Question

Prove that

(i) the three points A(1, 4, 2), B(3, 2, 4) and C(5, 0, 6) are collinear,

(ii) the four points P(2, ;1, 1), Q(1, 3, ;2), R((2, 1, ;3) and S(3, 2, 0) are coplanar

Answer 1

$\left(i\right)\:If\:three\:direction\:ratios\:of\:AB\:and\:BC\:are\:proportional,$

$then\:the\:points\:A,\:B\:and\:C\:are\:collinear.$

$Direction\:ratios\:are:\:$

$\vec{AB}\:=B-A$

$=\left(3-1\right),\left(2-4\right),\left(4-2\right)$

$=2,-2,2$

$so\:a_1=2\:\:b_1=-2\:c_1=2$

$\vec{BC}=C-B$

$=\left(5-3\right),\left(0-2\right),\left(6-4\right)$

$=2,-2,2$

$so\:a_2=2,\:b_2=-2\:c_2=2$

$\frac{a_2}{a_1}=\frac{2}{2}=1\:$

$\frac{b_2}{b_1}=\frac{-2}{-2}=1$

$\frac{c_2}{c_1}=\frac{2}{2}=1$

$since\:\frac{a_2}{a_1}=\frac{b_2}{b_1}=\frac{c_2}{c_1}=1$

$Therefore\:A,B\:and\:C\:are\:collinear.$

$\left(ii\right)Given\:that\:points\:A\left(2,2,1\right),\:Q\left(1,3,2\right),\:R\left(2,1,3\right)and\:S\left(3,2,0\right)$

$\vec{AQ}=Q-A=\left(-i+2j+k\right)$

$\vec{AR}=R-A=\left(2k\right)$

$\vec{AS}=S-A=\left(i+j-k\right)$

$Therefore\:\left|\vec{AQ}\:\vec{AR}\:\vec{AS}\right|=\begin{pmatrix}-1&2&1\\ 0&0&2\\ 1&1&-1\end{pmatrix}$

$=\:-1\left(0-2\right)+2\left(2-0\right)+1\left(0-0\right)$

$=2+4+0=6\:$

$\:Hence,\:the\:points\:are\:not\:coplanar.$