vectors question #4

• Oct 1st 2007, 04:51 AM
afeasfaerw23231233
vectors question #4
• Oct 1st 2007, 05:58 AM
topsquark
Quote:

Originally Posted by afeasfaerw23231233

-Dan
• Oct 1st 2007, 06:41 AM
afeasfaerw23231233
i cannot find the value of 'lambda'
it seems that you cannot see my image, would you please click on the following link to get my question?

my question is here
• Oct 1st 2007, 06:49 AM
afeasfaerw23231233
• Oct 1st 2007, 08:25 AM
topsquark
Quote:

Originally Posted by topsquark

-Dan

Sorry. The first attachment didn't load when I first visited this thread.

-Dan
• Oct 1st 2007, 08:39 AM
topsquark
Quote:

Originally Posted by afeasfaerw23231233

a)
OC = OA + AC = a + 2b

BC = BO + OC = (-b) + (a + 2b) = a + b

OQ = OB + BQ = OB + (1/3)BC = b + (1/3)(a + b) = (1/3)a + (4/3)b

b)
i) OR = OP + PR = (1/2)OA + hPQ

Now, PQ = PO + OQ = -(1/2)OA + OQ = -(1/2)a + (1/3)a + (4/3)b = -(1/6)a + (4/3)b

So
OR = (1/2)OA + hPQ = (1/2)a + h[-(1/6)a + (4/3)b]
(I'll let you simplify that.)

ii)
OR = k OC
(1/2)a + h[-(1/6)a + (4/3)b] = k[a + 2b]

From here equate the coefficients of a on each side of the equation and the coefficients of b on each side of the equation. Two equations in two unknowns, h and k, so you can solve this.

I have to go. See if you can carry on from here.

-Dan