In a quadrilateral ABCD, the point P divides | vedclass

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(A) Given the expression: $\overrightarrow{AB}+2\overrightarrow{AD}+\overrightarrow{BC}-2\overrightarrow{DC}$.Using the triangle law of vector additio

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-$6$

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(A) Given the expression: $\overrightarrow{AB}+2\overrightarrow{AD}+\overrightarrow{BC}-2\overrightarrow{DC}$.Using the triangle law of vector addition, $\overrightarrow{AB}+\overrightarrow{BC} = \overrightarrow{AC}$.So, the expression becomes $\overrightarrow{AC}+2\overrightarrow{AD}-2\overrightarrow{DC}$.Since $\overrightarrow{AD} = \overrightarrow{AC}+\overrightarrow{CD}$, we have:$\overrightarrow{AC}+2(\overrightarrow{AC}+\overrightarrow{CD})-2\overrightarrow{DC} = \overrightarrow{AC}+2\overrightarrow{AC}+2\overrightarrow{CD}+2\overrightarrow{CD} = 3\overrightarrow{AC}+4\overrightarrow{CD} = 3\overrightarrow{AC}-4\overrightarrow{DC}$.Given $Q$ is the midpoint of $AC$, $\overrightarrow{AC} = 2\overrightarrow{QC}$.Given $P$ divides $DC$ in ratio $1:2$, $\overrightarrow{DP} = \frac{1}{3}\overrightarrow{DC}$ and $\overrightarrow{PC} = \frac{2}{3}\overrightarrow{DC}$, so $\overrightarrow{DC} = \frac{3}{2}\overrightarrow{PC}$.Substituting these: $3(2\overrightarrow{QC}) - 4(\frac{3}{2}\overrightarrow{PC}) = 6\overrightarrow{QC} - 6\overrightarrow{PC} = 6(\overrightarrow{QC}+\overrightarrow{CP}) = 6\overrightarrow{QP}$.Since $\overrightarrow{QP} = -\overrightarrow{PQ}$, we have $6\overrightarrow{QP} = -6\overrightarrow{PQ}$.Comparing with $k\overrightarrow{PQ}$, we get $k = -6$.

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 $f(x)=\begin{cases} -3ax^2-2, & x < 1 \\ bx+a^2, & x \geq 1 \end{cases}$ 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 $(3-3\omega+2\omega^2)^{4n+3} + (2+3\omega-3\omega^2)^{4n+3} + (-3+2\omega+3\omega^2)^{4n+3}=0$,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)	$(S)$ $4$
	$(T)$ $5$

In a quadrilateral $ABCD$, the point $P$ divides $DC$ in the ratio $1:2$ and $Q$ is the midpoint of $AC$. If $\overrightarrow{AB}+2\overrightarrow{AD}+\overrightarrow{BC}-2\overrightarrow{DC}=k\overrightarrow{PQ}$, then $k$ is equal to

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