Question

Let u = (3, 2, 3, (3, 2, 3, -3) and ✓ = (2, 1,4, – 3). A vector perpendicular to ū and ū is A Hint: If two vectors are perpendicular, their dot product will be 0.

Answer 1

I am telling you two way of solving this method one by matrix and other by using pseufo code
- If u (21, 41, 21, tı), v = (x2, Y2, 22, t2) and (X3, Y3, 23, t3), then the vector: W = ei e2 e3 e4 21 Yi 21 ti X = X2 Y2 22 t2 X3 Y3 23 t3 is orthogonal to u, V and w. Here, {e; }4–1 is the canonical basis for R4. If you want a vector orthogonal to the first d three, with length d, consider -X. Given n 1 vectors in R", the above gives you one last vector orthogonal to all the given vectors.

7.3 4D Pseudocode // Compute the hypercross product using the formal determinant: hcross = det{{eo, el,e2, e3},{x0, x1, x2, x3},{yo, yl, y2, y3},{zo, zl , z2, z3}} where el = (1,0,0,0). el = (0,1,0,0), e2 = (0,0,1,0). e3 = (0,0,0,1). v0 = (x0, x1, x2, x3), v1 = (yo, yl, y2, y3), and v2 = (zo, zl , 22, 23). 10 template <typename Real Vector4<Real HyperCross (Vector 4<Real> const& vo, Vector 4<Real> const& vi, Vector 4<Real> const& v2) { Real xy01 = vO[0]* v1 (1) Real xy02 = vO (O)* v1 [2] Real xy03 = vo[0] + v1 [3] Real xy12 = vO[1]* v1 (2) Real xy13 = vo[1]* v1 [3] Real xy23 = V0 [2] * v1 [3] return Vector 4 <Real> VO[1]* v1 [0]: v0[2] + v1[0] VO [3]* v1 (0 vo [2]* v1 (1 VO[3]* v1 v0[3] + v1 +xy 23*v2 (1 xy13*v2 [2] + xy 12*v2 [3] – xy23*v2 10j + xy03*v212 - xy02*v2 13j. +xy13*v210 xy03*v2 [1] + xy01*v2 [3] -xy12*v2 10j + xy02*v211j - xy01*v212) ); Compute a right-handed orthonormal basis for the orthogonal complement of the input vectors. The function returns the smallest length of the un normalized vectors computed during the process. If this value is nearly zero, it is possible that the inputs are linearly dependent (within numerical round-off errors). On input, numInputs must be 1, 2, or 3 and v[0] through vinuminputs - 1) must be initialized On output, the vectors Il v[0] through v[3] form an orthonormal set. template <typename Real > Real Compute Orthogonal Complement (int numinputs, Vector 4<Real> v[4]) if (num Inputs = 1) { int maxlndex = 0; Real maxAbsValue = fabs (v[0][0]); for (int i = 1; i < 4; Hi) { Real absValue = fabs (v[0][i]); if (absValue > maxAbsValue) { maxIndex = i; maxAbsValue = absValue; } } if (maxIndex < 2) { v[1][0] = -v[0][1]: v[1][1] = +v[0][0]; v[1][2] = (Real)0; v[1]131 = (Real)0; else { v[1][0 v[1][1 = (Real)0; (Real) 0; +v[O] [3 -V[0][2] v[1][3] } numinputs = 2; } if (num Inputs 2) Real det [6] = { v[0][0]* V[1][1] vojoj*v1][2] vo][0]*V[1] [3] v[0][1]*v [1][2] v[0][1]*v [1][3 v [2]*V 1][3 LILIT v [1][0]*V[] vijfo*vo12 v [1][0 vo v [1][1 v[1][1 v [1][2 };

int maxindex = 0; Real maxAbsValue = fabs (det [0]); for (int i = 1; i < 6; Hi) Real absValue = fabs ( det[i]); if (absValue > maxAbsValue) maxIndex = i; maxAbs Value = abs Value; } } if (maxIndex = 0) { v[2] = Vector4<Real>(-det (4]. +det(2]. (Real), -det [0]); else if (maxIndex <= 2) } { v [2] = Vector 4 <Real >(+det(5](Real)o, -det [2]+det [1]); else { } v[2] = Vector4<Real >((Real)o, -det [5]. +det [4]. -det [3]); numinputs = 3; } if (numinputs 3) v[3] = HyperCross (v[0], v[1], [2]); return Orthonormalize <4, Real >(4, v); } return (Real)0; Create a rotation matrix. The input vectors are unit-length and perpendicular, angle01 is in radians and is rotation angle for plane of input vectors, angle23 // is in radians and is rotation angle for plane of orthogonal complement of input // vectors Matrix 4x4<Real Create Rotation (Vector 4<Real> uo, Vector4<Real> ul, float angle0i, float angle23 ) { v{i} Vector 4<Real> v[4]; = 0; = ul: Compute Orthogonal Complement (2, v); float cs0 = cos( angle01), sn0 = sin(angle01); float csl = cos angle23 ), snl = sin(angle23 ); Matrix4x4<Real> U(v10] [1] [2] [3]); Matrix 4x4 <Real> M ( cs0 -sno, (Real), (Real), sno, CSO Realjo, Realjo, (Real), (Real)0, csl , -snl, (Real), (Real)o, snl, cs1 ); Matrix 4x4 <Real> R= U*M* Transpose (U); return R: } 12