Linear Algebra's Questions.?

Question

Plz guide me in the following questions, step by step & plz write all important formulas, since my Linear Algebra is not good:
Q1-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)
Q2-Use the cross product to find the sine of the angle between the vectors u =(2,3,-6) & v= (2,3,6). Ans.12√13/49
Q5-Distance between the point P and the line through A & B: P(-3,1,2),A(1,1,0),B(0,2,-1). Ans.2√141/√29
Thank u in advance, and if u could provide me ur E-mail address to have future contanct in order to ask questions of MATHEMATICS. Thank u again.

Answer 1

Q1:

The vectors you are looking for are in the plane (u,v). They are then obtained by xu+yv, and you are looking for x and y. You have the following condition:
x(u.w)+y(v.w)=0 so 3x-y=0 then you are looking for the vector
x(u+3v) and x is given by the condition that it is a unit vector..

so x²(6²+3²+4²)=61x²=1.. then x=+1/sqrt(61) or x=-1/sqrt(61)

And you obtain the results (there is 2 vectors..)

Q2:

You better use directly the vector product, it is a lot faster
the vector product of u and v is (-36,24,0) with length 12sqrt(13)
It is equal to the product of the sine by the product of the length of the vector u and v which is 49..

Q5 :

Take M as a running point on A,B.. given by OM=OA+xAB
Write the condition that the cross product between PM and AB is 0 this gives x and then PM
PM=(4-x,x,-x-2) gives x=2/3
Then the result PM=13/3 different from what you give..

If I am not wrong..

But I am wrong indeed.. Captain Mephisto got it sqrt(168)/3

No e-mail sorry.. just passing by..

Answer 2

Q1. Define V as a general vector in the plane.
V = xA + yB where:
A = (3,0,1) and B = (1,-1,1)
V = (3x+y,-y,x+y)

N = (1,2,0) We want those V that are perpendicular to N in which case the dot product of the vectors will be 0.

Dot product is N.V = 0
(1,2,0) .(3x+y,-y,x+y) = 3x + y -2y = 3x - y
y = 3x

The vector V is then (6x,-3x,4x) = (6,-3,4)x
To get a unit vector divide by the magnitude
Magnitude = SQRT(36 + 9 + 16)x = SQRT(61)x
So V = (6,-3,4)/SQRT(61)

Q2. |UxV| = |U||V|sin(Angle)
UxV = (36,-24,0) and |UxV| = 12SQRT(13)
|U| = |V| = SQRT(4 + 9 + 36) =SQRT(49) and |U||V| = 49
So sin(Angle) = 12SQRT(13)/49

Q5. The points A, B and P form a triangle and the distance of P from the line AB is then the height of the triangle having AB as it's base. If we form the vectors AB and AP the sin of the angle between them can be found by the cross product and the distance from the line is then this sin times the triangle side AP.

Vector AB = (-1,1,-1) and |AB| = SQRT(3)
Vector AP = (-4,0,2) = SQRT(20) = 2SQRT(5)

Take the cross product to get the sin of the angle between.
sin(Angle) = |ABxAP| = |AB||AP|sin(Angle)
sin(Angle) = |ABxAP| / |AB||AP|
|ABxAP| = |(-2,-6,-4)| = SQRT(56)

The distance D from P to the line is then sin(Angle) = D/|AP|

D = |ABxAP| /|AB| = SQRT(56) /SQRT(3) = 2SQRT(14)/SQRT(3) = 4.32

I think your answer is incorrect or perhaps the points are incorrect or maybe my answer is incorrect. Actually they are very close since mine is 2SQRT(140)/SQRT(30) and yours 2SQRT(141)/SQRT(29). Also if I multiply top and bottom by SQRT(3) I get SQRT(168)/3 which is very close to SQRT(169)/3 or 13/3 which is the other answer that someone else came up with. So all these are very close to each other.

Answer 3

Q1;♠ the vector in question is
s=x*u +y*v = (3x,0,x) +(y,-y,y)=
= (3x+y, -y, x+y);
♣ meanwhile dot product (s·w)=0, as s ┴ w; that is
(3x+y)*1 –y*2 +0= 0, hence y=3x; and
s=(3x+3x, -3x, x+3x) =(6, -3, 4)*x;
♣ unit vector in question e=s/|s| =
=(6, -3, 4) / √(6^2 +3^2 +4^2) = (6, -3, 4)/√61;
Q2; Q5 after response;