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