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Vector Algebra — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVector Algebra1 Mark
MCQ
If $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{c}}=-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}},$ then a unit vector normal to the vectors $\vec{\text{a}}+\vec{\text{b}}$ and $\vec{\text{b}}-\vec{\text{c}}$ is :
  • ✓
    $\hat{\text{i}}$
  • B
    $\hat{\text{j}}$
  • C
    $\hat{\text{k}}$
  • D
    None of these
✓

Answer

Correct option: A.
$\hat{\text{i}}$
$\vec{\text{a}}+\vec{\text{b}}=0\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}-\vec{\text{c}}=0\hat{\text{i}}-0\hat{\text{j}}+3\hat{\text{k}}$
$\big(\vec{\text{a}}\times\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\0&3&1\\0&0&3 \end{vmatrix}$
$=9\hat{\text{i}}$
$\big|\big(\vec{\text{a}}+\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|=9|\hat{\text{i}}|$
$=9(1)$
$=9$
Unit vector perpendicular to both $\vec{\text{a}}+\vec{\text{b}}$ and $\vec{\text{b}}-\vec{\text{c}}=\frac{\big(\vec{\text{a}}+\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)}{\big|\big(\vec{\text{a}}+\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|}$
$=\frac{9\hat{\text{i}}}{9}$
$=\hat{\text{i}}$

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