MCQ
જો $\vec a\, = \,\vec i - 2\hat j + 3\hat k,\,\,\,\vec b = 2\vec i + 3\hat j - \hat k$ અને $\vec c = \lambda \vec i + \hat j + (2\lambda - 1\hat k)$ એ સમતલીય સદીશ હોય તો $\lambda $ મેળવો.
- ✓$0$
- 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.,$\begin{array}{*{20}{c}}
1&2&\lambda \\
{ - 2}&3&1\\
3&{ - 1}&{2\lambda - 1}
\end{array} = 0$
$\Rightarrow \quad 1(6 \lambda-2)-2(-4 \lambda-1)+\lambda(-7)=0$
$\Rightarrow \quad(6 \lambda-2)+8 \lambda+2+2+2 \lambda-9 \lambda=0$
$\Rightarrow \quad 7 \lambda=0 \Rightarrow \lambda=0$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
વિધાન $1$:કોઇક $c\; \in R$ માટે, $f\left( c \right) = \frac{1}{3}$
વિધાન $2$:$0 < f\left( x \right) < \frac{1}{{2\sqrt 2 }}\;,\forall x\; \in R$