MHT CET · Physics · Mathematics in Physics
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?
- A \(\overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{C}}, \mathrm{B}^{2}=\mathrm{A}^{2}+\mathrm{C}^{2}\)
- B \(\overrightarrow{\mathrm{A}}=\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}, \mathrm{B}^{2}=\mathrm{A}^{2}-\mathrm{C}^{2}\)
- C \(\overrightarrow{\mathrm{C}}=\overrightarrow{\mathrm{A}}+\overrightarrow{\mathrm{B}}, \mathrm{C}^{2}=\mathrm{A}^{2}+\mathrm{B}^{2}\)
- D \(\overrightarrow{\mathrm{A}}=\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}, \mathrm{B}^{2}=\mathrm{A}^{2}+\mathrm{C}^{2}\)
Answer & Solution
Correct Answer
(D) \(\overrightarrow{\mathrm{A}}=\overrightarrow{\mathrm{B}}+\overrightarrow{\mathrm{C}}, \mathrm{B}^{2}=\mathrm{A}^{2}+\mathrm{C}^{2}\)
Step-by-step Solution
Detailed explanation
\(
\begin{array}{l}
\vec{A}=3 \hat{i}-2 \hat{j}+\hat{k} \\
\vec{B}=\hat{\imath}-3 \hat{\jmath}+5 \hat{k} \\
\vec{C}=2 \hat{\imath}+\hat{\jmath}-4 \hat{k}
\end{array}
\)
Here,
\(\vec{A} =(\hat{\imath}-3 \hat{\jmath}+5 \hat{k})+(2 \hat{\imath}+\hat{\jmath}-4 \hat{k}) \)
\( =(\vec{B}+\vec{c}) \)
\( =3 \hat{\imath}-2 \hat{\jmath}+\hat{k} \)
\( \text { Satisfied } \vec{A}=\vec{B}+\vec{C}\)
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,
\(\vec{A} =(\hat{\imath}-3 \hat{\jmath}+5 \hat{k})+(2 \hat{\imath}+\hat{\jmath}-4 \hat{k}) \)
\( =(\vec{B}+\vec{c}) \)
\( =3 \hat{\imath}-2 \hat{\jmath}+\hat{k} \)
\( \text { Satisfied } \vec{A}=\vec{B}+\vec{C}\)
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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