MCQ
If $\vec a\, = \,\vec i - 2\hat j + 3\hat k,\,\,\,\vec b = 2\vec i + 3\hat j - \hat k$ and $\vec c = \lambda \vec i + \hat j + (2\lambda - 1\hat k)$ are coplanar vectors, then $\lambda $ is equal to
- ✓$0$
- B$-1$
- C$2$
- D$1$
$\vec{c}=\lambda \hat{i}+\hat{j}+(2 \lambda-1 \hat{k})$ are coplanar
therefore $[\bar{a} \vec{b} \vec{c}]=0$
i.e.,$\begin{array}{*{20}{c}}
1&2&\lambda \\
{ - 2}&3&1\\
3&{ - 1}&{2\lambda - 1}
\end{array} = 0$
$\Rightarrow \quad 1(6 \lambda-2)-2(-4 \lambda-1)+\lambda(-7)=0$
$\Rightarrow \quad(6 \lambda-2)+8 \lambda+2+2+2 \lambda-9 \lambda=0$
$\Rightarrow \quad 7 \lambda=0 \Rightarrow \lambda=0$
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f(x) > 0 for all $\text{x}\in\text{R}.$
f(x) > 0 for all $\text{x}\in\text{R}.$
f(x) is invertible.
None of these.