- ✓$0 $ and $ - (\alpha + \beta + \gamma )$
- B$0$ and $(\alpha + \beta + \gamma )$
- C$1$ and $(\alpha - \beta - \gamma )$
- D$0 $ and $({\alpha ^2} + {\beta ^2} + {\gamma ^2})$
$[{C_1} \to {C_1} + ({C_2} + {C_3})]$
or $(x + \alpha + \beta + \gamma )\,\left| {\,\begin{array}{*{20}{c}}1&\beta &\gamma \\1&{x + \beta }&\alpha \\1&\beta &{x + \gamma }\end{array}\,\,} \right|\, = 0$
or $(x + \alpha + \beta + \gamma )\,\left| {\,\begin{array}{*{20}{c}}1&\beta &\gamma \\0&x&{\alpha - \gamma }\\0&0&x\end{array}\,} \right|\, = \,0$,
$\left[ \begin{array}{l}{R_2} \to {R_2} - {R_1}\\{R_3} \to {R_3} - {R_1}\end{array} \right]$
or $(x + \alpha + \beta + \gamma )[{x^2} - 0] = 0$
or ${x^2}(x + \alpha + \beta + \gamma ) = 0$
$\therefore $ $x = 0$ or $x = - (\alpha + \beta + \gamma )$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
If $F^{\prime}(4)=\frac{\alpha e^{\beta}-224}{\left(e^{\beta}-4\right)^{2}}$, then $\alpha+\beta$ is equal to $....$
$\overrightarrow{ r }=\hat{ i }+\lambda(-\hat{ i }+2 \hat{ j }+2 \hat{ k }), \lambda \in R \text { and }$
$\overrightarrow{ r }=\mu(2 \hat{ i }-\hat{ j }+2 \hat{ k }), \mu \in R$
respectively. If $L _3$ is a line which is perpendicular to both $L _1$ and $L _2$ and cuts both of them, then which of the following options describe(s) $L _3$ ?
$(1)$ $\overrightarrow{ r }=\frac{1}{3}(2 \hat{ i }+\hat{ k })+ t (2 \hat{ i }+2 \hat{ j }-\hat{ k }), t \in R$
$(2)$ $\overrightarrow{ i }=\frac{2}{9}(2 \hat{ i }-\hat{ j }+2 \hat{ k })+ t (2 \hat{ i }+2 \hat{ j }-\hat{ k }), t \in R$
$(3)$ $\overrightarrow{ r }=t(2 \hat{ i }+2 \hat{ j }-\hat{ k }), t \in R$
$(4)$ $\overrightarrow{ r }=\frac{2}{9}(4 \hat{ i }+\hat{ j }+\hat{ k })+ t (2 \hat{ i }+2 \hat{ j }-\hat{ k }), t \in R$