- A$AB$
- B$A + B$
- ✓$A'B'$
- D$B'A'$
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$(A)$ $\vec{b}=(\vec{b} \cdot \vec{z})(\vec{z}-\vec{x})$
$(B)$ $\vec{a}=(\vec{a} \cdot \vec{y})(\vec{y}-\vec{z})$
$(C)$ $\vec{a} \cdot \vec{b}=-(\vec{a} \cdot \vec{y})(\vec{b} \cdot \vec{z})$
$(D)$ $\vec{a}=(\vec{a} \cdot \vec{y})(\vec{z}-\vec{y})$
Then which of the following options is/are correct?
$(1)$ For $x =1$, there exists a unit vector $\alpha \hat{ i }+\beta \hat{ j }+\gamma \hat{ k }$ for which $R \left[\begin{array}{l}\alpha \\ \beta \\ \gamma\end{array}\right]=\left[\begin{array}{l}0 \\ 0 \\ 0\end{array}\right]$
$(2)$ There exists a real number $x$ such that $P Q=Q P$
$(3)$ $\operatorname{det} R=\operatorname{det}\left[\begin{array}{lll}2 & x & x \\ 0 & 4 & 0 \\ x & x & 5\end{array}\right]+8$, for all $x \in R$
$(4)$ For $x=0$, if $R\left[\begin{array}{l}1 \\ a \\ b\end{array}\right]=6\left[\begin{array}{l}1 \\ a \\ b\end{array}\right]$, then $a+b=5$