Question:
If \(\bar{a}, \bar{b}, \bar{c}\) are three unit vectors such that \(|\bar{a} + \bar{b}|^2 + |\bar{a} + \bar{c}|^2 = 8\), then \(|\bar{a} + 3\bar{b}|^2 + |\bar{a} + 3\bar{c}|^2 =\)
If \(\bar{a}, \bar{b}, \bar{c}\) are three unit vectors such that \(|\bar{a} + \bar{b}|^2 + |\bar{a} + \bar{c}|^2 = 8\), then \(|\bar{a} + 3\bar{b}|^2 + |\bar{a} + 3\bar{c}|^2 =\)
Show Hint
Whenever squared magnitudes of vector sums are given, convert them into dot products. It makes the calculation much easier.
Updated On: May 14, 2026
- \(26\)
- \(32\)
- \(22\)
- \(36\)
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The Correct Option is B
Solution and Explanation
Concept:
For any vector \(\vec{u}\), \[ |\vec{u}|^2 = \vec{u}\cdot\vec{u} \] If \(\bar{a},\bar{b},\bar{c}\) are unit vectors, then: \[ |\bar{a}|=|\bar{b}|=|\bar{c}|=1 \] ip
Step 1: Expand the given condition.
\[ |\bar{a}+\bar{b}|^2 = |\bar{a}|^2+|\bar{b}|^2+2\bar{a}\cdot\bar{b} \] \[ =1+1+2\bar{a}\cdot\bar{b}=2+2\bar{a}\cdot\bar{b} \] Similarly, \[ |\bar{a}+\bar{c}|^2=2+2\bar{a}\cdot\bar{c} \] Given: \[ |\bar{a}+\bar{b}|^2+|\bar{a}+\bar{c}|^2=8 \] So, \[ (2+2\bar{a}\cdot\bar{b})+(2+2\bar{a}\cdot\bar{c})=8 \] \[ 4+2(\bar{a}\cdot\bar{b}+\bar{a}\cdot\bar{c})=8 \] \[ \bar{a}\cdot\bar{b}+\bar{a}\cdot\bar{c}=2 \] ip
Step 2: Expand the required expression.
\[ |\bar{a}+3\bar{b}|^2 = |\bar{a}|^2+9|\bar{b}|^2+6\bar{a}\cdot\bar{b} \] \[ =1+9+6\bar{a}\cdot\bar{b}=10+6\bar{a}\cdot\bar{b} \] Similarly, \[ |\bar{a}+3\bar{c}|^2=10+6\bar{a}\cdot\bar{c} \] So, \[ |\bar{a}+3\bar{b}|^2+|\bar{a}+3\bar{c}|^2 =20+6(\bar{a}\cdot\bar{b}+\bar{a}\cdot\bar{c}) \] ip
Step 3: Substitute the earlier result.
Since \[ \bar{a}\cdot\bar{b}+\bar{a}\cdot\bar{c}=2 \] we get \[ 20+6(2)=20+12=32 \] ip Hence, the correct answer is:
\[ \boxed{(B)\ 32} \]
For any vector \(\vec{u}\), \[ |\vec{u}|^2 = \vec{u}\cdot\vec{u} \] If \(\bar{a},\bar{b},\bar{c}\) are unit vectors, then: \[ |\bar{a}|=|\bar{b}|=|\bar{c}|=1 \] ip
Step 1: Expand the given condition.
\[ |\bar{a}+\bar{b}|^2 = |\bar{a}|^2+|\bar{b}|^2+2\bar{a}\cdot\bar{b} \] \[ =1+1+2\bar{a}\cdot\bar{b}=2+2\bar{a}\cdot\bar{b} \] Similarly, \[ |\bar{a}+\bar{c}|^2=2+2\bar{a}\cdot\bar{c} \] Given: \[ |\bar{a}+\bar{b}|^2+|\bar{a}+\bar{c}|^2=8 \] So, \[ (2+2\bar{a}\cdot\bar{b})+(2+2\bar{a}\cdot\bar{c})=8 \] \[ 4+2(\bar{a}\cdot\bar{b}+\bar{a}\cdot\bar{c})=8 \] \[ \bar{a}\cdot\bar{b}+\bar{a}\cdot\bar{c}=2 \] ip
Step 2: Expand the required expression.
\[ |\bar{a}+3\bar{b}|^2 = |\bar{a}|^2+9|\bar{b}|^2+6\bar{a}\cdot\bar{b} \] \[ =1+9+6\bar{a}\cdot\bar{b}=10+6\bar{a}\cdot\bar{b} \] Similarly, \[ |\bar{a}+3\bar{c}|^2=10+6\bar{a}\cdot\bar{c} \] So, \[ |\bar{a}+3\bar{b}|^2+|\bar{a}+3\bar{c}|^2 =20+6(\bar{a}\cdot\bar{b}+\bar{a}\cdot\bar{c}) \] ip
Step 3: Substitute the earlier result.
Since \[ \bar{a}\cdot\bar{b}+\bar{a}\cdot\bar{c}=2 \] we get \[ 20+6(2)=20+12=32 \] ip Hence, the correct answer is:
\[ \boxed{(B)\ 32} \]
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