R 3 R 3 T : R 3 R 3 T:R^(3)rarrR^(3)T: \mathbb{R}^3 \rightarrow \mathbb{R}^3T:R3R3 be a linear operator and suppose the matrix of the operator with respect to the ordered basis
B = { [ 1 0 0 ] , [ 1 0 1 ] , [ 0 1 0 ] } B = 1 0 0 , 1 0 1 , 0 1 0 B={[[1],[0],[0]],[[1],[0],[1]],[[0],[1],[0]]}\boldsymbol{B}=\left\{\left[\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right],\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right],\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right]\right\}B={[100],[101],[010]}
is [ 1 0 1 0 1 1 1 0 1 ] 1 0 1 0 1 1 1 0 1 [[1,0,-1],[0,1,1],[1,0,1]]\left[\begin{array}{ccc}1 & 0 & -1 \\ 0 & 1 & 1 \\ 1 & 0 & 1\end{array}\right][101011101]. Find the matrix of the linear transformation with respect to the basis
B = { [ 1 0 1 ] , [ 1 0 1 ] , [ 0 1 0 ] } B = 1 0 1 , 1 0 1 , 0 1 0 B^(‘)={[[1],[0],[1]],[[1],[0],[-1]],[[0],[1],[0]]}\boldsymbol{B}^{\prime}=\left\{\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right],\left[\begin{array}{c} 1 \\ 0 \\ -1 \end{array}\right],\left[\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\right]\right\}B={[101],[101],[010]}
b) Show that W = { ( x , 4 x , 3 x ) R 2 x R } W = ( x , 4 x , 3 x ) R 2 x R W={(x,4x,3x)inR^(2)∣x inR}W=\left\{(x, 4 x, 3 x) \in \mathbb{R}^2 \mid x \in \mathbb{R}\right\}W={(x,4x,3x)R2xR} is a subspace of R 3 R 3 R^(3)\mathbb{R}^3R3. Also find a basis for subspace U U UUU of R 3 R 3 R^(3)\mathbb{R}^3R3 which satisfies W U = R 3 W U = R 3 W o+U=R^(3)W \oplus U=\mathbb{R}^3WU=R3.
  1. a) Find the eigenvalues and eigenvectors of the matrix B = [ 1 1 0 1 3 0 1 1 1 ] B = 1      1      0 1      3      0 1      1      1 B=[[1,1,0],[-1,3,0],[1,-1,1]]B=\left[\begin{array}{rrr}1 & 1 & 0 \\ -1 & 3 & 0 \\ 1 & -1 & 1\end{array}\right]B=[110130111]. Is the matrix diagonalisable? Justify your answer.
b) Find Adj ( A ) Adj ( A ) Adj(A)\operatorname{Adj}(A)Adj(A) where A = [ 3 2 2 1 1 0 3 0 1 ] A = 3 2 2 1 1 0 3 0 1 A=[[3,2,2],[-1,1,0],[3,0,1]]A=\left[\begin{array}{ccc}3 & 2 & 2 \\ -1 & 1 & 0 \\ 3 & 0 & 1\end{array}\right]A=[322110301]. Hence find A 1 A 1 A^(-1)A^{-1}A1.
  1. a) Solve the folowing set of simultaneous equations using Cramer’s rule:
x + 2 y + z = 3 2 x y + 2 z = 1 3 x + y + z = 0 x + 2 y + z = 3 2 x y + 2 z = 1 3 x + y + z = 0 {:[x+2y+z=3],[2x-y+2z=1],[3x+y+z=0]:}\begin{aligned} x+2 y+z & =3 \\ 2 x-y+2 z & =1 \\ 3 x+y+z & =0 \end{aligned}x+2y+z=32xy+2z=13x+y+z=0
b) Find the minimal polynomial of the matrix
[ 2 1 0 1 1 0 0 1 2 2 1 3 0 0 0 1 ] 2      1      0      1 1      0      0      1 2      2      1      3 0      0      0      1 [[2,1,0,1],[-1,0,0,1],[-2,-2,-1,3],[0,0,0,1]]\left[\begin{array}{rrrr} 2 & 1 & 0 & 1 \\ -1 & 0 & 0 & 1 \\ -2 & -2 & -1 & 3 \\ 0 & 0 & 0 & 1 \end{array}\right][2101100122130001]
Part C (30 marks)
  1. a) Let V V VVV be the vector space of all real valued functions that are twice differentiable in R R R\mathbb{R}R and
S = { cos x , sin x , x cos x , x sin x } . S = { cos x , sin x , x cos x , x sin x } . S={cos x,sin x,x cos x,x sin x}.S=\{\cos x, \sin x, x \cos x, x \sin x\} .S={cosx,sinx,xcosx,xsinx}.
Check that S S SSS is a linearly independent set over R R R\mathbb{R}R. (Hint: Consider the equation
a 0 cos x + a 1 sin x + a 2 x cos x + a 3 x sin x . a 0 cos x + a 1 sin x + a 2 x cos x + a 3 x sin x . a_(0)cos x+a_(1)sin x+a_(2)x cos x+a_(3)x sin x.a_0 \cos x+a_1 \sin x+a_2 x \cos x+a_3 x \sin x .a0cosx+a1sinx+a2xcosx+a3xsinx.
(Put x = 0 , π , π 2 , π 4 x = 0 , π , π 2 , π 4 x=0,pi,(pi)/(2),(pi)/(4)x=0, \pi, \frac{\pi}{2}, \frac{\pi}{4}x=0,π,π2,π4, etc. and find a i a i a_(i)a_iai.)
b) Consider the linear operator T : C 3 C 3 T : C 3 C 3 T:C^(3)rarrC^(3)T: \mathbb{C}^3 \rightarrow \mathbb{C}^3T:C3C3, defined by
T ( z 1 , z 2 , z 3 ) = ( z 1 i z 2 , i z 1 2 z 2 + i z 3 , i z 2 + z 3 ) . T z 1 , z 2 , z 3 = z 1 i z 2 , i z 1 2 z 2 + i z 3 , i z 2 + z 3 . T(z_(1),z_(2),z_(3))=(z_(1)-iz_(2),iz_(1)-2z_(2)+iz_(3),-iz_(2)+z_(3)).T\left(z_1, z_2, z_3\right)=\left(z_1-i z_2, i z_1-2 z_2+i z_3,-i z_2+z_3\right) .T(z1,z2,z3)=(z1iz2,iz12z2+iz3,iz2+z3).
i) Compute T T T^(**)T^*T and check whether T T TTT is self-adjoint.
ii) Check whether T T TTT is unitary.
  1. a) Let ( x 1 , x 2 , x 3 ) x 1 , x 2 , x 3 (x_(1),x_(2),x_(3))\left(x_1, x_2, x_3\right)(x1,x2,x3) and ( y 1 , y 2 , y 3 ) y 1 , y 2 , y 3 (y_(1),y_(2),y_(3))\left(y_1, y_2, y_3\right)(y1,y2,y3) represent the coordinates with respect to the bases B 1 = { ( 1 , 0 , 0 ) , ( 1 , 1 , 0 ) , ( 0 , 0 , 1 ) } , B 2 = { ( 1 , 0 , 0 ) , ( 0 , 1 , 1 ) , ( 0 , 0 , 1 ) } B 1 = { ( 1 , 0 , 0 ) , ( 1 , 1 , 0 ) , ( 0 , 0 , 1 ) } , B 2 = { ( 1 , 0 , 0 ) , ( 0 , 1 , 1 ) , ( 0 , 0 , 1 ) } B_(1)={(1,0,0),(1,1,0),(0,0,1)},B_(2)={(1,0,0),(0,1,1),(0,0,1)}B_1=\{(1,0,0),(1,1,0),(0,0,1)\}, B_2=\{(1,0,0),(0,1,1),(0,0,1)\}B1={(1,0,0),(1,1,0),(0,0,1)},B2={(1,0,0),(0,1,1),(0,0,1)}. If
Q ( X ) = x 1 2 4 x 1 x 2 + 2 x 2 x 3 + x 2 2 + x 3 2 Q ( X ) = x 1 2 4 x 1 x 2 + 2 x 2 x 3 + x 2 2 + x 3 2 Q(X)=x_(1)^(2)-4x_(1)x_(2)+2x_(2)x_(3)+x_(2)^(2)+x_(3)^(2)Q(X)=x_1^2-4 x_1 x_2+2 x_2 x_3+x_2^2+x_3^2Q(X)=x124x1x2+2x2x3+x22+x32
find the representation of Q Q QQQ in terms of ( y 1 , y 2 , y 3 ) y 1 , y 2 , y 3 (y_(1),y_(2),y_(3))\left(y_1, y_2, y_3\right)(y1,y2,y3).
b) Find the orthogonal canonical reduction of the quadratic form x 2 + y 2 + z 2 + 4 x y + 4 x z x 2 + y 2 + z 2 + 4 x y + 4 x z -x^(2)+y^(2)+z^(2)+4xy+4xz-x^2+y^2+z^2+4 x y+4 x zx2+y2+z2+4xy+4xz. Also, find its principal axes.
  1. Which of the following statements are true and which are false? Justify your answer with a short proof or a counterexample.
    i) If W 1 W 1 W_(1)W_1W1 and W 2 W 2 W_(2)W_2W2 are proper subspaces of a non-zero, finite dimensional, vector space V V VVV and dim ( W 1 ) > dim ( V ) 2 , dim ( W 2 ) > dim ( V ) 2 dim W 1 > dim ( V ) 2 , dim W 2 > dim ( V ) 2 dim(W_(1)) > (dim(V))/(2),dim(W_(2)) > (dim(V))/(2)\operatorname{dim}\left(W_1\right)>\frac{\operatorname{dim}(V)}{2}, \operatorname{dim}\left(W_2\right)>\frac{\operatorname{dim}(V)}{2}dim(W1)>dim(V)2,dim(W2)>dim(V)2, the W 1 W 2 { 0 } W 1 W 2 { 0 } W_(1)nnW_(2)!={0}W_1 \cap W_2 \neq\{0\}W1W2{0}.
ii) If V V VVV is a vector space and S = { v 1 , v 2 , , v n } V , n 3 S = v 1 , v 2 , , v n V , n 3 S={v_(1),v_(2),dots,v_(n)}sub V,n >= 3S=\left\{v_1, v_2, \ldots, v_n\right\} \subset V, n \geq 3S={v1,v2,,vn}V,n3, is such that v i v j v i v j v_(i)!=v_(j)v_i \neq v_jvivj if i j i j i!=ji \neq jij, then S S SSS is a linearly independent set.
iii) If T 1 , T 2 : V V T 1 , T 2 : V V T_(1),T_(2):V rarr VT_1, T_2: V \rightarrow VT1,T2:VV are linear operators on a finite dimensional vector space V V VVV and T 1 T 2 T 1 T 2 T_(1)@T_(2)T_1 \circ T_2T1T2 is invertible, T 2 T 1 T 2 T 1 T_(2)@T_(1)T_2 \circ T_1T2T1 is also invertible.
iv) If an n × n n × n n xx nn \times nn×n square matrix, n 2 n 2 n >= 2n \geq 2n2 is diagonalisable then it has the same minimal polynomial and characteristic polynomial.
v) If T 1 , T 2 : V V T 1 , T 2 : V V T_(1),T_(2):V rarr VT_1, T_2: V \rightarrow VT1,T2:VV are self adjoint operators on a finite dimensional inner product space V V VVV, then T 1 + T 2 T 1 + T 2 T_(1)+T_(2)T_1+T_2T1+T2 is also a self adjoint operator.
\(cos\:2\theta =2\:cos^2\theta -1\)

BMTE-141 Sample Solution 2024

bmte-141-sample-solution-00121e82-06b0-4090-b5ab-7e9d6642f92d

bmte-141-sample-solution-00121e82-06b0-4090-b5ab-7e9d6642f92d

  1. i) Find the angle between the vectors 2 i + 2 j + 2 k 2 i + 2 j + 2 k sqrt2i+2j+2k\sqrt{2} \mathbf{i}+2 \mathbf{j}+2 \mathbf{k}2i+2j+2k and i + 2 j + 2 k i + 2 j + 2 k i+sqrt2j+sqrt2k\mathbf{i}+\sqrt{2} \mathbf{j}+\sqrt{2} \mathbf{k}i+2j+2k.
Answer:
To find the angle between two vectors, we use the dot product formula. The dot product of two vectors A A A\mathbf{A}A and B B B\mathbf{B}B is given by A B = | A | | B | cos ( θ ) A B = | A | | B | cos ( θ ) A*B=|A||B|cos(theta)\mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos(\theta)AB=|A||B|cos(θ), where θ θ theta\thetaθ is the angle between the vectors, and | A | | A | |A||\mathbf{A}||A| and | B | | B | |B||\mathbf{B}||B| are the magnitudes of the vectors.
The vectors given are A = 2 i + 2 j + 2 k A = 2 i + 2 j + 2 k A=sqrt2i+2j+2k\mathbf{A} = \sqrt{2} \mathbf{i} + 2 \mathbf{j} + 2 \mathbf{k}A=2i+2j+2k and B = i + 2 j + 2 k B = i + 2 j + 2 k B=i+sqrt2j+sqrt2k\mathbf{B} = \mathbf{i} + \sqrt{2} \mathbf{j} + \sqrt{2} \mathbf{k}B=i+2j+2k.
  1. Calculate the Dot Product A B A B A*B\mathbf{A} \cdot \mathbf{B}AB:
    A B = ( 2 i + 2 j + 2 k ) ( i + 2 j + 2 k ) A B = ( 2 i + 2 j + 2 k ) ( i + 2 j + 2 k ) A*B=(sqrt2i+2j+2k)*(i+sqrt2j+sqrt2k)\mathbf{A} \cdot \mathbf{B} = (\sqrt{2}\mathbf{i} + 2\mathbf{j} + 2\mathbf{k}) \cdot (\mathbf{i} + \sqrt{2}\mathbf{j} + \sqrt{2}\mathbf{k})AB=(2i+2j+2k)(i+2j+2k)
    = 2 1 + 2 2 + 2 2 = 2 1 + 2 2 + 2 2 =sqrt2*1+2*sqrt2+2*sqrt2= \sqrt{2} \cdot 1 + 2 \cdot \sqrt{2} + 2 \cdot \sqrt{2}=21+22+22
  2. Calculate the Magnitudes of A A A\mathbf{A}A and B B B\mathbf{B}B:
    • Magnitude of A A A\mathbf{A}A, | A | | A | |A||\mathbf{A}||A|: | A | = ( 2 ) 2 + 2 2 + 2 2 | A | = ( 2 ) 2 + 2 2 + 2 2 |A|=sqrt((sqrt2)^(2)+2^(2)+2^(2))|\mathbf{A}| = \sqrt{(\sqrt{2})^2 + 2^2 + 2^2}|A|=(2)2+22+22
    • Magnitude of B B B\mathbf{B}B, | B | | B | |B||\mathbf{B}||B|: | B | = 1 2 + ( 2 ) 2 + ( 2 ) 2 | B | = 1 2 + ( 2 ) 2 + ( 2 ) 2 |B|=sqrt(1^(2)+(sqrt2)^(2)+(sqrt2)^(2))|\mathbf{B}| = \sqrt{1^2 + (\sqrt{2})^2 + (\sqrt{2})^2}