MCQ
If $\left[ {a \times b\;\;b \times c\;\;c \times a} \right] = \alpha \;{\left[ {a\;\;b\;\;c} \right]^2}$ then $\lambda$ is equal to
- A$0$
- ✓$1$
- C$2$
- D$3$
$=(\bar{a} \times \bar{b}) \cdot[(\vec{b} \times \vec{c}) \times(\vec{c} \times \vec{a})]$
$=(\vec{a} \times \vec{b}) \cdot[(\vec{b} \times \vec{c} \cdot \vec{a}) \vec{c}-(\vec{b} \times \vec{c} \cdot \vec{c}) \vec{a}]$
$ = (\vec a \times \vec b) \cdot [[\bar b\bar c\bar a]\vec c]\quad [\vec b \times \vec c.\vec c = 0]$
$=[\bar{a} \bar{b} \bar{c}] \cdot(\bar{a} \times \bar{b} \cdot \bar{c})=[\bar{a} \bar{b} \bar{c}]^{2}$
$[\vec{a} \times \vec{b} \vec{b} \times \vec{c} \quad \vec{c} \times \vec{a}]=[\bar{a} \bar{b} \bar{c}]^{2}$
So $\lambda=1$
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Which of the following is the general solution of $\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-2\frac{\text{d}\text{y}}{\text{d}\text{x}}+\text{y}=0?$
$\text{y}=(\text{Ax}+\text{B})\text{e}^{\text{x}}$
$\text{y}=(\text{Ax}+\text{B})\text{e}^{-\text{x}}$
$\text{y}=\text{Ax}\text{e}^{\text{x}}+\text{B}\text{e}^{\text{x}}$
$\text{y}=\text{A}\cos\text{x}+\text{B}\sin\text{x}$