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Home  >>  CBSE XII  >>  Math  >>  Vector Algebra
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If $\overrightarrow a = 2\hat i + 2\hat j + 3\hat k,\: \overrightarrow b = − \hat i + 2\hat j + \hat k$ and $ \overrightarrow c = 3\hat i + \hat j$ are such that $\overrightarrow a + λ\overrightarrow b $ is perpendicular to $ \overrightarrow c $ , then find the value of $λ.$

$\begin{array}{1 1}(A) \lambda=8 \\ (B) \lambda=-8 \\(C) \lambda=\large\frac{1}{8} \\(D) \lambda=-\large\frac{1}{8} \end{array} $

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1 Answer

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Toolbox:
  • $(a_1\hat i+a_2\hat j+a_3\hat k).(b_1\hat i+b_2\hat j+b_3\hat k)=a_1b_1+a_2b_2+a_3b_3$
  • If $\overrightarrow a\perp\overrightarrow b$ then $\overrightarrow a.\overrightarrow b=0$
Step 1:
\(\overrightarrow a = 2\hat i + 2\hat j + 3\hat k,\: \overrightarrow b = − \hat i + 2\hat j + \hat k\) and \( \overrightarrow c = 3\hat i + \hat j\)
Also $\overrightarrow a+\lambda\overrightarrow b$ is perpendicular to $\overrightarrow c$
$\overrightarrow a+\lambda\overrightarrow b=2\hat i+2\hat j+3\hat k+\lambda(-\hat i+2\hat j+\hat k)$
$\qquad\;\;\;\;\;=(2-\lambda)\hat i+(2+2\lambda)\hat j+(3+\lambda)\hat k$
Step 2:
$\overrightarrow c=3\hat i+\hat j$
Since $(\overrightarrow a+\lambda\overrightarrow b)\perp \overrightarrow c$
$(\overrightarrow a+\lambda\overrightarrow b) . \overrightarrow c=0$
$\Rightarrow [(2-\lambda)\hat i+(2+2\lambda)\hat j+(3+\lambda)\hat k].(3\hat i+\hat j)=0$
Since $(a_1\hat i+a_2\hat j+a_3\hat k).(b_1\hat i+b_2\hat j+b_3\hat k)=a_1b_1+a_2b_2+a_3b_3$
Step 3:
$(2-\lambda).3+(2+2\lambda).1=0$
$\Rightarrow 6-3\lambda+2+2\lambda=0$
$-\lambda=-8$
$\lambda=8$
answered May 21, 2013 by sreemathi.v
 

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