Question about Cross product in “Vectors in 2-space & 3-space?

Plz help me out in the following questions %26amp; Plz plz plz at least write step by step formulas of all the following questions.


Q1-Find all unit vectors parallel to the yz-plane that are perpendicular to the vector (3,-1,2). Answer:(0,2/√5,1/√5).


Q2-Find all unit vectors in the plane determined by u=(3,0,1) and v = (1,-1,1) that are perpendicular to the vector w= (1,2,0). Ans. (6/√61,-3/√61,4/√61)


Q3-Use the cross product to find the sine of the angle between the vectors u =(2,3,-6) %26amp; v= (2,3,6). Ans.12√13/49


Q4-Find a vector n perpendicular to the plane determined by the points a(0,-2,1) b(1,-1,-2) %26amp; c(-1,1,0). Ans. Not written in book.


Q5-Distance between the point P and the line through A %26amp; B: P(-3,1,2),A(1,1,0),B(0,2,-1). Ans.2√141/√29


Plz plz plz at least write step by step formulas of all the above questions.


And also give me some good websites related to Linear Algebra %26amp; if u r kind enough %26amp; wants to give me support in it then give ur E-mail address.Thank u in advance.

Question about Cross product in “Vectors in 2-space %26amp; 3-space?
Q1. for the vector to be parallel to the yz plane (the plane x=0), x=0.


so the vector is (0,y,z)


scalar product with (3,-1,2) is to be 0 because perpendicular.


so -y+2z = 0.


so vector is (0,2,1) except that is not a unit vector.


unit vector is (0,2/sqrt(5),1/sqrt(5))


woohoo





Q2.


vector is a(3,0,1)+b(1,-1,1),


that is, (3a+b,-b,a+b)


scalar product with w is zero so


(3a+b-2b) = 0 so 3a = b





so vector is (3a+3a,-3a,4a),


that is, some multiple of (6,-3,4).


unit vector is


(6/sqrt(61),


-3/sqrt(61),


4/sqrt(61))


woohoo





Q3.


a×b =


[a2b3 - a3b2,


a3b1 - a1b3,


a1b2 - a2b1]


uxv=


[3*6--6*3,-6*2-2*6,2*3-3*2]


=[36,-24,0]


u×v = |u||v| sin(θ) n


uxv =


12sqrt(13)*


[3/sqrt(13),


-2/sqrt(13),0]


|u| = 7 (sqrt(2^2+3^2+6^2))


|v| = 7


sin(theta) = (12sqrt(13))/49


woohoo





Q4. a-b = (-1,-1,3)


a-c = (1,-3,1)





plane = p(-1,-1,3) + q(1,-3,1)


= (-p+q,-p-3q,3p+q)





let vector n be (x,y,z)





scalar product of (-p+q,-p-3q,3p+q)


and (x,y,z) is to be 0.





(-p+q)x +(-p-3q)y +(3p+q)z = 0


p(-x-y+3z) +q(-x-3y+z) = 0


this true for all values of p,q


so x+y=3z


also x+3y = z


so -2y=2z


y=-z


x=z-3y=4z


so vector is (4z,-z,z)


so (4,-1,1) is a vector n.








Q5.


Point on line AB has coordinates


L(1,1,0) +(1-L)(0,2,-1)


So point on line AB has coordinates


(L,2-L,L-1), Call this point Q.


line PQ is perpendicular to AB


vector for line PQ is (L+3,1-L,L-3)


vector for line AB is (1,-1,1)


so scalar product is 0


so L+3 + L-1 + L-3 = 0


so L = 1/3


so Q is (1/3,5/3,-2/3)


so PQ distance squared is


(10/3)^2 + (2/3)^2 +(8/3)^2=


168/9


so PQ is sqrt(168)/3 = sqrt(56/3)





to check,


line PQ is (L+3,1-L,L-3)


distance squared =


(L+3)^2 + (1-L)^2 + (L-3)^2


=


L^2+6L+9+


L^2-2L+1+


L^2-6L+9





=3L^2-2L+19





d(distance squared)/dL =


6L-2


which is 0 if L=1/3


so L=1/3 should yield the minimum distance


squared


3(1/3)^2 - 2(1/3) + 19


= 56/3


so distance should be sqrt((56)/3)





extra checks:


A (1,1,0)


B (0,2,-1)


P(-3,1,2)


Q(1/3,5/3,-2/3) (Q is point on AB closest to P)





can show that AQ + QB = AB as it should be if


AQB is a straight line


AQ^2 = 12/9


QB^2 = 1/3


AB^2 = 3


sqrt(12/9) + sqrt(1/3) = sqrt(3) because


2/sqrt(3) + 1/sqrt(3) = sqrt(3)





can also show that AP^2 = AQ^2 + QP^2


because


AQ^2 = (2/3)^2 + (2/3)^2 +(2/3)^2 = 12/9 and


AP^2 = 4^2 + 2^2 = 20 and


QP^2 = (3+(1/3))^2 + (2/3)^2 + (2+(2/3))^2 =


(100/9) +(4/9) + (64/9) = 168/9 = 56/3





woohoo





wikipedia is good


write


mmullins207@optusnet.com.au


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