Question

Let \( \overrightarrow a \) , \( \overrightarrow b \) and \( \overrightarrow c \) be three unit vectors such that
\( \overrightarrow a + \vec b + \overrightarrow c = \overrightarrow 0 \). If \( \lambda = \overrightarrow a .\vec b + \vec b.\overrightarrow c + \overrightarrow c .\overrightarrow a \) and
\( \overrightarrow d = \overrightarrow a \times \vec b + \vec b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a \), then the ordered pair, \( \left( {\lambda ,\overrightarrow d } \right) \) is equal to :

• A. \( \left( {{3 \over 2},3\overrightarrow a \times \overrightarrow c } \right) \)
• B. \( \left( { - {3 \over 2},3\overrightarrow c \times \overrightarrow b } \right) \)
• C. \( \left( { - {3 \over 2},3\overrightarrow a \times \overrightarrow b } \right) \)
• D. \( \left( {{3 \over 2},3\overrightarrow b \times \overrightarrow c } \right) \)

Hint:

Square the relation \(a+b+c=0\) for \(\lambda\), and cross it with each vector to show all three cross products are equal.

Answer 1

The Correct Option is C

Step 1: \((a+b+c)\cdot(a+b+c)=0\Rightarrow3+2\lambda=0\Rightarrow\lambda=-\tfrac32\).
Step 2: Substitute \(\vec c=-\vec a-\vec b\): \(\vec b\times\vec c=-\vec b\times\vec a=\vec a\times\vec b\) and \(\vec c\times\vec a=-\vec b\times\vec a=\vec a\times\vec b\).
Step 3: \(\vec d=3(\vec a\times\vec b)\).
Answer: (C) \(\left(-\tfrac32,\,3\vec a\times\vec b\right)\)