Maharashtra BoardEnglish MediumSTD 12 ScienceMathsREPEATER 20243 Marks
Question
Are the four points $A(1,-1,1), B(-1,1,1), C(1,1,1$,$) and D(2,-3,4)$ co-planar? Justify your answer.
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Answer
The position vectors $\bar{a}, \bar{b}, \bar{c}, \bar{d}$ of the points $A, B, C, D$ are $\bar{a}=\hat{\imath}-\hat{\jmath}+\hat{k}, \quad \bar{b}=-\hat{\imath}+\hat{\jmath}+\hat{k}, \quad \bar{c}=\hat{\imath}+\hat{\jmath}+\hat{k}, \quad \bar{d}=2 \hat{\imath}-3 \hat{\jmath}+4 \hat{k}$ $\therefore \quad \overline{A B}=\bar{b}-\bar{a}=(-\hat{\imath}+\hat{\jmath}+\hat{k})-(\hat{\imath}-\hat{\jmath}+\hat{k})=-2 \hat{\imath}+2 \hat{\jmath}$ $\therefore \quad \overline{A C}=\bar{c}-\bar{a}=(\hat{\imath}+\hat{\jmath}+\hat{k})-(\hat{\imath}-\hat{\jmath}+\hat{k})=2 \hat{\jmath}$ $\therefore \quad \overline{A D}=\bar{b}-\bar{a}=(2 \hat{\imath}-3 \hat{\jmath}+4 \hat{k})-(\hat{\imath}-\hat{\jmath}+\hat{k})=\hat{\imath}-2 \hat{\jmath}+3 k$ If $A, B, C, D$ are coplanar, then there exist non-zero scalars $x, y$ such that $\overline{A B}=x \cdot \overline{A C}+y \cdot \overline{A D}$ $\therefore \quad-2 \hat{\imath}+2 \hat{\jmath}=x \cdot(2 \hat{\jmath})+y \cdot(\hat{\imath}-2 \hat{\jmath}+3 k)$ $\therefore \quad-2 \hat{\imath}+2 \hat{\jmath}=2 x \hat{\jmath}+y \hat{\imath}-2 y \hat{\jmath}+3 y k$ $\therefore \quad-2 \hat{\imath}+2 \hat{\jmath}=y \hat{\imath}+(2 x-2 y) \hat{\jmath}+3 y k$ By equality of vectors. $\begin{aligned}y & =-2\ldots\ldots (1) \\2 x-2 y & =2\ldots\ldots (2) \\3 y & =0\ldots\ldots (3)\end{aligned}$ From (1), $\quad y=-2$ From (3), $\quad y=0$ This is not possible. Hence, the points A, B, C, D are not coplanar.
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