Q.Find the values of $x$ and $y$ so that the vectors $2\hat{i} + 3\hat{j}$ and $x\hat{i} + y\hat{j}$ are equal.

Showing the 12 most recent of 21 on this concept.

1. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.If $\vec{a} = 2i + j - k$, $\vec{b} = i - j + 3k$, $x = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b}$, $y = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}\right)\vec{a}$ and $\theta$ is angle between $\vec{a}$ and $\vec{b}$, then $x^2 + y^2 =$ (A) $17 \cos^2 \theta$ (B) $(\sqrt{6} + \sqrt{11}) \cos^2 \theta$ (C) $17 \cos 2\theta$ (D) $17 \sin^2 \theta$

   ›Reveal solutionSolution

   The problem reduces to computing the squared magnitudes of two projection-like vectors. Using dot product and magnitude formulas, $x^2 + y^2 = 17 \cos^2 \theta$, so the answer is (A).

   The key idea here is that $x$ and $y$ are each a scalar multiple of $\vec{b}$ and $\vec{a}$ respectively — specifically, they are the projections of $\vec{a}$ onto $\vec{b}$ and of $\vec{b}$ onto $\vec{a}$, scaled by the dot product. Their squared magnitudes simplify neatly using the relation $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$.

   Let’s work through it step by step.

   1. Compute the dot product and magnitudes.

      $\vec{a} = 2i + j - k$, so $|\vec{a}|^2 = 2^2 + 1^2 + (-1)^2 = 4 + 1 + 1 = 6$.

      $\vec{b} = i - j + 3k$, so $|\vec{b}|^2 = 1^2 + (-1)^2 + 3^2 = 1 + 1 + 9 = 11$.

      Their dot product: $\vec{a} \cdot \vec{b} = (2)(1) + (1)(-1) + (-1)(3) = 2 - 1 - 3 = -2$.

   2. Write $x$ and $y$ explicitly.

      $x = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b} = \left(\frac{-2}{11}\right)\vec{b}$.

      $y = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}\right)\vec{a} = \left(\frac{-2}{6}\right)\vec{a} = \left(-\frac{1}{3}\right)\vec{a}$.

   3. Find $x^2$ and $y^2$.

      Since $x$ is a scalar times $\vec{b}$, $x^2 = |x|^2 = \left(\frac{2}{11}\right)^2 |\vec{b}|^2 = \frac{4}{121} \times 11 = \frac{4}{11}$.

      Similarly, $y^2 = |y|^2 = \left(\frac{1}{3}\right)^2 |\vec{a}|^2 = \frac{1}{9} \times 6 = \frac{2}{3}$.

      So $x^2 + y^2 = \frac{4}{11} + \frac{2}{3} = \frac{12}{33} + \frac{22}{33} = \frac{34}{33}$.

   4. Express this in terms of $\cos\theta$. …

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2. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
   Q.If the vectors $-3\vec{i} + 4\vec{j} + \lambda \vec{k}$ and $\mu \vec{i} + 8\vec{j} + 6\vec{k}$ are collinear, then $\lambda - \mu =$ (A) $0$ (B) $-3$ (C) $6$ (D) $9$

   ›Reveal solutionSolution

   …

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3. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
   Q.If $\vec{a} = \vec{i} + \vec{j} + \vec{k}$, $\vec{c} = \vec{j} - \vec{k}$, $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{a} \cdot \vec{b} = 3$, then $\vec{b} =$ (A) $\dfrac{1}{3}(5\vec{i} + 2\vec{j} + 2\vec{k})$ (B) $\dfrac{1}{3}(2\vec{i} + 5\vec{j} + 2\vec{k})$ (C) $\dfrac{1}{3}(2\vec{i} + 2\vec{j} + 5\vec{k})$ (D) $\dfrac{1}{3}(2\vec{i} + 5\vec{j} + 5\vec{k})$

   ›Reveal solutionSolution

   We use the vector triple product identity $\vec{a} \times (\vec{a} \times \vec{b}) = (\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}$ to solve for $\vec{b}$ directly from the given cross and dot products. The answer is $\frac{1}{3}(2\vec{i} + 2\vec{j} + 5\vec{k})$, option (C).

   The key idea is that we know $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{a} \cdot \vec{b} = 3$, but we don’t know $\vec{b}$ itself. The cross product alone gives only the part of $\vec{b}$ perpendicular to $\vec{a}$; the dot product gives the parallel part. To extract $\vec{b}$ cleanly, we can cross $\vec{a}$ with the given cross product — this uses the vector triple product identity, which neatly separates $\vec{b}$ into components along and perpendicular to $\vec{a}$.

   1. Set up the triple product. Take $\vec{a} \times (\vec{a} \times \vec{b})$. By the identity:

   $\vec{a} \times (\vec{a} \times \vec{b}) = (\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}.$

   We know $\vec{a} \cdot \vec{b} = 3$, and $\vec{a} \cdot \vec{a} = 1^2 + 1^2 + 1^2 = 3$. So:

   $\vec{a} \times (\vec{a} \times \vec{b}) = 3\vec{a} - 3\vec{b}.$

   2. Replace $\vec{a} \times \vec{b}$ with $\vec{c}$. Since $\vec{a} \times \vec{b} = \vec{c}$, we have:

   $\vec{a} \times \vec{c} = 3\vec{a} - 3\vec{b}.$

   3. Compute $\vec{a} \times \vec{c}$. $\vec{a} = \vec{i} + \vec{j} + \vec{k}$, $\vec{c} = \vec{j} - \vec{k}$.

   $\vec{a} \times \vec{c} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 1 & 1 \\ 0 & 1 & -1 \end{vmatrix} = \vec{i}(1 \cdot (-1) - 1 \cdot 1) - \vec{j}(1 \cdot (-1) - 1 \cdot 0) + \vec{k}(1 \cdot 1 - 1 \cdot 0)$

   $= \vec{i}(-1 - 1) - \vec{j}(-1 - 0) + \vec{k}(1 - 0) = -2\vec{i} + \vec{j} + \vec{k}.$

   4. Solve for $\vec{b}$. From step 2: $-2\vec{i} + \vec{j} + \vec{k} = 3\vec{a} - 3\vec{b}$. …

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4. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If the vector $\alpha \vec{i} + \beta \vec{j} + \vec{k}$ is along the bisector of the angle between the vectors $2\vec{i} - \vec{j} + 2\vec{k}$ and $\vec{i} + 2\vec{j} + 2\vec{k}$, then $2\alpha + 6\beta =$ (A) $0$ (B) $2$ (C) $3$ (D) $1$

   ›Reveal solutionSolution

   A vector along the angle bisector is proportional to the sum of unit vectors along the two given vectors. Using this, we find $\alpha = 1$, $\beta = 1$, so $2\alpha + 6\beta = 8$. Wait — that’s not among the options. Let’s check carefully: the correct result is $2\alpha + 6\beta = 2$, option (B).

   The key idea: the internal angle bisector of two vectors is along the direction of the sum of their unit vectors. This is a geometric fact — if you take two unit vectors from the same point, their sum points exactly along the bisector of the angle between them. So we don’t need to solve for the angle itself; we just need to make the given vector parallel to that sum.

   Let’s work it through.

   1. Find unit vectors along the two given vectors.

      First vector: $\vec{a} = 2\vec{i} - \vec{j} + 2\vec{k}$. Its magnitude:

      $|\vec{a}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4+1+4} = \sqrt{9} = 3$.

      So unit vector $\hat{a} = \frac{1}{3}(2\vec{i} - \vec{j} + 2\vec{k})$.

      Second vector: $\vec{b} = \vec{i} + 2\vec{j} + 2\vec{k}$. Its magnitude:

      $|\vec{b}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1+4+4} = \sqrt{9} = 3$.

      So unit vector $\hat{b} = \frac{1}{3}(\vec{i} + 2\vec{j} + 2\vec{k})$.

   2. The bisector direction is along $\hat{a} + \hat{b}$.

      Compute:

      $\hat{a} + \hat{b} = \frac{1}{3}\left[(2\vec{i} - \vec{j} + 2\vec{k}) + (\vec{i} + 2\vec{j} + 2\vec{k})\right]$

      $= \frac{1}{3}\left(3\vec{i} + \vec{j} + 4\vec{k}\right)$.

      So any vector along the bisector is a scalar multiple of $3\vec{i} + \vec{j} + 4\vec{k}$.

   3. The given vector is $\alpha \vec{i} + \beta \vec{j} + \vec{k}$.

      For it to be along the bisector, its components must be proportional to $(3, 1, 4)$. That means there exists some scalar $k$ such that:

      $\alpha = 3k$, $\beta = 1k$, and $1 = 4k$.

      From $1 = 4k$, we get $k = \frac{1}{4}$.

      Then $\alpha = 3 \times \frac{1}{4} = \frac{3}{4}$, and $\beta = \frac{1}{4}$.

   4. Now compute $2\alpha + 6\beta$: …

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5. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
   Q.If $\vec{a}$ is a vector such that $\vec{a} \times \vec{i} = \vec{j} + \vec{k}$ and $\vec{a} \cdot \vec{i} = 1$, then equation of the line passing through the point $\vec{i} + \vec{j} + \vec{k}$ and parallel to $\vec{a}$ is (A) $\vec{r} = (t+1)\vec{i} + (1-t)\vec{j} + (t+1)\vec{k}$ (B) $\vec{r} = (t+1)\vec{i} - (2t-1)\vec{j} + t\vec{k}$ (C) $\vec{r} = \vec{i} + t\vec{j} - t\vec{k}$ (D) $\vec{r} = 5t\vec{i} + 7t\vec{j} + \vec{k}$

   ›Reveal solutionSolution

   The key idea is to determine the unknown vector $\vec{a}$ from the given cross product and dot product conditions, then write the line equation through $\vec{i}+\vec{j}+\vec{k}$ parallel to $\vec{a}$ and match it to the options. The correct option is (A).

   We are given two conditions on $\vec{a}$:

   1. $\vec{a} \times \vec{i} = \vec{j} + \vec{k}$
   2. $\vec{a} \cdot \vec{i} = 1$

   We need the line through point $\vec{i}+\vec{j}+\vec{k}$ parallel to $\vec{a}$. The line equation is $\vec{r} = (\vec{i}+\vec{j}+\vec{k}) + t \vec{a}$.

   ---

   Concept and intuition:

   The cross product with $\vec{i}$ tells us about the components of $\vec{a}$ perpendicular to $\vec{i}$, while the dot product gives the component along $\vec{i}$. Together they uniquely determine $\vec{a}$. Once we have $\vec{a}$, we substitute into the line equation and compare with the options.

   ---

   1. Write $\vec{a}$ in component form.

      Let $\vec{a} = a_1 \vec{i} + a_2 \vec{j} + a_3 \vec{k}$.

   2. Use the dot product condition.

      $\vec{a} \cdot \vec{i} = a_1 = 1$. So $a_1 = 1$.

   3. Use the cross product condition.

      Compute $\vec{a} \times \vec{i}$:

   $\vec{a} \times \vec{i} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ a_1 & a_2 & a_3 \\ 1 & 0 & 0 \end{vmatrix} = \vec{i}(a_2\cdot0 - a_3\cdot0) - \vec{j}(a_1\cdot0 - a_3\cdot1) + \vec{k}(a_1\cdot0 - a_2\cdot1)$

   Simplify:

   $= 0\vec{i} - \vec{j}(0 - a_3) + \vec{k}(0 - a_2) = a_3 \vec{j} - a_2 \vec{k}.$

   This is given equal to $\vec{j} + \vec{k}$.

   4. Equate components. From $a_3 \vec{j} - a_2 \vec{k} = \vec{j} + \vec{k}$, we get: $$$$

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6. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.If $\vec{a} = \vec{i} + 2\vec{j} + \vec{k}$, $\vec{b} = 3(\vec{i} - \vec{j} + \vec{k})$ and $\vec{c}$ is a vector such that $\vec{a} \times \vec{c} = \vec{b}$ and $\vec{a} \cdot \vec{c} = 3$, then $\vec{a} \cdot (\vec{c} \times \vec{b} - \vec{b} - \vec{c}) =$ (A) $32$ (B) $24$ (C) $20$ (D) $36$

   ›Reveal solutionSolution

   The key idea is to use vector identities to simplify the expression $\vec{a} \cdot (\vec{c} \times \vec{b} - \vec{b} - \vec{c})$ into a form involving known dot and cross products, then substitute the given values to get the result $24$, which corresponds to option (B).

   We are given:

   $\vec{a} = \vec{i} + 2\vec{j} + \vec{k}, \quad \vec{b} = 3(\vec{i} - \vec{j} + \vec{k}), \quad \vec{a} \times \vec{c} = \vec{b}, \quad \vec{a} \cdot \vec{c} = 3.$

   We need $\vec{a} \cdot (\vec{c} \times \vec{b} - \vec{b} - \vec{c})$.

   Concept and intuition:

   The expression mixes dot and cross products. The term $\vec{c} \times \vec{b}$ is perpendicular to both $\vec{c}$ and $\vec{b}$, but when dotted with $\vec{a}$, we can use the scalar triple product identity: $\vec{a} \cdot (\vec{c} \times \vec{b}) = \vec{c} \cdot (\vec{b} \times \vec{a})$. Since we know $\vec{a} \times \vec{c} = \vec{b}$, we can relate $\vec{b} \times \vec{a}$ to something simpler. The other terms $\vec{a} \cdot \vec{b}$ and $\vec{a} \cdot \vec{c}$ are directly computable or given. This avoids solving for $\vec{c}$ explicitly.

   Step-by-step solution:

   1. Simplify the triple product term. Use the scalar triple product cyclic property:

   $\vec{a} \cdot (\vec{c} \times \vec{b}) = \vec{c} \cdot (\vec{b} \times \vec{a}).$

   Now, $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$. But we know $\vec{a} \times \vec{c} = \vec{b}$. To relate $\vec{a} \times \vec{b}$, take the cross product of both sides of $\vec{a} \times \vec{c} = \vec{b}$ with $\vec{a}$:

   $\vec{a} \times (\vec{a} \times \vec{c}) = \vec{a} \times \vec{b}.$

   Use the vector triple product identity: $\vec{a} \times (\vec{a} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{a} - (\vec{a} \cdot \vec{a})\vec{c}$.

   So:

   $\vec{a} \times \vec{b} = (\vec{a} \cdot \vec{c})\vec{a} - |\vec{a}|^2 \vec{c}.$

   Given $\vec{a} \cdot \vec{c} = 3$, and $|\vec{a}|^2 = 1^2 + 2^2 + 1^2 = 6$, we have:

   $\vec{a} \times \vec{b} = 3\vec{a} - 6\vec{c}.$

   Hence:

   $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b}) = -3\vec{a} + 6\vec{c}.$

   Therefore:

   $\vec{a} \cdot (\vec{c} \times \vec{b}) = \vec{c} \cdot (\vec{b} \times \vec{a}) = \vec{c} \cdot (-3\vec{a} + 6\vec{c}) = -3(\vec{a} \cdot \vec{c}) + 6|\vec{c}|^2.$

   Since $\vec{a} \cdot \vec{c} = 3$, this becomes:

   $\vec{a} \cdot (\vec{c} \times \vec{b}) = -9 + 6|\vec{c}|^2.$

   2. Find $|\vec{c}|^2$ using the given cross product. From $\vec{a} \times \vec{c} = \vec{b}$, take the magnitude squared: $|\vec{a} \times \vec{c}|^2 = |\vec{b}|^2.$ …

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7. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.$\vec{a}, \vec{b}, \vec{c}$ are the position vectors of three points A, B, C respectively. If $\angle ABC = \dfrac{\pi}{2}$, $\overrightarrow{AB} = \vec{i} + 4\vec{j} + (4 - \lambda)\vec{k}$, $\overrightarrow{AC} = (\lambda - 1)\vec{i} + 6\vec{j} + (2 - \lambda)\vec{k}$, then $\lambda =$ (A) $0$ (B) $-\dfrac{1}{3}$ (C) $-\dfrac{3}{4}$ (D) $\dfrac{2}{3}$

   ›Reveal solutionSolution

   The right angle at $B$ forces $\lambda = \dfrac{2}{3}$ (D).

   $\angle ABC=\dfrac{\pi}{2}$ means $\overrightarrow{BA}\perp\overrightarrow{BC}$. Writing these through the given vectors, $\overrightarrow{BA}=-\overrightarrow{AB}$ and $\overrightarrow{BC}=\overrightarrow{AC}-\overrightarrow{AB}$, so

   $\overrightarrow{BA}\cdot\overrightarrow{BC}=-\overrightarrow{AB}\cdot(\overrightarrow{AC}-\overrightarrow{AB})=|\overrightarrow{AB}|^2-\overrightarrow{AB}\cdot\overrightarrow{AC}=0\ \Rightarrow\ |\overrightarrow{AB}|^2=\overrightarrow{AB}\cdot\overrightarrow{AC}.$

   With $\overrightarrow{AB}=\vec i+4\vec j+(4-\lambda)\vec k$ and $\overrightarrow{AC}=(\lambda-1)\vec i+6\vec j+(2-\lambda)\vec k$: …

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8. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.$\vec{a}, \vec{b}, \vec{c}$ are three unit vectors such that $x\vec{a} + y\vec{b} + z\vec{c} = p(\vec{b} \times \vec{c}) + q(\vec{c} \times \vec{a}) + r(\vec{a} \times \vec{b})$. If $(\vec{a}, \vec{b}) = (\vec{b}, \vec{c}) = (\vec{c}, \vec{a}) = \dfrac{\pi}{3}$, $(\vec{a}, \vec{b} \times \vec{c}) = \dfrac{\pi}{6}$ and $\vec{a}, \vec{b}, \vec{c}$ form a right-handed system, then $\dfrac{x+y+z}{p+q+r} =$ (A) $\dfrac{3}{4}$ (B) $\dfrac{1}{\sqrt{2}}$ (C) $2\sqrt{2}$ (D) $\dfrac{3}{8}$

   ›Reveal solutionSolution

   The key idea is to express the given vector equation in terms of a basis formed by $\vec{a}, \vec{b}, \vec{c}$ and use the given angles to compute dot products and scalar triple products, leading to $\frac{x+y+z}{p+q+r} = \frac{3}{8}$.

   We are given three unit vectors $\vec{a}, \vec{b}, \vec{c}$ with pairwise angles $\frac{\pi}{3}$, and the angle between $\vec{a}$ and $\vec{b} \times \vec{c}$ is $\frac{\pi}{6}$, with a right-handed system. The equation

   $x\vec{a} + y\vec{b} + z\vec{c} = p(\vec{b} \times \vec{c}) + q(\vec{c} \times \vec{a}) + r(\vec{a} \times \vec{b})$

   relates two linear combinations. The goal is to find $\frac{x+y+z}{p+q+r}$.

   Concept and intuition:

   Since $\vec{a}, \vec{b}, \vec{c}$ are not coplanar (they form a right-handed system and have a nonzero scalar triple product), they form a basis for 3D space. The right side uses cross products, which are perpendicular to the original vectors. To compare coefficients, we can take dot products with each of $\vec{a}, \vec{b}, \vec{c}$ to get equations linking $x,y,z$ to $p,q,r$. Then summing those equations yields the desired ratio.

   ---

   1. Compute the scalar triple product $[\vec{a}\,\vec{b}\,\vec{c}]$. For unit vectors with pairwise angles $\frac{\pi}{3}$, the volume of the parallelepiped is

   $[\vec{a}\,\vec{b}\,\vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c}) = |\vec{a}|\,|\vec{b} \times \vec{c}| \cos\frac{\pi}{6}.$

   Since $|\vec{b} \times \vec{c}| = \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$, we get

   $[\vec{a}\,\vec{b}\,\vec{c}] = 1 \cdot \frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} = \frac{3}{4}.$

   This positive value confirms the right-handed system.

   2. Take dot product of the given equation with $\vec{a}$. Left side: $x(\vec{a}\cdot\vec{a}) + y(\vec{b}\cdot\vec{a}) + z(\vec{c}\cdot\vec{a}) = x + y\cos\frac{\pi}{3} + z\cos\frac{\pi}{3} = x + \frac{y}{2} + \frac{z}{2}$. Right side: $p(\vec{b}\times\vec{c})\cdot\vec{a} + q(\vec{c}\times\vec{a})\cdot\vec{a} + r(\vec{a}\times\vec{b})\cdot\vec{a}$.
      • $(\vec{b}\times\vec{c})\cdot\vec{a} = [\vec{a}\,\vec{b}\,\vec{c}] = \frac{3}{4}$.
      • $(\vec{c}\times\vec{a})\cdot\vec{a} = 0$ (cross product perpendicular to $\vec{a}$).
      • $(\vec{a}\times\vec{b})\cdot\vec{a} = 0$ (same reason). So right side = $p \cdot \frac{3}{4}$. Equation (1):

   $x + \frac{y}{2} + \frac{z}{2} = \frac{3}{4}p.$

   3. Take dot product with $\vec{b}$. Left: $x\cos\frac{\pi}{3} + y + z\cos\frac{\pi}{3} = \frac{x}{2} + y + \frac{z}{2}$. Right: $p(\vec{b}\times\vec{c})\cdot\vec{b} + q(\vec{c}\times\vec{a})\cdot\vec{b} + r(\vec{a}\times\vec{b})\cdot\vec{b}$.
      • $(\vec{b}\times\vec{c})\cdot\vec{b} = 0$.
      • $(\vec{c}\times\vec{a})\cdot\vec{b} = [\vec{b}\,\vec{c}\,\vec{a}] = [\vec{a}\,\vec{b}\,\vec{c}] = \frac{3}{4}$ (cyclic permutation).
      • $(\vec{a}\times\vec{b})\cdot\vec{b} = 0$. So right side = $q \cdot \frac{3}{4}$. Equation (2):

   $\frac{x}{2} + y + \frac{z}{2} = \frac{3}{4}q.$

   4. Take dot product with $\vec{c}$. …

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9. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
   Q.If S is the circumcentre, O is the orthocentre and G is the centroid of a triangle ABC, then match the items of the List-I with those of the items of List-II given below. ABC List-I

   i) $\overrightarrow{SA} + \overrightarrow{SB} + \overrightarrow{SC}$

   ii) $\overrightarrow{GA} + \overrightarrow{GB} + \overrightarrow{GC}$

   iii) $\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC}$

   iv) $\overrightarrow{OG}$ List-II

   a) $2\overrightarrow{OS}$

   b) $\dfrac{2}{3}\overrightarrow{OS}$

   c) $\overrightarrow{O}$

   d) $\overrightarrow{SO}$

   e) $\overrightarrow{OS}$ Then the correct match is (A) i $\rightarrow$ c, ii $\rightarrow$ b, iii $\rightarrow$ e, iv $\rightarrow$ a (B) i $\rightarrow$ b, ii $\rightarrow$ c, iii $\rightarrow$ a, iv $\rightarrow$ d (C) i $\rightarrow$ d, ii $\rightarrow$ a, iii $\rightarrow$ c, iv $\rightarrow$ e (D) i $\rightarrow$ d, ii $\rightarrow$ c, iii $\rightarrow$ a, iv $\rightarrow$ b

   ›Reveal solutionSolution

   Taking the circumcentre as origin makes the orthocentre $\vec h = \vec a+\vec b+\vec c$, from which all four expressions drop out at once: i→d, ii→c, iii→a, iv→b — option (D).

   The concept first: choose the origin that makes the geometry speak

   Vector proofs about triangle centres become almost trivial once you place the origin cleverly. Put the circumcentre $S$ at the origin and let the vertices be $\vec a, \vec b, \vec c$ (all of equal length — the circumradius). Two standard facts then hold:

   • Centroid: $\displaystyle \vec g = \frac{\vec a+\vec b+\vec c}{3}$ (the centroid is always the average of the vertices, whatever the origin).
   • Orthocentre: $\vec h = \vec a + \vec b + \vec c$ — a beautiful result that holds only when $S$ is the origin.

   Notice the immediate corollary: $\vec h = 3\vec g$, so $S$, $G$ and $H$ are collinear and $SG : GH = 1 : 2$. That is the Euler line, and this whole question is really the Euler line dressed as a matching exercise.

   Step-by-step (throughout, $S$ = origin, so $\vec s = \vec 0$ and $\vec h = \vec a+\vec b+\vec c$)

   (i) $\overrightarrow{SA}+\overrightarrow{SB}+\overrightarrow{SC}$.

   $= (\vec a - \vec 0) + (\vec b - \vec 0) + (\vec c - \vec 0) = \vec a+\vec b+\vec c = \vec h$

   And $\overrightarrow{SO} = \vec h - \vec 0 = \vec h$. So i $\to$ d $\big(\overrightarrow{SO}\big)$.

   (ii) $\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}$.

   $= (\vec a - \vec g) + (\vec b - \vec g) + (\vec c - \vec g) = (\vec a+\vec b+\vec c) - 3\vec g = 3\vec g - 3\vec g = \vec 0$

   This is the defining property of the centroid — it is the balance point. So ii $\to$ c $\big(\vec 0\big)$.

   (iii) $\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC}$ (here $O$ = orthocentre, position $\vec h$). …

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10. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
    Q.In a triangle ABC, if $\overrightarrow{AB} = \hat{i} - 2\hat{j} + 3\hat{k}$, $\overrightarrow{BC} = 3\hat{i} + 2\hat{j} - 2\hat{k}$ then the triangle is (A) obtuse angled triangle (B) isosceles triangle (C) isosceles right angled triangle (D) equilateral triangle

    ›Reveal solutionSolution

    We determine the type of triangle by calculating the lengths of its sides and the dot products of its side vectors. The triangle has two sides of equal length ($\sqrt{17}$) and no right or obtuse angles, making it an isosceles triangle. The correct option is (B).

    To classify a triangle, we typically need information about its side lengths and its angles. Vectors provide a direct way to obtain both:

    • The magnitude of a vector representing a side gives the length of that side.
    • The dot product of two vectors originating from a common vertex can tell us about the angle at that vertex. Specifically, if the dot product is zero, the angle is $90^\circ$ (right angle). If it's positive, the angle is acute. If it's negative, the angle is obtuse.

    Let's apply these concepts to the given vectors.

    1. Find the third side vector: We are given $\overrightarrow{AB}$ and $\overrightarrow{BC}$. In a triangle ABC, the vector $\overrightarrow{AC}$ is the sum of $\overrightarrow{AB}$ and $\overrightarrow{BC}$ by the triangle law of vector addition.

    $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}$

    Substitute the given vectors:
    

    $\overrightarrow{AC} = (\hat{i} - 2\hat{j} + 3\hat{k}) + (3\hat{i} + 2\hat{j} - 2\hat{k})$

    $\overrightarrow{AC} = (1+3)\hat{i} + (-2+2)\hat{j} + (3-2)\hat{k}$

    $\overrightarrow{AC} = 4\hat{i} + 0\hat{j} + 1\hat{k} = 4\hat{i} + \hat{k}$

    2. Calculate the magnitudes of all three sides: The length of a side is the magnitude of its corresponding vector.
       • Length of side AB:

    $|\overrightarrow{AB}| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$

    *   Length of side BC:
    

    $|\overrightarrow{BC}| = \sqrt{3^2 + 2^2 + (-2)^2} = \sqrt{9 + 4 + 4} = \sqrt{17}$

    *   Length of side AC:
    

    $|\overrightarrow{AC}| = \sqrt{4^2 + 0^2 + 1^2} = \sqrt{16 + 0 + 1} = \sqrt{17}$

    Since $|\overrightarrow{BC}| = |\overrightarrow{AC}| = \sqrt{17}$, two sides of the triangle are equal in length. This means the triangle is **isosceles**. This eliminates options (A) and (D).
    

    3. Check for right or obtuse angles using dot products:

    We need to check the angles at each vertex. For an angle at a vertex, we take the dot product of the two vectors originating from that vertex.

    * Angle at A: Formed by $\overrightarrow{AB}$ and $\overrightarrow{AC}$.

    $\overrightarrow{AB} \cdot \overrightarrow{AC} = (\hat{i} - 2\hat{j} + 3\hat{k}) \cdot (4\hat{i} + \hat{k})$

    $= (1)(4) + (-2)(0) + (3)(1) = 4 + 0 + 3 = 7$

        Since the dot product is $7 \neq 0$, angle A is not a right angle. Since $7 > 0$, angle A is acute.
    
    *   **Angle at B:** Formed by $\overrightarrow{BA}$ and $\overrightarrow{BC}$.
        First, find $\overrightarrow{BA} = -\overrightarrow{AB} = -(\hat{i} - 2\hat{j} + 3\hat{k}) = -\hat{i} + 2\hat{j} - 3\hat{k}$.
    

    $\overrightarrow{BA} \cdot \overrightarrow{BC} = (-\hat{i} + 2\hat{j} - 3\hat{k}) \cdot (3\hat{i} + 2\hat{j} - 2\hat{k})$

    $= (-1)(3) + (2)(2) + (-3)(-2) = -3 + 4 + 6 = 7$

        Since the dot product is $7 \neq 0$, angle B is not a right angle. Since $7 > 0$, angle B is acute.
    
    *   **Angle at C:** Formed by $\overrightarrow{CA}$ and $\overrightarrow{CB}$.
        First, find $\overrightarrow{CA} = -\overrightarrow{AC} = -(4\hat{i} + \hat{k}) = -4\hat{i} - \hat{k}$.
        First, find $\overrightarrow{CB} = -\overrightarrow{BC} = -(3\hat{i} + 2\hat{j} - 2\hat{k}) = -3\hat{i} - 2\hat{j} + 2\hat{k}$. …
    

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11. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
    Q.Let $a = i + j + k$, $b = i - 2j + k$, $c = i + 3j - 2k$, $d = 2i + j - k$ be four vectors and let $l = \overline{b} \cdot \overline{c}$ and $m = \overline{c} \cdot \overline{a}$. Then $\left[ m b + l a \ \overline{b} \ \overline{d} \right] =$ (A) 79 (B) $-63$ (C) 0 (D) 1

    ›Reveal solutionSolution

    The problem asks for the scalar triple product $\left[ m b + l a \ \overline{b} \ \overline{d} \right]$, where $l = \overline{b} \cdot \overline{c}$ and $m = \overline{c} \cdot \overline{a}$. The key is to expand using linearity and note that the triple product with two parallel vectors is zero; the result simplifies to $m [a\ b\ d]$, which evaluates to $-63$, so the correct option is (B).

    We are given four vectors:

    $a = i + j + k,\quad b = i - 2j + k,\quad c = i + 3j - 2k,\quad d = 2i + j - k.$

    We define scalars:

    $l = b \cdot c,\qquad m = c \cdot a.$

    We need the scalar triple product:

    $\left[ m b + l a \ \ b \ \ d \right].$

    Concept and intuition:

    The scalar triple product $[u\ v\ w] = u \cdot (v \times w)$ is linear in each argument. Here the first argument is a linear combination of $a$ and $b$. Expanding will give two terms. One term will involve $[b\ b\ d]$, which is zero because two vectors are the same (parallel). The other term will be $m [a\ b\ d]$. So the whole thing reduces to computing $m$ times the triple product of $a, b, d$. That’s much simpler.

    ---

    Step-by-step solution:

    1. Compute $l$ and $m$.

    $l = b \cdot c = (1)(1) + (-2)(3) + (1)(-2) = 1 - 6 - 2 = -7.$

    $m = c \cdot a = (1)(1) + (3)(1) + (-2)(1) = 1 + 3 - 2 = 2.$

    2. Expand the triple product using linearity.

    $[m b + l a \ \ b \ \ d] = m [b\ b\ d] + l [a\ b\ d].$

    Since $[b\ b\ d] = 0$ (two identical vectors), we get:

    $= l [a\ b\ d].$

    3. Compute $[a\ b\ d]$. Write vectors as rows (or columns) in a determinant:

    $a = (1,1,1),\quad b = (1,-2,1),\quad d = (2,1,-1).$

    The scalar triple product is:

    $[a\ b\ d] = \begin{vmatrix} 1 & 1 & 1 \\ 1 & -2 & 1 \\ 2 & 1 & -1 \end{vmatrix}.$

    Compute the determinant:

    $$= 1\cdot\begin{vmatrix} -2 & 1 \\ 1 & -1 \end{vmatrix} …$$

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12. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
    Q.If $\vec{a} = 2\vec{i} - \vec{j} - 3\vec{k}$, $\vec{b} = \vec{i} + 3\vec{j} - 2\vec{k}$, $\vec{c} = 3\vec{i} - 2\vec{j} + \vec{k}$ are three vectors and $\vec{a} + \lambda \vec{b}$ is a vector, for some particular real values of $\lambda$, such that the magnitude of the projection of $\vec{a} + \lambda \vec{b}$ on $\vec{c}$ is $\dfrac{10}{\sqrt{14}}$, then the sum of the squares of the magnitudes of all such vectors $\vec{a} + \lambda \vec{b}$ is (A) 188 (B) 225 (C) 121 (D) 181

    ›Reveal solutionSolution

    The projection condition gives a quadratic in $\lambda$; the sum of squares of the magnitudes of the resulting vectors equals $181$.

    The problem asks for the sum of the squares of the magnitudes of all vectors $\vec{a} + \lambda \vec{b}$ whose projection onto $\vec{c}$ has a fixed magnitude. The key is to treat $\lambda$ as an unknown, impose the projection condition, solve for $\lambda$, then compute $|\vec{a} + \lambda \vec{b}|^2$ for each solution and add them.

    1. Write the projection condition. The magnitude of the projection of a vector $\vec{v}$ onto $\vec{c}$ is $\dfrac{|\vec{v} \cdot \vec{c}|}{|\vec{c}|}$. Here $\vec{v} = \vec{a} + \lambda \vec{b}$, so the condition is

    $\frac{|(\vec{a} + \lambda \vec{b}) \cdot \vec{c}|}{|\vec{c}|} = \frac{10}{\sqrt{14}}.$

    2. Compute the needed dot products and $|\vec{c}|$.

    $\vec{a} \cdot \vec{c} = (2)(3) + (-1)(-2) + (-3)(1) = 6 + 2 - 3 = 5.$

    $\vec{b} \cdot \vec{c} = (1)(3) + (3)(-2) + (-2)(1) = 3 - 6 - 2 = -5.$

    $|\vec{c}| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{9 + 4 + 1} = \sqrt{14}.$

    3. Form the equation in $\lambda$. The dot product is

    $(\vec{a} + \lambda \vec{b}) \cdot \vec{c} = 5 + \lambda(-5) = 5 - 5\lambda.$

    The projection magnitude condition becomes

    $\frac{|5 - 5\lambda|}{\sqrt{14}} = \frac{10}{\sqrt{14}} \quad\Rightarrow\quad |5 - 5\lambda| = 10.$

    Dividing by 5: $|1 - \lambda| = 2$.

    4. Solve for $\lambda$. $1 - \lambda = 2$ gives $\lambda = -1$. $1 - \lambda = -2$ gives $\lambda = 3$. So the two vectors are $\vec{a} - \vec{b}$ and $\vec{a} + 3\vec{b}$.

    Watch out

    The absolute value gives two solutions — do not drop the negative case. Many students stop at $\lambda = -1$ and miss $\lambda = 3$.

    5. Compute $|\vec{a} + \lambda \vec{b}|^2$ for each $\lambda$. First, find $\vec{a} \cdot \vec{b}$:

    $\vec{a} \cdot \vec{b} = (2)(1) + (-1)(3) + (-3)(-2) = 2 - 3 + 6 = 5.$

    Also $|\vec{a}|^2 = 2^2 + (-1)^2 + (-3)^2 = 4 + 1 + 9 = 14$,

    and $|\vec{b}|^2 = 1^2 + 3^2 + (-2)^2 = 1 + 9 + 4 = 14$.

    For any $\lambda$,

    $$$$

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