MCQ
Let $\vec{a}=2 \hat{i}+5 \hat{j}-\hat{k}, \vec{b}=2 \hat{i}-2 \hat{j}+2 \hat{k}$ and $\overrightarrow{ c }$ be three vectors such that $(\vec{c}+\hat{i}) \times(\vec{a}+\vec{b}+\hat{i})=\vec{a} \times(\vec{c}+\hat{i}) \cdot \vec{a} \cdot \vec{c}=-29$, then $\overrightarrow{ c } \cdot(-2 \hat{ i }+\hat{ j }+\hat{ k })$ is equal to :
  • A
    10
  • B
    5
  • C
    15
  • D
    12

Answer

Let's assume $\vec{v}=\vec{a}+\vec{b}+\hat{i}$$
=5 \hat{i}+3 \hat{j}+\hat{k}
$and $\overrightarrow{ c }+\hat{ i }=\overrightarrow{ p }$
So,
$\overrightarrow{ p } \times \overrightarrow{ v }=\overrightarrow{ a } \times \overrightarrow{ p }$
$\overrightarrow{ p } \times \overrightarrow{ v }+\overrightarrow{ p } \times \overrightarrow{ a }=\overrightarrow{0}$
$\overrightarrow{ p } \times(\overrightarrow{ v }+\overrightarrow{ a })=\overrightarrow{0}$
$\Rightarrow \overrightarrow{ p }=\lambda(\overrightarrow{ v }+\overrightarrow{ a })$
$\overrightarrow{ c }+ i =\lambda(7 \hat{ i }+8 \hat{ j })$
$\overline{ a } \cdot \overline{ c }+\overline{ a } \cdot \hat{ i }=\lambda \overline{ a } \cdot(7 \hat{ i }+8 \hat{ j })$
$-29+2=\lambda(14+40)$
$\lambda=-\frac{1}{2}$
$\overrightarrow{ c } \cdot(-2 \hat{ i }+\hat{ j }+\hat{ k })+\hat{ i } \cdot(-2 \hat{ i }+\hat{ j }+\hat{ k })=\lambda(7 \hat{ i }+8 \hat{ j }) \cdot(-2 \hat{ i }+\hat{ j }+\hat{ k })$ $=-\frac{1}{2}(-14+8)+2=5$

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