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STD 12 - 10. vector algebra — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 10. vector algebra1 Mark
MCQ
Let $\vec a = 2\hat i + \hat j - 2\hat k,\,\vec b = \hat i + \hat j$ . If $\vec c$ is a vector such that $\vec a.\vec c + 2\left| {\vec c} \right| = 0$ and $\left| {\vec c - \vec a} \right| = \sqrt {14} $ and angle between $\vec a \times \vec b$ and $\vec c$ is $30^o$ , then $\left| {\left( {\vec a \times \vec b} \right) \times \vec c} \right|$ is
  • ✓
    $\frac{3}{2}$
  • B
    $\frac{2}{3}$
  • C
    $2$
  • D
    $\frac{{\sqrt 3 }}{2}$
✓

Answer

Correct option: A.
$\frac{3}{2}$
a
$|\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}|=\sqrt{14}$

$ \Rightarrow |\overrightarrow c {|^2} + |\overrightarrow a {|^2} - 2\overrightarrow c  \cdot \overrightarrow a  = 14$         ........$(1)$

$\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}+2|\overrightarrow{\mathrm{c}}|=0$

$\Rightarrow \quad|\overrightarrow{\mathrm{a}}| \cdot|\overrightarrow{\mathrm{c}}| \cdot \cos \theta+2|\overrightarrow{\mathrm{c}}|=0$

$ \Rightarrow |\overrightarrow c | \cdot (|\overrightarrow {\rm{a}} | \cdot \cos \theta  + 2) = 0$

$\Rightarrow \quad \cos \theta=-\frac{2}{3},$ given $|\vec{a}|=3$

from $(i)$

$\Rightarrow \quad|\overrightarrow{\mathrm{c}}|^{2}+9-2|\overrightarrow{\mathrm{c}}| \cdot|\overrightarrow{\mathrm{a}}| \cdot\left(-\frac{2}{3}\right)-14=0$

$\Rightarrow \quad|\overrightarrow{\mathrm{c}}|^{2}+4|\overrightarrow{\mathrm{c}}|-5=0 \Rightarrow|\overrightarrow{\mathrm{c}}|=1,-5$

$\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\left|\begin{array}{ccc}{\hat{\mathrm{i}}} & {\hat{\mathrm{j}}} & {\hat{\mathrm{k}}} \\ {2} & {1} & {-2} \\ {1} & {1} & {0}\end{array}\right|=2 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}}$

$|(\vec{a} \times \vec{b}) \times \vec{c}|=|(\vec{a} \times \vec{b})| \cdot|\vec{c}| \cdot \sin \theta$

$=3.1 \times \frac{1}{2}=\frac{3}{2}$

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