Let OA = a, OB = b be two non-collinear vecto | vedclass

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(A) Given $OA = a, OB = b$. $A^{\prime}O = OA \implies OA^{\prime} = -a$. $B^{\prime}O = OB \implies OB^{\prime} = -b$.For point $P$, $OP = -\frac{3}{

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$P$ lies inside the $	riangle A^{\prime}OB$ and $Q$ lies outside the $	riangle AOB$

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(A) Given $OA = a, OB = b$. $A^{\prime}O = OA \implies OA^{\prime} = -a$. $B^{\prime}O = OB \implies OB^{\prime} = -b$.For point $P$, $OP = -\frac{3}{4}a + \frac{7}{4}b = \frac{7b - 3a}{4}$. Since the coefficients are $x_1 = -\frac{3}{4}$ and $y_1 = \frac{7}{4}$, and $x_1 + y_1 = 1$, $P$ lies on the line $AB$. Specifically, $P$ divides $AB$ externally in the ratio $7:3$. In the coordinate system defined by vectors $a$ and $b$, $P$ lies in the region where $x  0$, which corresponds to the interior of $	riangle A^{\prime}OB$.For point $Q$, $OQ = \frac{1}{3}a + \frac{5}{3}b = 2(\frac{1}{6}a + \frac{5}{6}b)$. Since the sum of coefficients $\frac{1}{3} + \frac{5}{3} = 2 > 1$, $Q$ lies outside the $	riangle AOB$.

Let $OA = a, OB = b$ be two non-collinear vectors, $OP = x_1 a + y_1 b, OQ = x_2 a + y_2 b$ and $A^{\prime}O = OA, B^{\prime}O = OB$. If $x_1 = -\frac{3}{4}, x_2 = \frac{1}{3}, y_1 = \frac{7}{4}, y_2 = \frac{5}{3}$, then

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