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STD 12 - 10. vector algebra — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 10. vector algebra4 Marks
MCQ
If $\vec a = \hat i - \hat j - \hat k$ and $\vec b = \lambda \hat i - 3\hat j + \hat k$ and the orthogonal projection of  $\vec b$ on $\vec a$ is $\frac{4}{3}\left( {\hat i - \hat j - \hat k} \right)$, then $\lambda$ is equal to
  • A
    $0$
  • ✓
    $2$
  • C
    $12$
  • D
    $-1$
✓

Answer

Correct option: B.
$2$
b
$\frac{{(\overrightarrow b  \cdot \overrightarrow a )\overrightarrow a }}{{|\overrightarrow a {|^2}}} = \frac{4}{3}(\widehat {\rm{i}} - \widehat {\rm{j}} - \widehat {\rm{k}})$

$ \Rightarrow \frac{{\{ (\lambda \widehat i - 3\widehat j + \widehat k) \cdot (\widehat i - \widehat j - \widehat k)\} (\widehat i - \widehat j - \widehat k)}}{{(1 + 1 + 1)}}$

$=\frac{4}{3}(\hat{i}-\hat{j}-\hat{k})$

$\Rightarrow(\lambda+3-1)(\hat{i}-\hat{j}-\hat{k})=4(\hat{i}-\hat{j}-\hat{k})$

$\Rightarrow(\lambda+2)(\hat{i}-\hat{j}-\hat{k})=4(\hat{i}-\hat{j}-\hat{k})$

On equating the coefficient of $\widehat i$, we get

$\lambda+2=4 \Rightarrow \lambda=2$

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