Answer to Question #212257 in Analytic Geometry for Moe

Question

Let the plane P pass through the 3 points R = (1, 1, 6), S= (2,5,4) and T = (1,2,3).

Fill in the following:

1. RS = (__,__,__)
2. RT = (__,__,__)
3. RS x RT = (__,__,__)

Answer 1

R = (1, 1, 6)

S = (2, 5, 4)

T = (1, 2, 3)

To check if the points are collinear

$\begin{vmatrix} 1 & 1& 6\\ 2& 5& 4\\1&2&3 \end{vmatrix}=-1$

$\begin{vmatrix} (x- 1 )&( y-1)& (z-6)\\ (1-2)& (1-5)& (6-4)\\(1-1)&(1-2)&(6-3) \end{vmatrix}=0$

$\begin{vmatrix} (x- 1 )&( y-1)& (z-6)\\ -1& -4& 2\\0&-1&3 \end{vmatrix}= 0$

On solving,

$-14(x-1)-((-3)(y-1))+1(z-6)= 0$

$-14x+14+3y-3+z-6 =0$

$14x-3y-z = 5$ is the required equation.

R = 1i + 1j + 6k

S = 2i + 5j + 4k

T = 1i + 2i + 3k

1. $\overrightarrow{RS}= (x_2 – x_1) \hat{i} + (y_2 – y_1) \hat{j} + (z_2 – z_1) \hat{k}$

$\overrightarrow{RS}= (2 – 1) \hat{i} + (5–1) \hat{j} + (4– 6) \hat{k} = \hat{i}+ 4\hat{j}-2 \hat{k}= (1,4,-2)$

2.

$\overrightarrow{RT}= (x_3– x_1) \hat{i} + (y_3– y_1) \hat{j} + (z_3 – z_1) \hat{k}$

$\overrightarrow{RT}= (1– 1) \hat{i} + (2–1) \hat{j} + (3– 6) \hat{k} = 0\hat{i}+ \hat{j}-3\hat{k}= (0,1,-3)$

3.

$\overrightarrow{RS} ×\overrightarrow{ RT }= (\hat{i}+ 4\hat{j}-2 \hat{k})×(0\hat{i}+ \hat{j}-3\hat{k}) =$

$\begin{vmatrix} i & j& k\\ 1& 4& -2\\0&1&-3 \end{vmatrix}=$

$i(-12+2) -j(-3-0)+k(1-0)=-10\hat{i}+ 3\hat{j}+1\hat{k}$

$\therefore\overrightarrow{RS} ×\overrightarrow{ RT }= (-10 ,3,1)$