Question

Let \(\overline{\mathrm{a}}, \overline{\mathrm{b}}\) and \(\overline{\mathrm{c}}\) be vectors of magnitude 2,3 and 4 respectively. If \(\bar{a}\) is perpendicular to \((\overline{\mathrm{b}}+\overline{\mathrm{c}}), \overline{\mathrm{b}}\) is perpendicular to ( \(\overline{\mathrm{c}}+\overline{\mathrm{a}}\) ) and \(\overline{\mathrm{c}}\) is perpendicular to \((\bar{a}+\bar{b})\), then the magnitude of \(\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}\) is

• A. 29
• B. \(\sqrt{28}\)
• C. \(\sqrt{29}\)
• D. 28

Hint:

Expand the square of the sum and use the three perpendicularity conditions to find the dot products.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Perpendicular vectors have zero dot product, so \(\bar a\cdot(\bar b+\bar c)=0\), \(\bar b\cdot(\bar c+\bar a)=0\), \(\bar c\cdot(\bar a+\bar b)=0\).

Step 2: Add the conditions
Adding gives \(2(\bar a\cdot\bar b+\bar b\cdot\bar c+\bar c\cdot\bar a)=0\).

Step 3: Expand the square
\(|\bar a+\bar b+\bar c|^{2}=|\bar a|^{2}+|\bar b|^{2}+|\bar c|^{2}+2(\bar a\cdot\bar b+\bar b\cdot\bar c+\bar c\cdot\bar a)\).
\(=4+9+16+0=29\).

Step 4: Result
The magnitude is \(\sqrt{29}\).

Final Answer:
The magnitude is the square root of 29. \[ \boxed{\text{(C) }\sqrt{29}} \]