Question

If \(\overset{⃗}{a},\overset{⃗}{b},\overset{⃗}{c}\) are three vectors such that \(\overset{⃗}{a}⟂(\overset{⃗}{b}+\overset{⃗}{c}),\overset{⃗}{b}⟂(\overset{⃗}{c}+\overset{⃗}{a}),\) and \(\overset{⃗}{c}⟂(\overset{⃗}{a}+\overset{⃗}{b})\) and \(|\overset{⃗}{a}| = 1,|\overset{⃗}{b}| = 2,|\overset{⃗}{c}| = 3\), then \(|\overset{⃗}{a}+\overset{⃗}{b}+\overset{⃗}{c}|\) is...

• A. \(\sqrt{8}\)
• B. \(8\)
• C. \(14\)
• D. \(\sqrt{14}\)

Hint:

Adding the three perpendicularity conditions shows that every pairwise dot product is zero.

Answer 1

The Correct Option is D

Step 1: Translate the conditions
Perpendicular vectors have zero dot product: \(\vec{a}\cdot(\vec{b} + \vec{c}) = 0\), \(\vec{b}\cdot(\vec{c} + \vec{a}) = 0\), \(\vec{c}\cdot(\vec{a} + \vec{b}) = 0\). Writing \(p = \vec{a}\cdot\vec{b}\), \(q = \vec{b}\cdot\vec{c}\), \(r = \vec{c}\cdot\vec{a}\), we have \(p + r = 0\), \(q + p = 0\), \(r + q = 0\).

Step 2: Solve
Adding all three gives \(2(p + q + r) = 0\). Subtracting the pairs gives \(p = q = r = 0\). So the vectors are mutually perpendicular.

Step 3: Use the square of the magnitude
\[ |\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(p + q + r) \]

Step 4: Result
\(= 1 + 4 + 9 + 0 = 14\), so the magnitude is \(\sqrt{14}\), option (D). The value 14 is the square of the magnitude, not the magnitude.

Final Answer:
The magnitude is sqrt(14). This is option (D). \[ \boxed{\text{(D) }\sqrt{14}} \]