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10. vector algebra — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths10. 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 = \left| {\vec c} \right|,\left| {\vec c - \vec a} \right| = 2\sqrt 2 $ and the angle between $\vec a \times \vec b$ and $\vec c$ is $30^o$, then $\left| {\left( {\vec a \times \vec b} \right) \times \vec c} \right|$ equals 
  • A
    $\frac{1}{2}$
  • B
    $\frac{{3\sqrt 3 }}{2}$
  • C
    $3$
  • ✓
    $\frac{3}{2}$
✓

Answer

Correct option: D.
$\frac{3}{2}$
d
${\vec{a}=2 \hat{i}+\hat{j}-2 \hat{k}, \vec{b}=\hat{i}+\hat{j}}$

${\Rightarrow|\vec{a}|=3}$

and $\vec a \times \vec b = \begin{array}{*{20}{c}}
{\hat i}&{\hat j}&{\hat k}\\
2&1&{ - 2}\\
1&1&0
\end{array} = 2\hat i - 2\hat j + \hat k$

$|\vec{a} \times \vec{b}|=\sqrt{4}+4+1=3$

$\text { Now, }|\vec{c}-\vec{a}|=2 \sqrt{2} \Rightarrow|\vec{c}-\vec{a}|^{2}=8$

$\Rightarrow|\vec{c}-\vec{a}| \cdot(\vec{c}-\vec{a})=8$

${\Rightarrow|\vec{c}|^{2}+|\vec{a}|^{2}-2 \vec{c} \cdot \vec{a}=8}$

${\Rightarrow|\vec{c}|^{2}+9-2|\vec{c}|=8} $

$ \Rightarrow {(\left| {\vec c} \right| - 1)^2} = 0 \Rightarrow |\vec c| = 1$

$\therefore|(\vec{a} \times \vec{b}) \times \vec{c}|=|\vec{a} \times \vec{b}||\vec{c}| \sin 30^{\circ}$

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

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