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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?
Options:
triangle then out of the following which one is satisfied?
Solution:
2993 Upvotes
Verified Answer
The correct answer is:
$\overrightarrow{\mathrm{A}}=\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}, \mathrm{B}^{2}=\mathrm{A}^{2}+\mathrm{C}^{2}$
$$
\begin{array}{l}
\vec{A}=3 \hat{i}-2 \hat{j}+\hat{k} \\
\vec{B}=\hat{\imath}-3 \hat{\jmath}+5 \hat{k} \\
\vec{C}=2 \hat{\imath}+\hat{\jmath}-4 \hat{k}
\end{array}
$$
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
$$
\begin{array}{l}
A^{2}=(\sqrt{14})^{2}=14 \\
B^{2}=35 \\
C^{2}=21 \\
\text { So, } B^{2}=A^{2}+C^{2}
\end{array}
$$
\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}
$$
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