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10. vector algebra — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths10. vector algebra1 Mark
MCQ
Let $\vec{a}=\hat{i}-2 \hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-\hat{j}+\hat{k}$ be two vectors. If $\overrightarrow{\mathrm{c}}$ is a vector such that $\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}$ and $\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}=0,$ then $\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{b}}$ is equal to
  • A
    $\frac{1}{2}$
  • B
    $-1$
  • ✓
    $-\frac{1}{2}$
  • D
    $-\frac{3}{2}$
✓

Answer

Correct option: C.
$-\frac{1}{2}$
c
$\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{0}$

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

$\overrightarrow{\mathrm{b}}=\lambda(\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}})\ldots(i)$

$\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=\lambda\left(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}}^{2}\right)$

$4=\lambda(0-6) \Rightarrow \lambda=\frac{-4}{6}=\frac{-2}{3}$

from $(\text {i})\;\; {\overrightarrow {\mathrm{b}}}=\frac{-2}{3}(\overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{a}})$

$\overrightarrow{\mathrm{c}}=\frac{-3}{2} \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{a}}=\frac{-1}{2}(\hat{i}+\hat{j}+\hat{\mathrm{k}})$

$\overrightarrow{\mathrm{b} \cdot \overrightarrow{\mathrm{c}}=\frac{-1}{2}}$

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