MCQ
Find the value of $\lambda$ for which the vectors $3 \hat{i}-6 \hat{j}+\hat{k}$ and $2 \hat{i}-4 \hat{j}+\lambda \hat{k}$ are parallel.
  • $\frac{2}{3}$
  • B
    $\frac{-3}{2}$
  • C
    $\frac{-2}{3}$
  • D
    $\frac{3}{2}$

Answer

Correct option: A.
$\frac{2}{3}$
(a) : $\vec{a}=3 \hat{i}-6 \hat{j}+\hat{k}$ and $\vec{b}=2 \hat{i}-4 \hat{j}+\lambda \hat{k}$
Since, $\vec{a}$ and $\vec{b}$ are parallel $\quad \therefore \quad \vec{a} \times \vec{b}=\overrightarrow{0}$
$
\begin{aligned}
\Rightarrow & \left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -6 & 1 \\
2 & -4 & \lambda
\end{array}\right|=\overrightarrow{0} \\
\Rightarrow & (-6 \lambda+4) \hat{i}-(3 \lambda-2) \hat{j}+(-12+12) \hat{k}=\overrightarrow{0} \\
\Rightarrow & (-6 \lambda+4) \hat{i}+(2-3 \lambda) \hat{j}=0 \hat{i}+0 \hat{j}
\end{aligned}
$
Comparing coefficients of $\hat{i}$ and $\hat{j}$, we get $-6 \lambda+4=0$ and $2-3 \lambda=0 \Rightarrow \lambda=2 / 3$

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