MATHEMATICS 9709 O/N 2010 33 Qn 6 Help!!

[gentleman]

Can anyone help me solving this qn... please?

 

[intel1993]

q6)
fIRST FIND THE EQUATION OF THE LINE.....AS TWO POINTS ARE GIVEN SO U CAN FIND THE DIRECTION VECTOR.....

Suppose the points from which the line 'l' is passing as A and B .....so now

AB= (5 8 1) -(-5 3 6).... solve it u ll get the direction vector of the line......

then take any of the point A or B as a fixed point and made the eqaution (r= c+ lamda d) where c= fixed point and d = direction vector and lamda as perameter...
the eq will be this r= (-5 3 6) +$(10 5 -5) $ DENOTING LAMBDA............ I HAVE TAKEN pOINT a AS FIXED POINT

now made the general eq of line...... ( -5+10$ 3+5$ 6-5$)

now putr the x y z of general eq in the eq of plane.......
u will find lamda....

now put the lamda in general eq
u ll get the points of intersection,...

b) cos(thita)= Dot product of (D.V * N.V)/ magnitude of (D.V * N.V) D.V= Direction vector (10 5 -5) N.V= Normal vector(2 -1 4)
u will get the anle using this formula

 

[gentleman]

please, can u calculate the value of the angle and tell me cuz im not getting the correct value from the marking scheme??

 

[intel1993]

k.......my mistakle........instead of cos(thita) , sin thita will be thre......
calculate now u ll get the ans........