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Algebra of Vectors — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsAlgebra of Vectors5 Marks
Question
Show that the points $\text{A}\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big),\ \text{B}\big(\hat{\text{i}}-3\hat{\text{j}}-5\hat{\text{k}}\big),$ $\text{C}\big(3\hat{\text{i}}-4\hat{\text{j}}-4\hat{\text{k}}\big)$ are the vertices of a right angled triangle.
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Answer

Given the points $\text{A}\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big),\ \text{B}\big(\hat{\text{i}}-3\hat{\text{j}}-5\hat{\text{k}}\big)$and $\text{C}\big(3\hat{\text{i}}-4\hat{\text{j}}-4\hat{\text{k}}\big)$. Then, $\overrightarrow{\text{AB}}=$ Position vector of B - Position vector of A $=\hat{\text{i}}-3\hat{\text{j}}-5\hat{\text{k}}-\big(2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$ $=\hat{\text{i}}-3\hat{\text{j}}-5\hat{\text{k}}-2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$ $=-\hat{\text{i}}-2\hat{\text{j}}-6\hat{\text{k}}$ $\overrightarrow{\text{BC}}=$ Position vector of C - Position vector of B $=3\hat{\text{i}}-4\hat{\text{j}}-4\hat{\text{k}}-\big(\hat{\text{i}}-3\hat{\text{j}}-5\hat{\text{k}}\big)$ $=3\hat{\text{i}}-4\hat{\text{j}}-4\hat{\text{k}}-\hat{\text{i}}+3\hat{\text{j}}+5\hat{\text{k}}$$=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
$\overrightarrow{\text{CA}}=$ Position vector of A - Position vector of C $=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}-\big(3\hat{\text{i}}-4\hat{\text{j}}-4\hat{\text{k}}\big)$ $=2\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}-3\hat{\text{i}}+4\hat{\text{j}}+4\hat{\text{k}}$ $=-\hat{\text{i}}+3\hat{\text{j}}+5\hat{\text{k}}$ Clearly, $\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}=\vec0$Now, $\Big|\overrightarrow{\text{AB}}\Big|=\sqrt{(-1)^2+(-2)^2+(-6)^2}$
$=\sqrt{1+4+36}$ $=\sqrt{41}$ $\Big|\overrightarrow{\text{BC}}\Big|=\sqrt{(2)^2+(-1)^2+(1)^2}$ $=\sqrt{4+1+1}$ $=\sqrt{6}$ $\Big|\overrightarrow{\text{CA}}\Big|=\sqrt{(-1)^2+(3)^2+(5)^2}$ $=\sqrt{1+9+25}$ $=\sqrt{35}$ Clearly, $\Big|\overrightarrow{\text{AB}}\Big|^2=\Big|\overrightarrow{\text{BC}}\Big|^2+\Big|\overrightarrow{\text{CA}}\Big|^2$$\Rightarrow\text{AB}^2=\text{BC}^2+\text{CA}^2$
So, A, B, C forms a right angled triangle.

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