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Vector Algebra — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVector Algebra1 Mark
MCQ
Choose the correct answer from the given four options. For any vector $\vec{\text{a}},$ the value of $(\vec{\text{a}}\times\hat{\text{i}})^2+(\vec{\text{a}}\times\hat{\text{j}})^2+(\vec{\text{a}}\times\hat{\text{k}})^2$ is :
  • A
    $\vec{\text{a}}^2$
  • B
    $3\vec{\text{a}}^2$
  • C
    $4\vec{\text{a}}^2$
  • ✓
    $2\vec{\text{a}}^2$
✓

Answer

Correct option: D.
$2\vec{\text{a}}^2$
Let $\vec{\text{a}}^2=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$
$\therefore\vec{\text{a}}^2=\text{x}^2+\text{y}^2+\text{z}^2$
$\therefore\ \vec{\text{a}}\times\vec{\text{i}}\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}} \\\text{x}&\text{y}&\text{z}\\1&0&0 \end{vmatrix}$
$=\hat{\text{i}}[0]-\hat{\text{j}}[-\text{z}]+\hat{\text{k}}[-\text{y}]$
$=\text{z}\hat{\text{j}}-\text{y}\hat{\text{k}}$
$\therefore\ (\vec{\text{a}}\times\hat{\text{i}})=(\text{z}\hat{\text{j}}-\text{y}\hat{\text{k}})(\text{z}\hat{\text{j}}-\text{y}\hat{\text{k}})$
$=\text{y}^2+\text{z}^2$
Similarly, $(\vec{\text{a}}\times\hat{\text{j}})^2=\text{x}^2+\text{z}^2$
And $(\vec{\text{a}}\times\hat{\text{k}})^2=\text{x}^2+\text{y}^2$
$\therefore\ (\vec{\text{a}}\times\hat{\text{i}})^2+(\vec{\text{a}}\times\hat{\text{j}})^2+(\vec{\text{a}}\times\hat{\text{k}})^2$
$=\text{y}^2+\text{z}^2+\text{x}^2+\text{z}^2+\text{x}^2+\text{y}^2$
$2(\text{x}^2+\text{y}^2+\text{z}^2)=2\vec{\text{a}}^2$

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