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VECTOR ALGEBRA — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVECTOR ALGEBRA1 Mark
Question
Choose the correct answer from the given four options.
The vectors from origin to the points A and B are $\vec{\text{a}}=2\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}$ and $\vec{\text{b}}=2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}},$ respectively, then the area of the triangle OAB is:
  1. $340$
  2. $\sqrt{25}$
  3. $\sqrt{229}$
  4. $\frac{1}{2}\sqrt{229}$
✓

Answer

  1. $\frac{1}{2}\sqrt{229}$

Solution:

$\therefore$ Area of $\triangle\text{OAB}=\frac{1}{2}|\overrightarrow{\text{OA}}\times\overrightarrow{\text{OB}}|$

$=\frac{1}{2}|(2\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}})\times(2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}})|$

$=\frac{1}{2}\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}} \\2&-3&2\\2&3&1 \end{vmatrix}$

$=\frac{1}{2}|[\hat{\text{i}}(-3-6)-\hat{\text{j}}(2-4)+\hat{\text{k}}(6+6)]|$

$=\frac{1}{2}|-9\hat{\text{i}}+2\hat{\text{j}}+12\hat{\text{k}}|$

$\therefore$ Area of $\triangle\text{OAB}=\frac{1}{2}\sqrt{81+4+144}$

$=\frac{1}{2}\sqrt{229}$

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