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STD 12 - 10. vector algebra — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 10. vector algebra1 Mark
MCQ
Let $\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$ and $\vec{b}$ be a vector such that $\vec{a} \times \vec{b}=2 \hat{i}-\hat{k}$ and $\vec{a} \cdot \vec{b}=3$. Then the projection of $\vec{b}$ on the vector $\vec{a}-\vec{b}$ is :-
  • ✓
    $\frac{2}{\sqrt{21}}$
  • B
    $2 \sqrt{\frac{3}{7}}$
  • C
    $\frac{2}{3} \sqrt{\frac{7}{3}}$
  • D
    $\frac{2}{3}$
✓

Answer

Correct option: A.
$\frac{2}{\sqrt{21}}$
a
$\vec{a}=\hat{i}-\hat{j}+2 \hat{k}$

$\vec{a} \times \vec{b}=2 \hat{i}-\hat{k}$

$\vec{a} \cdot \vec{b}=3$

$|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}=|\vec{a}|^{2} \cdot|\vec{b}|^{2}$

$5+9=6|\vec{b}|^{2}$

$|b|^{2}=\frac{7}{3}$

$|\vec{a}-\vec{b}|=\sqrt{|\vec{a}|^{2}+|\vec{b}|^{2}-2 \vec{a} \cdot \vec{b}}=\sqrt{\frac{7}{3}}$

projection of $\vec{b}$ on $\vec{a}-\vec{b}=\frac{\vec{b} \cdot(\vec{a}-\vec{b})}{|\vec{a}-\vec{b}|}$

$=\frac{\vec{b} \cdot \vec{a}-|\vec{b}|^{2}}{|\vec{a}-\vec{b}|}=\frac{3-\frac{7}{3}}{\sqrt{\frac{7}{3}}}$

$=\frac{2}{\sqrt{21}}$

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