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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 $\vec{a}, \vec{b}, \vec{c}$ be three vectors mutually perpendicular to each other and have same magnitude. If a vector $\overrightarrow{\mathrm{r}}$ satisfies.

$\overrightarrow{\mathrm{a}} \times\{(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{b}}) \times \overrightarrow{\mathrm{a}}\}+\overrightarrow{\mathrm{b}} \times\{(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{c}}) \times \overrightarrow{\mathrm{b}}\}+\overrightarrow{\mathrm{c}} \times\{(\overrightarrow{\mathrm{r}}-\overrightarrow{\mathrm{a}}) \times \overrightarrow{\mathrm{c}}\}=\overrightarrow{0}$

then $\overrightarrow{\mathrm{r}}$ is equal to:

  • A
    $\frac{1}{3}(\vec{a}+\vec{b}+\vec{c})$
  • B
    $\frac{1}{3}(2 \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}})$
  • ✓
    $\frac{1}{2}(\vec{a}+\vec{b}+\vec{c})$
  • D
    $\frac{1}{2}(\vec{a}+\vec{b}+2 \vec{c})$
✓

Answer

Correct option: C.
$\frac{1}{2}(\vec{a}+\vec{b}+\vec{c})$
c
Suppose $\overrightarrow{\mathrm{r}}=\mathrm{x} \overrightarrow{\mathrm{a}}+\mathrm{yb}+2 \overrightarrow{\mathrm{c}}$

and $|\overrightarrow{\mathrm{a}}|=|\overrightarrow{\mathrm{b}}|=|\overrightarrow{\mathrm{c}}|=\mathrm{k}$

$\vec{a} \times\{(\vec{r}-\vec{b}) \times \vec{a}\}+\vec{b} \times\{(\vec{r}-\vec{c}) \times \vec{b}\}+\vec{c} \times\{(\vec{r}-\vec{a}) \times \vec{c}\}=\overrightarrow{0}$

$\Rightarrow k^{2}(\vec{r}-\vec{b})-k^{2} x \vec{a}+k^{2}(\vec{r}-\vec{c})-k^{2} y \vec{b}+k^{2}(\vec{r}-\vec{a})-k^{2} z \vec{c}=\overrightarrow{0}$

$\Rightarrow 3 \vec{r}-(\vec{a}+\vec{b}+\vec{c})-\vec{r}=\overrightarrow{0}$

$\Rightarrow \vec{r}=\frac{\vec{a}+\vec{b}+\vec{c}}{2}$

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