Question

If \(\vec a,\vec b,\vec c\) are three vectors such that \[ |\vec a|=a,\qquad |\vec b|=b,\qquad |\vec c|=c \] and each one of them is perpendicular to the sum of the other two, then \[ |\vec a+\vec b+\vec c| \] equals

• A. \(a+b+c\)
• B. \(a^2+b^2+c^2\)
• C. \(\sqrt{a^2+b^2+c^2}\)
• D. \(\sqrt{a+b+c}\)

Hint:

If vectors are mutually perpendicular, then \[ |\vec a+\vec b+\vec c|^2 = |\vec a|^2+|\vec b|^2+|\vec c|^2 \] since all cross-product terms vanish.

Answer 1

The Correct Option is C

Concept: Given \[ \vec a\perp(\vec b+\vec c), \qquad \vec b\perp(\vec c+\vec a), \qquad \vec c\perp(\vec a+\vec b) \] Use dot products to establish relationships among the vectors.

Step 1: Write the perpendicularity conditions. \[ \vec a\cdot(\vec b+\vec c)=0 \] \[ \vec b\cdot(\vec c+\vec a)=0 \] \[ \vec c\cdot(\vec a+\vec b)=0 \] Thus, \[ \vec a\cdot\vec b+\vec a\cdot\vec c=0 \] \[ \vec a\cdot\vec b+\vec b\cdot\vec c=0 \] \[ \vec a\cdot\vec c+\vec b\cdot\vec c=0 \]

Step 2: Solve these equations. Subtracting the first two equations, \[ \vec a\cdot\vec c = \vec b\cdot\vec c \] Using the third equation, \[ 2(\vec b\cdot\vec c)=0 \] Hence, \[ \vec b\cdot\vec c=0 \] Similarly, \[ \vec a\cdot\vec b=0, \qquad \vec a\cdot\vec c=0 \] Thus the three vectors are mutually perpendicular.

Step 3: Find the magnitude of the sum. \[\begin{aligned} |\vec a+\vec b+\vec c|^2 &= (\vec a+\vec b+\vec c)\cdot (\vec a+\vec b+\vec c) \\ &= a^2+b^2+c^2 \end{aligned}\] Therefore, \[ |\vec a+\vec b+\vec c| = \sqrt{a^2+b^2+c^2} \] \[\begin{aligned} \boxed{\sqrt{a^2+b^2+c^2}} \end{aligned}\] Hence, option \(\mathbf{(C)}\) is correct.