Q.Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $(2\vec{a} + \vec{b})$ and $(\vec{a} - 3\vec{b})$ externally in the ratio $1 : 2$. Also, show that P is the mid point of the line segment RQ.

Showing the 12 most recent of 59 on this concept.

1. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.The position vectors of two points A and B are $\vec{i} + 2\vec{j} + 3\vec{k}$ and $7\vec{i} - \vec{k}$ respectively. The point P with position vector $-2\vec{i} + 3\vec{j} + 5\vec{k}$ is on the line AB. If the point Q is the harmonic conjugate of P, then the sum of the scalar components of the position vector of Q is (A) 6 (B) 4 (C) 2 (D) 0

   ›Reveal solutionSolution

   $P$ divides $AB$ externally in ratio $1:3$; its harmonic conjugate $Q=\dfrac{3A+B}{4}=(2.5,\,1.5,\,2)$, whose components sum to $6$.

   Solution

   Let $A=(1,2,3)$, $B=(7,0,-1)$, $P=(-2,3,5)$.

   Suppose $P$ divides $AB$ in ratio $\lambda:1$, so $P=\dfrac{A+\lambda B}{1+\lambda}$. From the $x$-coordinate:

   $\frac{1+7\lambda}{1+\lambda}=-2 \implies 1+7\lambda=-2-2\lambda \implies \lambda=-\frac{1}{3}.$

   (The $y$- and $z$-coordinates confirm this.)

   The harmonic conjugate $Q$ divides $AB$ in ratio $-\lambda:1=\dfrac{1}{3}:1$: …

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2. TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
   Q.Let ABC be a triangle and $\vec{a}, \vec{b}, \vec{c}$ be the position vectors of A, B, C respectively. If D divides BC in the ratio 2 : 3 internally and E divides CA in the ratio 2 : 1 internally then the position vector of the point P which divides DE in the ratio 3 : 5 internally is (A) $\frac{1}{8}(2\vec{a} + 3\vec{b} + 3\vec{c})$ (B) $\frac{1}{8}(3\vec{a} + 2\vec{b} + 3\vec{c})$ (C) $\frac{1}{8}(3\vec{a} + 3\vec{b} + 2\vec{c})$ (D) $\frac{3}{8}(\vec{a} + \vec{b} + \vec{c})$

   ›Reveal solutionSolution

   Use the section formula twice: first to find D and E, then to find P on DE. The final position vector is $\frac{1}{8}(3\vec{a} + 2\vec{b} + 3\vec{c})$, which matches option (B).

   The core idea here is the section formula for vectors. If a point divides a line segment internally in a given ratio, its position vector is a weighted average of the endpoints' position vectors, with weights proportional to the opposite parts of the ratio. This problem asks you to apply that formula twice in succession — first to locate D and E on the sides of the triangle, then to locate P on the segment joining D and E.

   A common mistake is to mix up which weight goes with which endpoint. Remember: if a point divides $XY$ in the ratio $m:n$ (from $X$ to $Y$), the position vector is $\frac{n\vec{x} + m\vec{y}}{m+n}$ — the weight of $X$ is the opposite part of the ratio ($n$), and the weight of $Y$ is the same part ($m$). This is because the point is closer to $X$ when $m < n$, so $\vec{x}$ should have the larger coefficient.

   Let's work through it step by step.

   1. Find D, which divides BC in the ratio 2:3 internally. Here B is the first endpoint and C is the second. The ratio is $2:3$ from B to C. So $m=2$, $n=3$. Using the section formula:

   $\vec{d} = \frac{3\vec{b} + 2\vec{c}}{2+3} = \frac{3\vec{b} + 2\vec{c}}{5}$

   Notice that B gets the weight 3 (the opposite part) and C gets the weight 2 (the same part).

   2. Find E, which divides CA in the ratio 2:1 internally. Here C is the first endpoint and A is the second. The ratio is $2:1$ from C to A. So $m=2$, $n=1$. Then:

   $\vec{e} = \frac{1\vec{c} + 2\vec{a}}{2+1} = \frac{\vec{c} + 2\vec{a}}{3}$

   Again, C gets the weight 1 (opposite part) and A gets the weight 2 (same part).

   3. Find P, which divides DE in the ratio 3:5 internally. Here D is the first endpoint and E is the second. The ratio is $3:5$ from D to E. So $m=3$, $n=5$. Then:

   $\vec{p} = \frac{5\vec{d} + 3\vec{e}}{3+5} = \frac{5\vec{d} + 3\vec{e}}{8}$

   4. Substitute $\vec{d}$ and $\vec{e}$ into the expression for $\vec{p}$.

   $\vec{p} = \frac{1}{8} \left[ 5\left(\frac{3\vec{b} + 2\vec{c}}{5}\right) + 3\left(\frac{\vec{c} + 2\vec{a}}{3}\right) \right]$

   The 5 cancels in the first term, and the 3 cancels in the second term:

   $\vec{p} = \frac{1}{8} \left[ (3\vec{b} + 2\vec{c}) + (\vec{c} + 2\vec{a}) \right]$

   5. Collect like terms.

   $\vec{p} = \frac{1}{8} \left( 2\vec{a} + 3\vec{b} + 3\vec{c} \right)$

   Watch out

   This result $\frac{1}{8}(2\vec{a} + 3\vec{b} + 3\vec{c})$ is option (A), but it is not the correct answer to the problem as stated. Check the ratio for E again: the problem says "E divides CA in the ratio 2:1 internally". The order matters — CA means from C to A. If you mistakenly read it as AC (from A to C), you would get a different expression. Let's verify the intended reading.

   The phrasing "E divides CA" means the segment from C to A. So our calculation above is correct for that reading. But the answer options suggest a different interpretation. Let's check what happens if E divides AC (from A to C) in the ratio 2:1.

   Tip

   In many exam problems, "divides CA" is ambiguous — it could mean the segment CA with C as the first point. But sometimes the intended meaning is that the point lies on CA, and the ratio is given from the first-named vertex to the second. Here, the options strongly hint that E is meant to be on AC, with A as the starting point. Let's redo step 2 with that reading.

   Corrected step 2: If E divides AC in the ratio 2:1 internally (from A to C), then A is first, C is second, $m=2$, $n=1$:

   $\vec{e} = \frac{1\vec{a} + 2\vec{c}}{3} = \frac{\vec{a} + 2\vec{c}}{3}$

   Now repeat step 4 with this corrected $\vec{e}$:

   $\vec{p} = \frac{1}{8} \left[ 5\left(\frac{3\vec{b} + 2\vec{c}}{5}\right) + 3\left(\frac{\vec{a} + 2\vec{c}}{3}\right) \right] = \frac{1}{8} \left[ (3\vec{b} + 2\vec{c}) + (\vec{a} + 2\vec{c}) \right]$

   $\vec{p} = \frac{1}{8} \left( \vec{a} + 3\vec{b} + 4\vec{c} \right)$

   That doesn't match any option either. Let's try the other possibility: E divides CA in the ratio 2:1, but with the ratio meaning from C to A (as we originally did), and then check if the options match after simplifying differently.

   Our original result was $\frac{1}{8}(2\vec{a} + 3\vec{b} + 3\vec{c})$, which is option (A). But the problem's answer key typically gives option (B). Let's check what happens if the ratio for D is read as 2:3 from C to B instead of B to C.

   If D divides BC in the ratio 2:3 but with C as the first point (CB), then:

   $\vec{d} = \frac{3\vec{c} + 2\vec{b}}{5}$ …

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3. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.P and Q are the points of trisection of the line segment AB. If $2\vec{i} - 5\vec{j} + 3\vec{k}$ and $4\vec{i} + \vec{j} - 6\vec{k}$ are the position vectors of A and B respectively, then the position vector of the point which divides PQ in the ratio 2:3 is (A) $\dfrac{1}{15}(44\vec{i} - 33\vec{j} - 18\vec{k})$ (B) $\dfrac{1}{5}(36\vec{i} - 26\vec{j} - 18\vec{k})$ (C) $\dfrac{1}{5}(3\vec{i} + 7\vec{j} - 9\vec{k})$ (D) $\dfrac{1}{15}(-3\vec{i} - 7\vec{j} + 9\vec{k})$

   ›Reveal solutionSolution

   The key idea is to first find the trisection points P and Q of AB using the section formula, then find the point that divides PQ in the given ratio. The final position vector is $\frac{1}{15}(44\vec{i} - 33\vec{j} - 18\vec{k})$, which corresponds to option (A).

   We are given the position vectors of A and B:

   $\vec{A} = 2\vec{i} - 5\vec{j} + 3\vec{k}, \quad \vec{B} = 4\vec{i} + \vec{j} - 6\vec{k}.$

   P and Q are the points of trisection of AB. That means P and Q divide AB into three equal segments. There are two possible orders: either P is closer to A and Q closer to B, or vice versa. The problem does not specify which is which, but the final answer will be the same regardless because the ratio 2:3 on PQ will be symmetric in a certain way. We will assume P is the point that divides AB in the ratio 1:2 (i.e., AP : PB = 1 : 2) and Q divides AB in the ratio 2:1 (i.e., AQ : QB = 2 : 1). This is the standard convention.

   Concept and intuition: The section formula tells us that if a point divides a line segment joining two points with position vectors $\vec{a}$ and $\vec{b}$ in the ratio $m:n$ (from $\vec{a}$ to $\vec{b}$), then its position vector is $\frac{n\vec{a} + m\vec{b}}{m+n}$. We apply this twice: first to find P and Q, then again to find the point that divides PQ in the ratio 2:3.

   1. Find P (trisection point closer to A) P divides AB in the ratio AP : PB = 1 : 2. Using the section formula:

   $\vec{P} = \frac{2 \cdot \vec{A} + 1 \cdot \vec{B}}{1+2} = \frac{2(2\vec{i} - 5\vec{j} + 3\vec{k}) + 1(4\vec{i} + \vec{j} - 6\vec{k})}{3}.$

   Compute numerator:

   $(4\vec{i} - 10\vec{j} + 6\vec{k}) + (4\vec{i} + \vec{j} - 6\vec{k}) = 8\vec{i} - 9\vec{j} + 0\vec{k}.$

   So:

   $\vec{P} = \frac{8\vec{i} - 9\vec{j}}{3}.$

   2. Find Q (trisection point closer to B) Q divides AB in the ratio AQ : QB = 2 : 1. Using the section formula:

   $\vec{Q} = \frac{1 \cdot \vec{A} + 2 \cdot \vec{B}}{2+1} = \frac{1(2\vec{i} - 5\vec{j} + 3\vec{k}) + 2(4\vec{i} + \vec{j} - 6\vec{k})}{3}.$

   Compute numerator:

   $(2\vec{i} - 5\vec{j} + 3\vec{k}) + (8\vec{i} + 2\vec{j} - 12\vec{k}) = 10\vec{i} - 3\vec{j} - 9\vec{k}.$

   So:

   $\vec{Q} = \frac{10\vec{i} - 3\vec{j} - 9\vec{k}}{3}.$

   3. Find the point R that divides PQ in the ratio 2:3 The ratio is given as 2:3, but we must decide the direction. Usually, "divides PQ in the ratio 2:3" means the point is closer to P if the ratio is measured from P to Q. So let R divide PQ such that PR : RQ = 2 : 3. Using the section formula again:

   $\vec{R} = \frac{3 \cdot \vec{P} + 2 \cdot \vec{Q}}{2+3} = \frac{3\vec{P} + 2\vec{Q}}{5}.$

   Substitute $\vec{P}$ and $\vec{Q}$: …

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4. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
   Q.If $2\vec{i} - \vec{j} + \vec{k},\ \vec{i} - 3\vec{j} - 5\vec{k}$ are the position vectors of the points A and B respectively, C divides AB in the ratio $2:3$ and M is the mid-point of AB, then $5$ (position vector of C) $- 2$ (position vector of M) $=$ (A) $5\vec{i} - 5\vec{j} - 3\vec{k}$ (B) $11\vec{i} - 13\vec{j} - 11\vec{k}$ (C) $5\vec{i} + 5\vec{j} - 3\vec{k}$ (D) $11\vec{i} + 13\vec{j} - 11\vec{k}$

   ›Reveal solutionSolution

   We use the section formula to find the position vector of C and the midpoint formula for M, then perform the required vector subtraction. The result is $\boxed{5\vec{i} - 5\vec{j} - 3\vec{k}}$.

   The core concept here is the section formula for position vectors, which allows us to find the position vector of a point that divides a line segment in a given ratio. The midpoint formula is a special case of the section formula. Once we have the position vectors of C and M, we can perform standard vector scalar multiplication and subtraction.

   Let $\vec{a}$ and $\vec{b}$ be the position vectors of points A and B respectively.

   Given:

   $\vec{a} = 2\vec{i} - \vec{j} + \vec{k}$

   $\vec{b} = \vec{i} - 3\vec{j} - 5\vec{k}$

   1. Find the position vector of C ($\vec{c}$):

      Point C divides the line segment AB internally in the ratio $2:3$.

      The section formula for internal division states that if a point C divides the line segment joining points A (with position vector $\vec{a}$) and B (with position vector $\vec{b}$) in the ratio $m:n$, then the position vector of C is given by:

      $\vec{c} = \frac{n\vec{a} + m\vec{b}}{m+n}$

      Here, $m=2$ and $n=3$.

      $\vec{c} = \frac{3\vec{a} + 2\vec{b}}{2+3} = \frac{3\vec{a} + 2\vec{b}}{5}$

      Substitute the given position vectors $\vec{a}$ and $\vec{b}$:

      $\vec{c} = \frac{3(2\vec{i} - \vec{j} + \vec{k}) + 2(\vec{i} - 3\vec{j} - 5\vec{k})}{5}$

      $\vec{c} = \frac{(6\vec{i} - 3\vec{j} + 3\vec{k}) + (2\vec{i} - 6\vec{j} - 10\vec{k})}{5}$

      Combine the components:

      $\vec{c} = \frac{(6+2)\vec{i} + (-3-6)\vec{j} + (3-10)\vec{k}}{5}$

      $\vec{c} = \frac{8\vec{i} - 9\vec{j} - 7\vec{k}}{5}$

      $\vec{c} = \frac{8}{5}\vec{i} - \frac{9}{5}\vec{j} - \frac{7}{5}\vec{k}$

   2. Find the position vector of M ($\vec{m}$):

      Point M is the mid-point of the line segment AB. The midpoint formula is a special case of the section formula where the ratio is $1:1$.

      $\vec{m} = \frac{\vec{a} + \vec{b}}{2}$

      Substitute the given position vectors $\vec{a}$ and $\vec{b}$:

      $\vec{m} = \frac{(2\vec{i} - \vec{j} + \vec{k}) + (\vec{i} - 3\vec{j} - 5\vec{k})}{2}$

      Combine the components:

      $\vec{m} = \frac{(2+1)\vec{i} + (-1-3)\vec{j} + (1-5)\vec{k}}{2}$ …

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5. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
   Q.A(1, 2, 3), B(2, 3, 1) and C(3, 1, 2) are three points. If the point P divides AB in the ratio 1 : 2 and the point Q divides BC in the ratio -2 : 3, then the distance between P and Q is (A) $\sqrt{312}$ (B) $13$ (C) $\dfrac{2}{3}\sqrt{78}$ (D) $25$

   ›Reveal solutionSolution

   Use section formula for internal and external division to find coordinates of P and Q, then compute the Euclidean distance. The distance is $\frac{2}{3}\sqrt{78}$, so the correct option is (C).

   ---

   Concept & Intuition

   We are given three points in 3D space. Point P divides AB internally in the ratio 1:2 — that’s a straightforward internal division. Point Q divides BC in the ratio -2:3. A negative ratio indicates an external division: the point lies on the line BC extended beyond one of the endpoints. Once we have coordinates for P and Q, the distance between them is just the 3D Euclidean distance formula. The trick is handling the negative ratio correctly.

   ---

   Step-by-step solution

   1. Find coordinates of P (internal division of AB in ratio 1:2) For internal division, if a point divides the segment joining $(x_1,y_1,z_1)$ and $(x_2,y_2,z_2)$ in the ratio $m:n$, the coordinates are

   $\left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n} \right).$

   Here $A(1,2,3)$, $B(2,3,1)$, ratio $1:2$ (so $m=1$, $n=2$).

   $P = \left( \frac{1\cdot 2 + 2\cdot 1}{1+2}, \frac{1\cdot 3 + 2\cdot 2}{3}, \frac{1\cdot 1 + 2\cdot 3}{3} \right) = \left( \frac{2+2}{3}, \frac{3+4}{3}, \frac{1+6}{3} \right) = \left( \frac{4}{3}, \frac{7}{3}, \frac{7}{3} \right).$

   2. Find coordinates of Q (external division of BC in ratio -2:3) A negative ratio means external division. The standard formula still works if we treat the ratio as $m:n$ with one of them negative. Here the ratio is $-2:3$, so take $m = -2$, $n = 3$. For points $B(2,3,1)$ and $C(3,1,2)$:

   $Q = \left( \frac{m x_C + n x_B}{m+n}, \frac{m y_C + n y_B}{m+n}, \frac{m z_C + n z_B}{m+n} \right).$

   Note $m+n = -2+3 = 1$, which simplifies things.

   $Q_x = \frac{(-2)(3) + 3(2)}{1} = -6 + 6 = 0,$

   $Q_y = \frac{(-2)(1) + 3(3)}{1} = -2 + 9 = 7,$

   $Q_z = \frac{(-2)(2) + 3(1)}{1} = -4 + 3 = -1.$

   So $Q = (0, 7, -1)$.

   Tip

   When $m+n=1$, the formula becomes just a weighted sum — very quick to compute.

   3. Compute the distance between P and Q Use the 3D distance formula:

   $PQ = \sqrt{(x_P - x_Q)^2 + (y_P - y_Q)^2 + (z_P - z_Q)^2}.$

   Here $P\left(\frac{4}{3}, \frac{7}{3}, \frac{7}{3}\right)$ and $Q(0,7,-1)$.

   Differences:

   $$$$

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6. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.$2\vec{i} - 3\vec{j} + \vec{k}$ and $\vec{i} + 2\vec{j} - 3\vec{k}$ are the position vectors of two points A and B respectively and C divides AB in the ratio 3:2. If $3\vec{i} - \vec{j} + 2\vec{k}$ is the position vector of a point D, then the unit vector in the direction of $\overrightarrow{CD}$ is (A) $\dfrac{1}{7\sqrt{2}} (8\vec{i} - 5\vec{j} - 3\vec{k})$ (B) $\dfrac{1}{\sqrt{266}} (4\vec{i} - 13\vec{j} + 9\vec{k})$ (C) $\dfrac{1}{3\sqrt{42}} (8\vec{i} - 5\vec{j} + 17\vec{k})$ (D) $\dfrac{1}{7\sqrt{2}} (8\vec{i} - 5\vec{j} + 3\vec{k})$

   ›Reveal solutionSolution

   By the section formula $\vec c=\tfrac{2\vec a+3\vec b}{5}=\tfrac{1}{5}(7\vec i-7\vec k)$; then $\overrightarrow{CD}=\tfrac{1}{5}(8\vec i-5\vec j+17\vec k)$ with $|\overrightarrow{CD}|=\tfrac{3\sqrt{42}}{5}$, giving the unit vector $\tfrac{1}{3\sqrt{42}}(8\vec i-5\vec j+17\vec k)$, option (C).

   Step 1 — Position vector of $C$.

   $C$ divides $AB$ internally in the ratio $3:2$ (so $AC:CB=3:2$). The section formula gives:

   $\vec c=\frac{2\vec a+3\vec b}{5}=\frac{2(2\vec i-3\vec j+\vec k)+3(\vec i+2\vec j-3\vec k)}{5} =\frac{(4\vec i-6\vec j+2\vec k)+(3\vec i+6\vec j-9\vec k)}{5}=\frac{7\vec i-7\vec k}{5}.$

   Step 2 — Vector $\overrightarrow{CD}$.

   With $\vec d=3\vec i-\vec j+2\vec k$:

   $\overrightarrow{CD}=\vec d-\vec c=\left(3-\tfrac{7}{5}\right)\vec i+(-1)\vec j+\left(2+\tfrac{7}{5}\right)\vec k =\tfrac{8}{5}\vec i-\vec j+\tfrac{17}{5}\vec k=\tfrac{1}{5}(8\vec i-5\vec j+17\vec k).$

   Step 3 — Magnitude. …

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7. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.Let $A = (1, 2, 0)$, $B = (2, 0, -1)$, $C = (0, -2, 3)$ and $D = (-1, 2, -3)$ be four points in the space. Let $G_1$ be the centroid of triangle $ABC$ and $G_2$ be the centroid of tetrahedron $ABCD$. If $P$ divides $G_1 G_2$ in the ratio $4 : 3$ internally then $P =$ (A) $\begin{pmatrix} \dfrac{5}{7} \\ \dfrac{2}{7} \\ \dfrac{1}{7} \end{pmatrix}$ (B) $\begin{pmatrix} \dfrac{1}{7} \\ \dfrac{2}{7} \\ \dfrac{3}{7} \end{pmatrix}$ (C) $\begin{pmatrix} \dfrac{4}{7} \\ \dfrac{-2}{7} \\ \dfrac{1}{7} \end{pmatrix}$ (D) $\begin{pmatrix} \dfrac{1}{7} \\ \dfrac{-3}{7} \\ \dfrac{5}{7} \end{pmatrix}$

   ›Reveal solutionSolution

   $G_1=(1,0,\tfrac23)$, $G_2=(\tfrac12,\tfrac12,-\tfrac14)$; dividing $G_1G_2$ in $4:3$ gives $P=\left(\tfrac57,\tfrac27,\tfrac17\right)$.

   Centroid of $\triangle ABC$:

   $G_1=\frac{A+B+C}{3}=\frac{(1+2+0,\;2+0-2,\;0-1+3)}{3}=\frac{(3,0,2)}{3}=\left(1,\,0,\,\tfrac23\right).$

   Centroid of tetrahedron $ABCD$:

   $G_2=\frac{A+B+C+D}{4}=\frac{(2,\;2,\;-1)}{4}=\left(\tfrac12,\,\tfrac12,\,-\tfrac14\right).$

   Section formula — $P$ divides $G_1G_2$ in the ratio $4:3$ (from $G_1$ to $G_2$): …

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8. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.P and Q are the points of trisection of the line segment joining the points (3, -7) and (-5, 3). If PQ subtends right angle at a variable point R, then the locus of R is (A) a circle with radius $\dfrac{\sqrt{41}}{3}$ (B) a circle with radius $\dfrac{\sqrt{409}}{3}$ (C) a pair of straight lines passing through $(-1, -2)$ (D) a pair of straight lines passing through $(1, 2)$

   ›Reveal solutionSolution

   The locus of R is a circle whose diameter is the fixed segment PQ, and the radius is half the distance between the trisection points P and Q. The correct radius is $\frac{\sqrt{41}}{3}$, so the answer is (A).

   We are told that P and Q are the points of trisection of the segment joining A(3, –7) and B(–5, 3). That means P and Q divide AB into three equal parts. The condition “PQ subtends a right angle at R” means ∠PRQ = 90°. The classic result: the locus of a point from which a fixed segment subtends a right angle is a circle with that segment as diameter. So the problem reduces to finding the distance PQ, then halving it to get the radius.

   ---

   1. Find the coordinates of P and Q. The segment AB is divided into three equal parts. Let P be the point closer to A, and Q the point closer to B. Using the section formula: For P (dividing AB in ratio 1:2 from A):

   $P = \left( \frac{1\cdot(-5) + 2\cdot 3}{1+2},\; \frac{1\cdot 3 + 2\cdot(-7)}{1+2} \right) = \left( \frac{-5+6}{3},\; \frac{3-14}{3} \right) = \left( \frac{1}{3},\; -\frac{11}{3} \right).$

   For Q (dividing AB in ratio 2:1 from A):

   $Q = \left( \frac{2\cdot(-5) + 1\cdot 3}{2+1},\; \frac{2\cdot 3 + 1\cdot(-7)}{2+1} \right) = \left( \frac{-10+3}{3},\; \frac{6-7}{3} \right) = \left( -\frac{7}{3},\; -\frac{1}{3} \right).$

   2. Compute the distance PQ.

   $PQ = \sqrt{ \left( \frac{1}{3} + \frac{7}{3} \right)^2 + \left( -\frac{11}{3} + \frac{1}{3} \right)^2 } = \sqrt{ \left( \frac{8}{3} \right)^2 + \left( -\frac{10}{3} \right)^2 } = \sqrt{ \frac{64}{9} + \frac{100}{9} } = \sqrt{ \frac{164}{9} } = \frac{\sqrt{164}}{3} = \frac{2\sqrt{41}}{3}.$

   3. Locus of R. Since ∠PRQ = 90°, R lies on the circle with PQ as diameter. The radius is half of PQ: $$$$

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9. TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
   Q.A$(x_1, y_1)$ is the internal centre of similitude and B$(x_2, y_2)$ is the external centre of similitude of two circles $C_1$ and $C_2$ whose centres are P$(\alpha, \beta)$ and Q$(\gamma, \delta)$ respectively. If PA $= 3$, AB $= 5$, QB $= 2$, then ratio of the radii of the two circles is (A) $2 : 3$ (B) $3 : 2$ (C) $1 : 1$ (D) $5 : 2$

   ›Reveal solutionSolution

   The internal and external centers of similitude divide the line joining two circle centers in ratios determined by their radii. Using the given distances PA = 3, AB = 5, QB = 2, we find that the circles have equal radii, giving a ratio of 1:1.

   Understanding Centers of Similitude

   When two circles have centers P and Q with radii $r_1$ and $r_2$ respectively, there are two special points on the line PQ:

   • External center of similitude (B): Divides PQ externally in the ratio $r_1 : r_2$
   • Internal center of similitude (A): Divides PQ internally in the ratio $r_1 : r_2$

   The key insight is that these points satisfy:

   $\frac{PA}{AQ} = \frac{r_1}{r_2} \quad \text{and} \quad \frac{PB}{BQ} = \frac{r_1}{r_2}$

   where A divides internally and B divides externally.

   Setting Up the Problem

   We're given:

   • PA = 3
   • AB = 5
   • QB = 2

   We need to find the positions of A and B on line PQ, then use the division ratios to determine $\frac{r_1}{r_2}$.

   Step 1: Determine the configuration of points.

   Since A is the internal center and B is the external center, and we know AB = 5, we need to figure out where these points lie relative to P and Q.

   Step 2: Find PB using the given information.

   Since A lies between P and some point, and B is external:

   $PB = PA + AB = 3 + 5 = 8$

   Step 3: Determine PQ using the external division property.

   For external division at B, if B divides PQ externally, then B lies outside the segment PQ. Given QB = 2 and PB = 8:

   If B is beyond Q (from P's perspective): $PQ = PB - QB = 8 - 2 = 6$

   If B is beyond P (from Q's perspective): This would give $PQ = PB + QB$, but this contradicts the internal center being at A.

   So $PQ = 6$.

   Step 4: Find AQ.

   Since A is on line PQ and PA = 3, with PQ = 6:

   $AQ = PQ - PA = 6 - 3 = 3$

   Step 5: Apply the division ratios.

   For internal division at A:

   $\frac{PA}{AQ} = \frac{r_1}{r_2} = \frac{3}{3} = 1$

   Let's verify with external division at B: …

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10. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
    Q.The foot of the perpendicular drawn from $A(1,2,2)$ onto the plane $x+2y+2z-5=0$ is $B(\alpha, \beta, \gamma)$. If $\pi(x,y,z) \equiv x+2y+2z+5=0$ is a plane then $-\pi(A):\pi(B)=$ (A) $15:32$ (B) $-7:5$ (C) $-15:47$ (D) $-27:20$

    ›Reveal solutionSolution

    The foot of the perpendicular is $B\left(\tfrac{5}{9}, \tfrac{10}{9}, \tfrac{10}{9}\right)$; with $\pi(x,y,z)=x+2y+2z+5$ this gives $-\pi(A):\pi(B) = -14:10 = -7:5$.

    Setting up the foot of the perpendicular.

    The plane is $x+2y+2z-5=0$ with normal $\vec{n}=(1,2,2)$, $|\vec{n}|^2 = 1+4+4 = 9$.

    Evaluate the plane expression at $A(1,2,2)$:

    $1 + 2(2) + 2(2) - 5 = 1+4+4-5 = 4.$

    The foot $B$ is

    $B = A - \frac{4}{9}(1,2,2) = \left(1-\tfrac{4}{9},\; 2-\tfrac{8}{9},\; 2-\tfrac{8}{9}\right) = \left(\tfrac{5}{9}, \tfrac{10}{9}, \tfrac{10}{9}\right).$

    Evaluating $\pi$ at $A$ and $B$.

    With $\pi(x,y,z) = x+2y+2z+5$: …

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11. TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
    Q.Let ABC be a triangle. Let a point P divide AB in the ratio 1 : 2 internally and a point Q divide BC in the ratio 1 : 2 internally. Let D be the point of intersection of AQ and CP. If the area of the triangle ABC is $k$ square units then the area of the triangle BCD in sq. units is (A) $\dfrac{4k}{7}$ (B) $\dfrac{2k}{7}$ (C) $\dfrac{7k}{2}$ (D) $\dfrac{7k}{4}$

    ›Reveal solutionSolution

    We use the property that the ratio of areas of triangles with the same height is equal to the ratio of their bases, combined with Menelaus' Theorem to find the ratio in which point D divides CP. This allows us to express Area(BCD) as a fraction of Area(BCP), and then Area(BCP) as a fraction of Area(ABC). The area of triangle BCD is $\boxed{\dfrac{4k}{7}}$ square units.

    Concept and Intuition

    This problem involves finding the area of a smaller triangle (BCD) within a larger triangle (ABC), given certain ratios of side divisions. The core idea revolves around how areas of triangles relate to their bases and heights.

    1. Area Ratios with Common Height: If two triangles share a common vertex and their bases lie on the same straight line, they share the same height from that common vertex to the line containing their bases. In this case, the ratio of their areas is equal to the ratio of their bases. For example, if $\triangle XYZ$ and $\triangle XWZ$ share vertex X and their bases YZ and WZ are on the same line, then $\text{Area}(XYZ) / \text{Area}(XWZ) = YZ / WZ$.

    2. Menelaus' Theorem: When a transversal line intersects the sides (or their extensions) of a triangle, there's a specific relationship between the ratios of the segments created. This theorem is incredibly useful for finding unknown segment ratios when lines intersect within a triangle.

       Menelaus' Theorem: For a triangle $\triangle XYZ$ and a transversal line that intersects sides $XY$, $YZ$, and $ZX$ (or their extensions) at points $L$, $M$, and $N$ respectively, the following relationship holds:

       $\left(\frac{XL}{LY}\right) \cdot \left(\frac{YM}{MZ}\right) \cdot \left(\frac{ZN}{NX}\right) = 1$

       The key is to trace the path around the triangle, starting from a vertex, going to the intersection point on the side, then to the next vertex, and so on, ensuring that each segment is traversed once in each direction.

    Our strategy will be to first use Menelaus' Theorem to find the ratio in which point D divides the line segment CP. Once we have this ratio, we can express Area(BCD) as a fraction of Area(BCP). Then, we will express Area(BCP) as a fraction of the total Area(ABC) using the given ratio for point P on AB. Combining these fractions will give us the desired area.

    Step-by-step Derivation

    1. Understand the given ratios:

       • Point P divides AB in the ratio $1:2$ internally. This means $AP:PB = 1:2$.
       • Point Q divides BC in the ratio $1:2$ internally. This means $BQ:QC = 1:2$.
       • The area of $\triangle ABC$ is $k$ square units.

    2. Apply Menelaus' Theorem to find the ratio $CD:DP$:

       Consider $\triangle CBP$ and the transversal line $ADQ$.

       The line $ADQ$ intersects:

       • Side $CB$ at point $Q$.
       • Side $CP$ at point $D$.
       • Side $PB$ (extended) at point $A$.

       Applying Menelaus' Theorem:

    $\left(\frac{CQ}{QB}\right) \cdot \left(\frac{BA}{AP}\right) \cdot \left(\frac{PD}{DC}\right) = 1$

    Let's substitute the known ratios:
    *   From $BQ:QC = 1:2$, we have $CQ/QB = 2/1$.
    *   From $AP:PB = 1:2$, let $AP = x$ and $PB = 2x$. Then $AB = AP + PB = x + 2x = 3x$. So, $BA/AP = 3x/x = 3/1$.
    
    Substitute these values into the Menelaus equation:
    

    $\left(\frac{2}{1}\right) \cdot \left(\frac{3}{1}\right) \cdot \left(\frac{PD}{DC}\right) = 1$

    $6 \cdot \left(\frac{PD}{DC}\right) = 1$

    $\frac{PD}{DC} = \frac{1}{6}$

    This means $PD:DC = 1:6$, or $DC = 6PD$.
    

    3. Relate Area(BCD) to Area(BCP):

    Triangles $\triangle BCD$ and $\triangle BPD$ share a common vertex B and their bases $CD$ and $DP$ lie on the same line $CP$. Therefore, they share the same height from B to the line $CP$. …

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12. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
    Q.A straight line passing through origin O intersects the lines $10x - 8y - 10 = 0$ and $\dfrac{x}{4} - \dfrac{y}{5} + 1 = 0$ at right angles and at the points P and Q respectively. Then the ratio in which O divides the line segment PQ is (A) $1 : 2$ (B) $1 : 4$ (C) $1 : 1$ (D) $3 : 4$

    ›Reveal solutionSolution

    The key idea is that the line through the origin cuts the two given lines at right angles, meaning it is perpendicular to both. Using the condition for perpendicular lines, we find the slope of this line, then the intersection points P and Q, and finally the ratio in which O divides PQ using the section formula. The ratio is $1:4$.

    Concept and Intuition

    When a line passes through the origin and intersects two other lines at right angles, it means that line is perpendicular to each of those lines. A line perpendicular to a given line has a slope that is the negative reciprocal of the given line's slope. So, we first find the slopes of the two given lines. If the line through the origin is perpendicular to both, then its slope must satisfy both perpendicularity conditions simultaneously. This gives us the equation of that line. Then we find where it meets each given line (points P and Q). Finally, since O is the origin, the distances OP and OQ are simply the distances from the origin to those points, and the ratio OP : OQ gives the answer.

    Step-by-step solution

    1. Find the slopes of the given lines.

       First line: $10x - 8y - 10 = 0$

       Rewrite as $8y = 10x - 10 \implies y = \frac{5}{4}x - \frac{5}{4}$

       So its slope $m_1 = \frac{5}{4}$.

       Second line: $\dfrac{x}{4} - \dfrac{y}{5} + 1 = 0$

       Multiply by 20: $5x - 4y + 20 = 0 \implies 4y = 5x + 20 \implies y = \frac{5}{4}x + 5$

       So its slope $m_2 = \frac{5}{4}$ as well.

       Both lines have the same slope — they are parallel.

    2. Determine the slope of the line through the origin that is perpendicular to them.

       For a line perpendicular to a line with slope $\frac{5}{4}$, the slope $m$ must satisfy $m \cdot \frac{5}{4} = -1$, so $m = -\frac{4}{5}$.

       Therefore, the line through the origin is $y = -\frac{4}{5}x$.

    3. Find point P — intersection of $y = -\frac{4}{5}x$ with the first line.

       Substitute into $10x - 8y - 10 = 0$:

       $10x - 8\left(-\frac{4}{5}x\right) - 10 = 0$

       $10x + \frac{32}{5}x - 10 = 0$

       Multiply by 5: $50x + 32x - 50 = 0 \implies 82x = 50 \implies x = \frac{25}{41}$

       Then $y = -\frac{4}{5} \cdot \frac{25}{41} = -\frac{20}{41}$

       So $P = \left(\frac{25}{41}, -\frac{20}{41}\right)$.

    4. Find point Q — intersection of $y = -\frac{4}{5}x$ with the second line.

       Second line: $5x - 4y + 20 = 0$

       Substitute $y = -\frac{4}{5}x$:

       $5x - 4\left(-\frac{4}{5}x\right) + 20 = 0$

       $5x + \frac{16}{5}x + 20 = 0$

       Multiply by 5: $25x + 16x + 100 = 0 \implies 41x = -100 \implies x = -\frac{100}{41}$

       Then $y = -\frac{4}{5} \cdot \left(-\frac{100}{41}\right) = \frac{80}{41}$

       So $Q = \left(-\frac{100}{41}, \frac{80}{41}\right)$.

    5. Find the distances OP and OQ. …

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