A-Level Mathematics Tuition Singapore/JC Maths/H2 Math Tuition and Tutor
Hi Crystal as per request
Q2i)
let vector a = (1,2,1) and vector b = (2,4,c)
a.b = lallblCosx
2+6+c = root(6).root(20+c^2)Cos0
……………
Solve for c
4c)
Vector eqn r = (0,0,1)lamda
Cartesian eqn : x=0, y=0, Z=lamda
Q6) P=(0,0,0) line r=(3,1,4)+t(-2,1,2)
Let N be the foot of perpendicular from point P to the line
Since N lies on the line
ON = (3-2t,1+t,4+2t)
PN = ON – OP
= (3-2t,1+t,4+2t)
Since PN is perpendicular to the line
(3-2t,1+t,4+2t).(-2,1,2)=0
Solve for t and sub back to ON
Cheers!
From A-Level Maths Tuition Singapore Tutor