Search any question & find its solution
Question:
Answered & Verified by Expert
Let $\overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$ be three mutually perpendicular vectors each
of unit magnitud. If $\overrightarrow{\mathrm{A}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}, \quad \overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}$ and
$\overrightarrow{\mathrm{C}}=\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}$, then which one of the following is correct?
Options:
of unit magnitud. If $\overrightarrow{\mathrm{A}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}, \quad \overrightarrow{\mathrm{B}}=\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}$ and
$\overrightarrow{\mathrm{C}}=\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}$, then which one of the following is correct?
Solution:
1874 Upvotes
Verified Answer
The correct answer is:
$|\overrightarrow{\mathrm{A}}|=|\overrightarrow{\mathrm{B}}|=|\overrightarrow{\mathrm{C}}|$
For simplicity let us take $\vec{a}, \vec{b}, \vec{c}$ as $\hat{i}, \hat{j}, \hat{k}$
Now magnitude of $\overrightarrow{\mathrm{A}}, \overrightarrow{\mathrm{B}}$ and $\overrightarrow{\mathrm{C}}$ will be $\sqrt{3}$.
Now magnitude of $\overrightarrow{\mathrm{A}}, \overrightarrow{\mathrm{B}}$ and $\overrightarrow{\mathrm{C}}$ will be $\sqrt{3}$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.