# Solve my vector Problems

• Jan 24th 2010, 11:05 AM
Salman91
Solve my vector Problems
a) Show that the points (1,3,1) ,(1,1,-1),(-1,1,1) (2,2,-1) are lying on the same plane

b) For any vector r prove that r = (r.i)i+(r.j)j+(r.k)k

c) If a*(b*c)=(a*b)*c then prove that (c*a)*b=0
• Jan 24th 2010, 11:21 AM
Jhevon
Quote:

Originally Posted by Salman91
a) Show that the points (1,3,1) ,(1,1,-1),(-1,1,1) (2,2,-1) are lying on the same plane

Let P = (1,3,1), Q = (1,1,-1), R = (-1,1,1) and S = (2,2,-1)

Show that
$\displaystyle \|\vec {PQ} \cdot (\vec{PR} \times \vec{PS}) \| = 0$

Quote:

b) For any vector r prove that r = (r.i)i+(r.j)j+(r.k)k

$\displaystyle \vec r = \left< r_i, r_j, r_k \right> = \left< r_i, 0, 0 \right> + \left< 0, r_j, 0 \right> + \left< 0, 0, r_k \right> = \cdots$

think you can finish up?

Quote:

c) If a*(b*c)=(a*b)*c then prove that (c*a)*b=0

what are you using * to mean?
• Jan 24th 2010, 11:32 AM
Salman91
* = multiply ( x )
• Jan 24th 2010, 11:46 AM
Jhevon
Quote:

Originally Posted by Salman91
* = multiply ( x )

i assume that a, b, c are vectors here. "multiply" makes no sense. do you mean "cross product"? and they are all cross products? you have to distinguish between the dot and cross product!
• Jan 24th 2010, 11:48 AM
Salman91
yes sir it is a cross-product

and can you please show me the answer of part b (step by step)
• Jan 24th 2010, 12:29 PM
Jhevon
Quote:

Originally Posted by Salman91
yes sir it is a cross-product

i will just solve this one...

recall these properties from your text: $\displaystyle a \times ( b \times c) = (a \cdot c)b - (a \cdot b)c$, $\displaystyle a \times b = -b \times a$ and $\displaystyle a \cdot b = b \cdot a$ for any vectors $\displaystyle a,b,c$ (of course, for cross-products we need 3-d vectors)

Note that $\displaystyle a \times (b \times c) = (a \cdot c)b - (a \cdot b)c$

$\displaystyle (a \times b) \times c = -c \times (a \times b) = (-c \cdot b)a - (-c \cdot a)b$ and

$\displaystyle (c \times a) \times b = -b \times (c \times a) = (-b \cdot a)c - (-b \cdot c)a$

So, assume $\displaystyle a \times (b \times c) = (a \times b) \times c$, then

$\displaystyle (a \cdot c)b - (a \cdot b)c = (-c \cdot b)a - (-c \cdot a)b$

$\displaystyle \Rightarrow (a \cdot c)b - (a \cdot b)c - (-c \cdot b)a + (-c \cdot a)b = 0$

$\displaystyle \Rightarrow (a \cdot c)b + (-b \cdot a)c - (-b \cdot c)a - (a \cdot c)b = 0$

$\displaystyle \Rightarrow (-b \cdot a)c - (-b \cdot c)a = 0$

$\displaystyle \Rightarrow (c \times a) \times b = 0$
Quote:

and can you please show me the answer of part b (step by step)
there is pretty much one step left, you can't finish it? try.