C4 vector q

The line L1 has vector equation

r = 8i + 12j + 14k + £(i + j – k) (£=greek symbol, where £ is a parametre)

Given that the point O is the origin, and that the point P lies on L1 such that OP is perpendicular to L1, find the coordinates of P.

Any help would be greatly appreciated...

Reply 1

KAISER_MOLE

7

OP has direction vector equivalent to position vector of P, so use the dot product, for lines to be perpendicular, OP . L1 = 0

P = ( xi + yj + zk )

Therefore

x + y - z = 0
=> x+y = z

Also

8 + £ = x
12 + £ = y
14 - £ = z

So 20 + 2£ = (x+y) = z = 14 - £

so 3£ = -6 <=> £ =-2

so
6 = x
10 = y
16 = z

Reply 2

evariste_3086

1

benfre

The line L1 has vector equation

r = 8i + 12j + 14k + £(i + j – k) (£=greek symbol, where £ is a parametre)

Given that the point O is the origin, and that the point P lies on L1 such that OP is perpendicular to L1, find the coordinates of P.

Any help would be greatly appreciated...

r=8i + 12j + 14k + s(i + j – k)
r=(8+s)i+(12+s)j+(14-s)k
since P lies on r it equals
(8+s)i+(12+s)j+(14-s)k for a value of s to be found
since its perpendicular to the line we have
[(8+s)i+(12+s)j+(14-s)k ].[(i + j – k)]=0
8+s+12+s-14+s=0
3s=-6
s=-2
giving
0P=(8-2)i+(12-2)j+(14-(-2))k
=6i+10j+16

Reply 3

benfre

OP

ahh thanks a lot, those really help.

Reply 4

ps1265A

14

Original post by evariste

r=8i + 12j + 14k + s(i + j – k)
r=(8+s)i+(12+s)j+(14-s)k
since P lies on r it equals
(8+s)i+(12+s)j+(14-s)k for a value of s to be found
since its perpendicular to the line we have
[(8+s)i+(12+s)j+(14-s)k ].[(i + j – k)]=0
8+s+12+s-14+s=0
3s=-6
s=-2
giving
0P=(8-2)i+(12-2)j+(14-(-2))k
=6i+10j+16

Why is it 14 MINUS "s"????

Reply 5

TenOfThem

18

Original post by ps1265A

Why is it 14 MINUS "s"????

Because s multiplied by -1 is -s

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C4 vector q

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