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VECTOR ALGEBRA — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVECTOR ALGEBRA1 Mark
Question
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:
  1. $\hat{\text{i}}$
  2. $\hat{\text{j}}$
  3. $\hat{\text{k}}$
  4. $\text{None of these}$
✓

Answer

  1. $\hat{\text{i}}$

Solution:

$\vec{\text{a}}+\vec{\text{b}}=0\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$

$\vec{\text{b}}-\vec{\text{c}}=0\hat{\text{i}}-0\hat{\text{j}}+3\hat{\text{k}}$

$\big(\vec{\text{a}}\times\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\0&3&1\\0&0&3 \end{vmatrix}$

$=9\hat{\text{i}}$

$\big|\big(\vec{\text{a}}+\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|=9|\hat{\text{i}}|$

$=9(1)$

$=9$

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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