MCQ
If $\vec a = 2 \hat i + 3 \hat j+ \hat k ,\vec b = \hat i - \hat j+ \hat k , \vec c = \hat i + \hat j+ \hat k$ and let $\vec d$  be such that $\vec a \times \vec b = \vec d \times \vec b, \vec d. \vec c = 8$, then value of $ \vec d. \vec b$ is -
  • $6$
  • B
    $-6$
  • C
    $3$
  • D
    $-3$

Answer

Correct option: A.
$6$
a
$(\vec{a}-\vec{d}) \times \vec{b}=0 \Rightarrow \vec{a}=\vec{d}+\lambda \vec{b}$

Dot with $\overrightarrow{\mathrm{c}}$

$2+3+1=8+\lambda(1-1+1)$

$\Rightarrow \lambda=-2$

$\therefore \overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}$

$=4 \hat{i}+\hat{j}+3 \hat{k}$

$\therefore \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{d}}=4-1+3=6$

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