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Vector Algebra — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVector Algebra1 Mark
MCQ
If $\big[2\vec{\text{a}}+4\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big],$ then $\lambda+\mu=$
  • ✓
    $6$
  • B
    $-6$
  • C
    $10$
  • D
    $8$
✓

Answer

Correct option: A.
$6$
We have
$\big[2\vec{\text{a}}+4\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]$
$\Rightarrow\big[\big(2\vec{\text{a}}+4\vec{\text{b}}\big]\times\vec{\text{c}}\big].\vec{\text{d}}=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big] ($By definition of scalar triple product$)$
$\Rightarrow\big[\big(2\vec{\text{a}}\times\vec{\text{c}}\big)+\big(4\vec{\text{b}}\times\vec{\text{c}}\big)\big].\vec{\text{d}}=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]$
$\Rightarrow\big(2\vec{\text{a}}\times\vec{\text{c}}\big).\vec{\text{d}}+\big(4\vec{\text{b}}\times\vec{\text{c}}\big).\vec{\text{d}}=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]$
$\Rightarrow\big[2\vec{\text{a}}\vec{\text{ c }}\vec{\text{d}}\big]+\big[4\vec{\text{b}}\vec{\text{ c }}\vec{\text{d}}\big]=\lambda\big[\vec{\text{a}}\vec{\text{ c }}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{ c }}\vec{\text{d}}\big]$
$\Rightarrow2\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+4\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]=\lambda\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{d}}\big]+\mu\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{d}}\big]$ $\big(\therefore\big[\lambda\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=\lambda\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$ for any scaler $\lambda\big)$
Comparing both sides, we get
$\lambda=2$
$\mu=4$
$\therefore\lambda+\mu=2+4=6$

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