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10. vector algebra — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths10. vector algebra1 Mark
MCQ
If $\vec a$ and $\vec b$ are perpendicular unit vectors and vector $\vec c$ is such that $\vec c = \vec a + \vec b$ , then $\left( {\vec a \times \vec b} \right).\left( {\vec b \times \vec c} \right) + \left( {\vec b \times \vec c} \right).\left( {\vec c \times \vec a} \right) + \left( {\vec c \times \vec a} \right).\left( {\vec a \times \vec b} \right)$ is
  • A
    $0$
  • B
    $1$
  • ✓
    $-1$
  • D
    $\vec b.\vec c + \vec c.\vec a$
✓

Answer

Correct option: C.
$-1$
c
$(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}})(\overrightarrow{\mathrm{b} .} \overrightarrow{\mathrm{c}})-(\overrightarrow{\mathrm{a} .} \cdot \overrightarrow{\mathrm{c}})(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{b}})+(\overrightarrow{\mathrm{b} .} \cdot \overrightarrow{\mathrm{c}})(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}})$

$-(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}})(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{c}})+(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}})((\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}})-(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}})(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}})-(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{b}})(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{a}})$

$ = 0 - (\vec a \cdot \vec c) + (\vec b \cdot \vec c)(\bar c \cdot \vec a) - 0 + 0 - (\vec c \cdot \vec b)$  $(\because \vec a \cdot \vec b = 0)$

$=(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}})(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}})-(\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}})-(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}})+1-1$

$=((\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}})-1)((\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}})-1)-1$

$=(1-1)(1-1)-1=-1(\because \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}})$

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