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
Let P = (1,3,1), Q = (1,1,-1), R = (-1,1,1) and S = (2,2,-1)
Show that $\displaystyle \|\vec {PQ} \cdot (\vec{PR} \times \vec{PS}) \| = 0$
$\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 $b) For any vector r prove that r = (r.i)i+(r.j)j+(r.k)k
think you can finish up?
what are you using * to mean?c) If a*(b*c)=(a*b)*c then prove that (c*a)*b=0
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
$\displaystyle (a \cdot c)b - (a \cdot b)c = (-c \cdot b)a - (-c \cdot a)b$
$\displaystyle \Rightarrow (a \cdot c)b - (a \cdot b)c - (-c \cdot b)a + (-c \cdot a)b = 0$
$\displaystyle \Rightarrow (a \cdot c)b + (-b \cdot a)c - (-b \cdot c)a - (a \cdot c)b = 0$
$\displaystyle \Rightarrow (-b \cdot a)c - (-b \cdot c)a = 0$
$\displaystyle \Rightarrow (c \times a) \times b = 0$
there is pretty much one step left, you can't finish it? try.and can you please show me the answer of part b (step by step)