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Algebra of Vectors — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Vectors3 Marks
Question
Show that the four points having position vectors $6\hat{\text{i}}-7\hat{\text{j}},\ 16\hat{\text{i}}-19\hat{\text{j}}-4\hat{\text{k}},\ 3\hat{\text{j}}-6\hat{\text{k}},\ 2\hat{\text{i}}-5\hat{\text{j}}+10\hat{\text{k}}$ are coplanar.
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Answer

Let the given four points be P, Q, R and S respectively. Three points are coplanar if the vectors $\overrightarrow{\text{PQ}},\ \overrightarrow{\text{PR}}\text{ and }\overrightarrow{\text{PS}}$ are coplanar. These vectors are coplanar if one of them can be expressed as a linear combination of the other two. So, let$\overrightarrow{\text{PQ}}=\text{x}\overrightarrow{\text{PR}}+\text{y}\overrightarrow{\text{PS}}$
$10\hat{\text{i}}-12\hat{\text{j}}-4\hat{\text{k}}\\=\text{x}\big(-6\hat{\text{i}}+10\hat{\text{j}}-6\hat{\text{k}}\big)+\text{y}\big(-4\hat{\text{i}}+2\hat{\text{j}}+10\hat{\text{k}}\big)$
$\Rightarrow10\hat{\text{i}}-12\hat{\text{j}}-4\hat{\text{k}}\\=\hat{\text{i}}\big(-6{\text{x}}-4\text{y}\big)+\hat{\text{j}}\big(10{\text{x}}+2{\text{y}}\big)+\hat{\text{k}}\big(-6\text{x}+10\text{y}\big)$
$\Rightarrow-6\text{x}-4\text{y}=10,\ 10\text{x}+2\text{y}=-12$ and $-6\text{x}+10\text{y}=-4$ [Equating coefficients of $\hat{\text{i}},\ \hat{\text{j}},\ \hat{\text{k}}$ on both sides]
Solving the first of these three equations, we get x = -1 and y = -1. These values also satisfy the thied equation. Hence, the given four points are coplanar.

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