Vectors Question, Please solve with working?

A Plane π has an equation r.(2i+3j)= -6.





a) find, in vector form, an equation for the line passing through the point P with position vector 2i+j+4k and normal to the plane π.





b) Find the position vector of the foot Q (could somebody please tell me what foot Q would be) of the perpendicular from P to the plane π.





c) Find the sine of the angle between OQ and the plane π. The plane π' has equation r.(i+j+k)=5. Find the position vector of the point A where the planes π, π' and the plane with the equation r.j=0 meet. Find also the position vector of the point B where the planes π, π' and the plane with the equation r.j=0 meets. Find, in vector form, an equation for the line passing through A and B.





(I know this is a lot, help with any one of these questions will be much appreciated!!) Thank You.

Vectors Question, Please solve with working?
a. the line has the same direction as the normal vector of the plane... thus the line has direction (2i + 3j).


Thus the line is r = (2i + 3j)t + (2i + j + 4k).








b. you need to intersect r from the two equations then...


{(2i + 3j)t + (2i + j + 4k)} · (2i+3j)= -6 ... we have to obtain such a t.





(4 + 9) t + 4 + 3 = -6


t = -1.


thus the point is (2i + 3j)(-1) + (2i + j + 4k) = -2j + 4k.











c. ok... so there are so many subquestions in this part...





you need to get the vector OQ. (O is the origin.)


note: the angle between the vectors A %26amp; B is θ such that


A·B = |A| |B| cosθ


the two vectors you have is the normal and the vector OQ. this gives you the complement of the angle you need. thus you get the cosine of those two vectors you have...








well... i expect that you need to transform these into cartesian (i dont know how you can solve these using the vector forms)


r·(2i+3j) = -6 becomes 2x + 3y = -6... and so on...


thus you have 3 linear equations with 3 variables... solve these and you now have your point.





(why do you have repeated questions to get points A %26amp; B?)











§
Reply:First of all right the equation of plane in nice way


r is a point xi + yj + zk





2x + 3y + 0z = -6


2x+3y = -6


normal to plane = 2i + 3j





a)





Eqn. of line r=a+tb


a = ( 2i + j +4k)


b is direction which will be normal of plane





r = ( 2i +j +4k) + t(2i + 3j)





b)


Find the point of intersection of line with the plane .





2x +3y = -6


r=( 2i +j +4k) + t(2i + 3j)


Q = (2+2t)i + (1 +3t)j + 4k





2( 2 +2t) + 3( 1 + 3t) = -6


4 +4t + 3 + 9t = -6


13t = -13


t = -1





Q (-2j + 4k)





c)


OQ has position vector = -2j + 4k





First of all we will find the angle between OQ and Normal .


CosΘ = (OQ).(n) / |OQ| |n|





cosΘ = (2x0) + (3x-2) + (4x0) / (√(2²+3²)(√(2²+4²) )


Θ = 68.15455236


Angle with plane = 90 - Θ = 21.854764





Sin(angle) = sin(21.854764) = 0.3721042038 .





Second part :-


First i will get equation of line which intersects with π and π'


then will get point A by getting the intersection of line with the third plane .





Equation of plane π 2x+3y = -6


Eqn of plane π' x+y+z =5


2x+3y = -6


x+y +z = 5 ( X-2)





2x+3y= -6


-2x -2y -2z = -10 ( add eqn)


y - 2z = -16


y = 1 , z=17/2 , x = -9/2


y = 2 , z=9 , x = -6





Eqn of line


r = (-6i + 2y + 9j) +t( 3/2i + -j + -1/2k )


Eqn of third plane .


y = 0


( 2 -t) = 0


t = 2





A ( -3i +8k )





For point B


Get the eqn of line which intersect with π' and third plane


then get point which intersect with π .


π' x+y+z = 5


third plane y = 0





x+y+z = 5


y=0 ( X -1)


x+y+z=5


-y = 0





x+z = 5


x = 1 , z = 4 , y =0


x = 2 , z =3 , y = 0





r = a +tb


r = ( i + 4k ) +t( i -k )


Now find the intersection with π


Eqn π 2x + 3y = -6





2( 1 +t) + 3(0) = -6


2 + 2t = -6


t = -4





B ( -3i + 8k )





So there is no line that is passing through A and B , Point A = Point B .





Cheers .
Reply:r.(2i+3j)= -6





what is the period between r and the parenthetic expression?

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