MCQ
Let $\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}, \quad \vec{b}=2 \hat{i}+3 \hat{j}-5 \hat{k} \quad$ and $\overrightarrow{\mathrm{c}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+\lambda \hat{\mathrm{k}}$ be three vectors. Let $\overrightarrow{\mathrm{r}}$ be a unit vector along $\vec{b}+\vec{c}$. If $\vec{r} . \vec{a}=3$, then $3 \lambda$ is equal to :
  • A
    $27$
  • $25$
  • C
    $30$
  • D
    $21$

Answer

Correct option: B.
$25$
b
$ \overrightarrow{\mathrm{r}}=\mathrm{k}(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}) $

$ \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}=3 $

$ \overrightarrow{\mathrm{r}} \cdot \overrightarrow{\mathrm{a}}=\mathrm{k}(\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}) $

$ 3=\mathrm{k}(2+6-15+3-2+3 \lambda) $

$ 3=\mathrm{k}(-6+3 \lambda) $ ...............($1$)

$ \overrightarrow{\mathrm{r}}=\mathrm{k}(5 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-(5-\lambda) \hat{\mathrm{k}}) $

$ |\overrightarrow{\mathrm{r}}|=\mathrm{k} \sqrt{25+4+25+\lambda^2-10 \lambda}=1 . $        ...............($2$)

$ \mathrm{k}=\frac{3}{-6+3 \lambda}=\frac{1}{-2+\lambda} \quad \text { put in }(2) $

$ 4+\lambda^2-4 \lambda=54+\lambda^2-10 \lambda $

$ 6 \lambda=50 $

$ 3 \lambda=25$

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