MCQ
Let $\vec a=j -k $ and $ \vec c= i -j -k $ Then the vector $\vec b$ satisfying $\vec a \times \vec b + \vec c = 0$ and $\vec a\cdot \vec b=3 $
- A$2i -j +2k$
- B$i -j -2k$
- C$i+j-2k$
- ✓$-i+j-2k$
${\text {given}, \vec{a} \cdot \vec{b}=3}$
${\Rightarrow y-z=3 \ldots \ldots(1)} $
${\text { Also, }(\vec{a} \times \vec{b})+\vec{c}=0} $
${\Rightarrow(z+y) \vec{i}-x \vec{j}-x \vec{k}+\vec{i}-\vec{j}-\vec{k}=0} $
${\Rightarrow(z+y+1) \vec{i}-(x+1) \vec{j}-(x+1) \vec{k}=0}$
${\Rightarrow x+1=0 \Rightarrow x=-1} $
${\text { and } z+y+1=0 \Rightarrow y+z=-1 \quad \cdots \cdots(2)} $
${\text { solving equation }(1) \text { and }(2) \text { we get, }}$
$y=1$ and $z=-2$
hence $\vec{b}=-1 \vec{i}+1 \vec{j}-2 \vec{k}$
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| Column-$I$ | Column-$II$ |
| $(A)$ In $R ^2$, if the magnitude of the projection vector of the vector $\alpha \hat{i}+\beta \hat{j}$ on $\sqrt{3} \hat{i}+\hat{j}$ is $\sqrt{3}$ and if $\alpha=2+\sqrt{3} \beta$, then possible value(s) of $|\alpha|$ is (are) | $(P)$ $1$ |
| $(B)$ Let $a$ and $b$ be real numbers such that the function $f(x)=\left\{\begin{array}{cc}-3 a x^2-2, & x < 1 \\ b x+a^2, & x \geq 1\end{array}\right.$ is differentiable for all $x \in R$. Then possible value(s) of a is (are) | $(Q)$ $2$ |
| $(C)$ Let $\omega \neq 1$ be a complex cube root of unity. If $\left(3-3 \omega+2 \omega^2\right)^{4 n+3}$ $+\left(2+3 \omega-3 \omega^2\right)^{4 \omega+3}+\left(-3+2 \omega+3 \omega^2\right)^{4 \omega+3}=0$, then possible value(s) of $n$ is (are) | $(R)$ $3$ |
| $(D)$ Let the harmonic mean of two positive real numbers $a$ and $b$ be $4$. If $q$ is a positive real number such that $a, 5, q, b$ is an arithmetic progression, then the value$(s)$ of $|q-a|$ is (are) | $(S)$ $4$ |
| $(T)$ $5$ |