If $\overrightarrow{\mathrm{A}}=3 \hat{\imath}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}},…

If $\overrightarrow{\mathrm{A}}=3 \hat{\imath}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}, \overrightarrow{\mathrm{B}}=\hat{\imath}-3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{C}}=2 \hat{\imath}+\hat{\jmath}-4 \hat{\mathrm{k}}$ form a right angled triangle then out of the following which one is satisfied?
  1. $\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{C}}, \mathrm{B}^{2}=\mathrm{A}^{2}+\mathrm{C}^{2}$
  2. $\overrightarrow{\mathrm{A}}=\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}, \mathrm{B}^{2}=\mathrm{A}^{2}-\mathrm{C}^{2}$
  3. $\overrightarrow{\mathrm{C}}=\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}, \mathrm{C}^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}$
  4. $\overrightarrow{\mathrm{A}}=\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}, \mathrm{B}^{2}=\mathrm{A}^{2}+\mathrm{C}^{2}$

Solution

$\begin{array}{l} \vec{A}=3 \hat{i}-2 \hat{j}+\hat{k} \\ \vec{B}=\hat{\imath}-3 \hat{\jmath}+5 \hat{k} \\ \vec{C}=2 \hat{\imath}+\hat{\jmath}-4 \hat{k} \end{array}$ Here, $\begin{aligned} \vec{A} &=(\hat{\imath}-3 \hat{\jmath}+5 \hat{k})+(2 \hat{\imath}+\hat{\jmath}-4 \hat{k}) \\ &=(\vec{B}+\vec{c}) \\ &=3 \hat{\imath}-2 \hat{\jmath}+\hat{k} \\ \text { Satisfied } \vec{A}=\vec{B}+\vec{C} \end{aligned}$ and $\begin{array}{l} A^{2}=(\sqrt{14})^{2}=14 \\ B^{2}=35 \\ C^{2}=21 \\ \text { So, } B^{2}=A^{2}+C^{2} \end{array}$

Asked in: MHT CET 2020 (13 Oct Shift 1)

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