MCQ
If $\left[\begin{array}{lll}\vec{a} \times \vec{b} & \vec{b} \times \vec{c} & \vec{c} \times a\end{array}\right]=\lambda[\vec{a} \vec{b} \vec{c}]^2$, then $\lambda$ is equal to
- A3
- B
- C1
- D2
$\begin{aligned} & \text {(c) : }\left[\begin{array}{lll}\vec{a} \times \vec{b} & \vec{b} \times \vec{c} & \vec{c} \times \vec{a}\end{array}\right] \\ & =(\vec{a} \times \vec{b})\{(\vec{b} \times \vec{c}) \times(\vec{c} \times \vec{a})\} \\ & =(\vec{a} \times \vec{b})\{(\vec{b} \times \vec{c} \cdot \vec{a}) \vec{c}-(\vec{b} \times \vec{c} \cdot \vec{c}) \vec{a}\} \\ & =(\vec{a} \times \vec{b})[\vec{a} \vec{b} \vec{c}] \vec{c}=[\vec{a} \vec{b} \vec{c}][\vec{a} \vec{b} \vec{c}]=[\vec{a} \vec{b} \vec{c}]^2 \\ & \therefore \quad \text { On comparison, } \lambda=1\end{aligned}$
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If the lines $\frac{x-k}{2}=\frac{y+1}{3}=\frac{z-1}{4} \text { and } \frac{x-3}{1}=\frac{y-\frac{9}{2}}{2}=\frac{z}{1}$ intersect, then the value of $k$ is