Question:
If $\bar{\mathrm{a}}$, $\bar{\mathrm{b}}$, $\bar{\mathrm{c}}$ are three vectors which are perpendicular to $\bar{\mathrm{b}} + \bar{\mathrm{c}}$, $\bar{\mathrm{c}} + \bar{\mathrm{a}}$ and $\bar{\mathrm{a}} + \bar{\mathrm{b}}$ respectively, such that $|\bar{a}| = 2$, $|\bar{b}| = 3$, $|\bar{c}| = 4$, then $|\bar{a} + \bar{b} + \bar{c}| =$
If $\bar{\mathrm{a}}$, $\bar{\mathrm{b}}$, $\bar{\mathrm{c}}$ are three vectors which are perpendicular to $\bar{\mathrm{b}} + \bar{\mathrm{c}}$, $\bar{\mathrm{c}} + \bar{\mathrm{a}}$ and $\bar{\mathrm{a}} + \bar{\mathrm{b}}$ respectively, such that $|\bar{a}| = 2$, $|\bar{b}| = 3$, $|\bar{c}| = 4$, then $|\bar{a} + \bar{b} + \bar{c}| =$
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Whenever vectors are mutually orthogonal or satisfy the condition $\bar{a}\cdot(\bar{b}+\bar{c}) = 0$, all cross-product and dot-product mixed terms drop out to zero. The problem simplifies beautifully into just squaring and adding the individual magnitudes: $\sqrt{2^2 + 3^2 + 4^2} = \sqrt{29}$!
Updated On: Jun 3, 2026
- $29$
- $3$
- $9$
- $\sqrt{29}$
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Question:
We are given the magnitudes of three individual vectors along with conditions stating that each vector is orthogonal to the sum of the remaining two. We need to find the absolute magnitude of their vector sum, $|\bar{a} + \bar{b} + \bar{c}|$.
Step 2: Key Formula or Approach:
We can expand the squared magnitude of the vector sum using the standard algebraic identity: $$ |\bar{a} + \bar{b} + \bar{c}|^2 = |\bar{a}|^2 + |\bar{b}|^2 + |\bar{c}|^2 + 2(\bar{a}\cdot\bar{b} + \bar{b}\cdot\bar{c} + \bar{c}\cdot\bar{a}) $$ The perpendicularity conditions mean that the dot product of each vector with its corresponding sum pair is exactly zero.
Step 3: Detailed Explanation:
Let's write down the mathematical implications of the three given perpendicularity conditions:
• $\bar{a} \perp (\bar{b} + \bar{c}) \implies \bar{a}\cdot(\bar{b} + \bar{c}) = 0 \implies \bar{a}\cdot\bar{b} + \bar{a}\cdot\bar{c} = 0$
• $\bar{b} \perp (\bar{c} + \bar{a}) \implies \bar{b}\cdot(\bar{c} + \bar{a}) = 0 \implies \bar{b}\cdot\bar{c} + \bar{b}\cdot\bar{a} = 0$
• $\bar{c} \perp (\bar{a} + \bar{b}) \implies \bar{c}\cdot(\bar{a} + \bar{b}) = 0 \implies \bar{c}\cdot\bar{a} + \bar{c}\cdot\bar{b} = 0$
Adding these three individual scalar equations together gives: $$ (\bar{a}\cdot\bar{b} + \bar{a}\cdot\bar{c}) + (\bar{b}\cdot\bar{c} + \bar{b}\cdot\bar{a}) + (\bar{c}\cdot\bar{a} + \bar{c}\cdot\bar{b}) = 0 $$ $$ 2(\bar{a}\cdot\bar{b} + \bar{b}\cdot\bar{c} + \bar{c}\cdot\bar{a}) = 0 \implies \bar{a}\cdot\bar{b} + \bar{b}\cdot\bar{c} + \bar{c}\cdot\bar{a} = 0 $$ Now, let's substitute this result along with the given magnitudes ($|\bar{a}| = 2$, $|\bar{b}| = 3$, $|\bar{c}| = 4$) into our main squared expansion formula: $$ |\bar{a} + \bar{b} + \bar{c}|^2 = (2)^2 + (3)^2 + (4)^2 + 0 $$ $$ |\bar{a} + \bar{b} + \bar{c}|^2 = 4 + 9 + 16 = 29 $$ Taking the square root of both sides to isolate the magnitude gives: $$ | \bar{a} + \bar{b} + \bar{c} | = \sqrt{29} $$
Step 4: Final Answer:
The magnitude of the vector sum is $\sqrt{29}$, which corresponds to option (D).
We are given the magnitudes of three individual vectors along with conditions stating that each vector is orthogonal to the sum of the remaining two. We need to find the absolute magnitude of their vector sum, $|\bar{a} + \bar{b} + \bar{c}|$.
Step 2: Key Formula or Approach:
We can expand the squared magnitude of the vector sum using the standard algebraic identity: $$ |\bar{a} + \bar{b} + \bar{c}|^2 = |\bar{a}|^2 + |\bar{b}|^2 + |\bar{c}|^2 + 2(\bar{a}\cdot\bar{b} + \bar{b}\cdot\bar{c} + \bar{c}\cdot\bar{a}) $$ The perpendicularity conditions mean that the dot product of each vector with its corresponding sum pair is exactly zero.
Step 3: Detailed Explanation:
Let's write down the mathematical implications of the three given perpendicularity conditions:
• $\bar{a} \perp (\bar{b} + \bar{c}) \implies \bar{a}\cdot(\bar{b} + \bar{c}) = 0 \implies \bar{a}\cdot\bar{b} + \bar{a}\cdot\bar{c} = 0$
• $\bar{b} \perp (\bar{c} + \bar{a}) \implies \bar{b}\cdot(\bar{c} + \bar{a}) = 0 \implies \bar{b}\cdot\bar{c} + \bar{b}\cdot\bar{a} = 0$
• $\bar{c} \perp (\bar{a} + \bar{b}) \implies \bar{c}\cdot(\bar{a} + \bar{b}) = 0 \implies \bar{c}\cdot\bar{a} + \bar{c}\cdot\bar{b} = 0$
Adding these three individual scalar equations together gives: $$ (\bar{a}\cdot\bar{b} + \bar{a}\cdot\bar{c}) + (\bar{b}\cdot\bar{c} + \bar{b}\cdot\bar{a}) + (\bar{c}\cdot\bar{a} + \bar{c}\cdot\bar{b}) = 0 $$ $$ 2(\bar{a}\cdot\bar{b} + \bar{b}\cdot\bar{c} + \bar{c}\cdot\bar{a}) = 0 \implies \bar{a}\cdot\bar{b} + \bar{b}\cdot\bar{c} + \bar{c}\cdot\bar{a} = 0 $$ Now, let's substitute this result along with the given magnitudes ($|\bar{a}| = 2$, $|\bar{b}| = 3$, $|\bar{c}| = 4$) into our main squared expansion formula: $$ |\bar{a} + \bar{b} + \bar{c}|^2 = (2)^2 + (3)^2 + (4)^2 + 0 $$ $$ |\bar{a} + \bar{b} + \bar{c}|^2 = 4 + 9 + 16 = 29 $$ Taking the square root of both sides to isolate the magnitude gives: $$ | \bar{a} + \bar{b} + \bar{c} | = \sqrt{29} $$
Step 4: Final Answer:
The magnitude of the vector sum is $\sqrt{29}$, which corresponds to option (D).
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