- ✓$\vec b = 3\vec a$
- B$\vec b = - 3\vec a$
- C$\vec b = 2\vec a$
- D$\vec b = - 2\vec a$
$ \Rightarrow \frac{{|\vec a + \lambda \vec a|}}{{|\vec a - \lambda \vec a|}} = 2\quad \Rightarrow \quad \frac{{1 + \lambda }}{{1 - \lambda }} = \pm 2$
$ \Rightarrow \lambda = 3,\frac{1}{3}\quad \Rightarrow \quad \vec b\left\langle {\begin{array}{*{20}{c}}
{3\vec a}\\
{\frac{{\vec a}}{3}}
\end{array}} \right.$
$ \Rightarrow \overrightarrow {\rm{b}} = 3\vec a\quad $ ($\because $ $|\overrightarrow {\rm{b}} | > |\overrightarrow {\rm{a}} |$)
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$(A)$ determinant of $\left( M ^2+ MN ^2\right)$ is $0$
$(B)$ there is a $3 \times 3$ non-zero matrix $U$ such that $\left( M ^2+ MN ^2\right) U$ is the zero matrix
$(C)$ determinant of $\left( M ^2+ MN ^2\right) \geq 1$
$(D)$ for a $3 \times 3$ matrix $U$, if $\left( M ^2+ MN ^2\right) U$ equals the zero matrix then $U$ is the zero matrix