Chapter: Vectors· Mathematics & Statistics Part 1
Yes, the four points A, B, C and D are coplanar, because the scalar triple product of \(\overrightarrow{AB}\), \(\overrightarrow{AC}\) and \(\overrightarrow{AD}\) is zero.
Rule: four points A, B, C, D are coplanar if and only if \([\overrightarrow{AB}\;\overrightarrow{AC}\;\overrightarrow{AD}] = 0\).
Step 1: Find the three vectors from A.
\(\overrightarrow{AB} = (2-1)\hat i + (-3-2)\hat j + (4-1)\hat k = \hat i - 5\hat j + 3\hat k\)
\(\overrightarrow{AC} = (3-1)\hat i + (4-2)\hat j + (-5-1)\hat k = 2\hat i + 2\hat j - 6\hat k\)
\(\overrightarrow{AD} = (2-1)\hat i + (3-2)\hat j + (-2-1)\hat k = \hat i + \hat j - 3\hat k\)
Step 2: Find the scalar triple product.
\[ [\overrightarrow{AB}\;\overrightarrow{AC}\;\overrightarrow{AD}] = \begin{vmatrix} 1 & -5 & 3 \\ 2 & 2 & -6 \\ 1 & 1 & -3 \end{vmatrix} \]
\[ = 1\big(2(-3) - (-6)(1)\big) - (-5)\big(2(-3) - (-6)(1)\big) + 3\big(2(1) - 2(1)\big) \]
\[ = 1(-6 + 6) + 5(-6 + 6) + 3(2 - 2) = 0 + 0 + 0 = 0 \]
Step 3: Conclusion. Since \([\overrightarrow{AB}\;\overrightarrow{AC}\;\overrightarrow{AD}] = 0\), the vectors \(\overrightarrow{AB}\), \(\overrightarrow{AC}\), \(\overrightarrow{AD}\) lie in one plane, so A, B, C and D are coplanar.
Quick check: notice that \(\overrightarrow{AC} = 2\,\overrightarrow{AD}\). So A, C and D lie on one straight line, and a line and a point (B) always lie in one plane. That is why the determinant came out zero (row 2 is twice row 3).