Q.Find the values of $a, b, c,$ and $d$ from the following equation: $\begin{bmatrix} 2a+b & a-2b \\ 5c-d & 4c+3d \end{bmatrix} = \begin{bmatrix} 4 & -3 \\ 11 & 24 \end{bmatrix}$

Showing the 12 most recent of 21 on this concept.

1. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
   Q.If $\left(\dfrac{9}{4}, \dfrac{5}{4}, \dfrac{15}{4}\right)$ is the centroid of a tetrahedron whose vertices are $(a, 2, 1)$, $(1, b, 4)$, $(4, 0, c)$ and $(1, 1, 7)$, then (A) $a = b = c$ (B) $a = b = c + 1$ (C) $b = c = a + 1$ (D) $a = c = b + 1$

   ›Reveal solutionSolution

   The centroid of a tetrahedron is the average of its four vertices. Equating the given centroid to the average yields three equations, which give $a=2$, $b=3$, $c=4$, so $b = c = a + 1$.

   The centroid of a tetrahedron is not the same as the centroid of a triangle — but the idea is similar. For a triangle, the centroid is the average of the three vertices. For a tetrahedron (a 3D solid with four vertices), the centroid is simply the average of the coordinates of its four vertices. That’s the key concept: if the vertices are $A$, $B$, $C$, $D$, then the centroid $G$ is

   $G = \left( \frac{x_A + x_B + x_C + x_D}{4}, \frac{y_A + y_B + y_C + y_D}{4}, \frac{z_A + z_B + z_C + z_D}{4} \right).$

   We are told this centroid equals $\left(\frac{9}{4}, \frac{5}{4}, \frac{15}{4}\right)$. So we just match coordinates and solve.

   1. Set up the x-coordinate equation. The x-coordinates of the vertices are $a$, $1$, $4$, $1$. Their sum is $a + 1 + 4 + 1 = a + 6$. The centroid’s x-coordinate is $\frac{a+6}{4}$, and this must equal $\frac{9}{4}$.

   $\frac{a+6}{4} = \frac{9}{4} \quad\Rightarrow\quad a+6 = 9 \quad\Rightarrow\quad a = 3.$

   2. Set up the y-coordinate equation. The y-coordinates are $2$, $b$, $0$, $1$. Their sum is $2 + b + 0 + 1 = b + 3$. The centroid’s y-coordinate is $\frac{b+3}{4}$, and this must equal $\frac{5}{4}$.

   $\frac{b+3}{4} = \frac{5}{4} \quad\Rightarrow\quad b+3 = 5 \quad\Rightarrow\quad b = 2.$

   3. Set up the z-coordinate equation. …

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2. 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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3. 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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4. 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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5. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
   Q.Let $\vec{a}, \vec{b}, \vec{c}$ be unit vectors such that $2\vec{a} + 3\vec{b} + 4\vec{c} = \vec{0}$. Then $|\vec{b} \times \vec{c}| =$ (A) $\dfrac{\sqrt{15}}{8}$ (B) $\dfrac{\sqrt{15}}{16}$ (C) $\dfrac{\sqrt{15}}{4}$ (D) $\dfrac{\sqrt{15}}{2}$

   ›Reveal solutionSolution

   By manipulating the given vector equation and using the properties of unit vectors, we first find the dot product $\vec{b} \cdot \vec{c}$. Then, using the identity relating the magnitude of the cross product to the dot product, we calculate $|\vec{b} \times \vec{c}|$. The result is $\boxed{\dfrac{\sqrt{15}}{8}}$.

   The problem asks for the magnitude of the cross product of two unit vectors, $|\vec{b} \times \vec{c}|$, given a linear relationship between three unit vectors. The core idea is to use the given vector equation $2\vec{a} + 3\vec{b} + 4\vec{c} = \vec{0}$ to find the dot product $\vec{b} \cdot \vec{c}$. Once we have this dot product, we can use a fundamental identity that connects the magnitude of the cross product, the magnitudes of the individual vectors, and their dot product. Since $\vec{b}$ and $\vec{c}$ are unit vectors, their magnitudes are 1, which simplifies the calculation significantly.

   To find $\vec{b} \cdot \vec{c}$ from the given equation, we can isolate the term involving $\vec{a}$ and then take the dot product of both sides with themselves. This eliminates $\vec{a}$ from the equation and introduces dot products of $\vec{b}$ and $\vec{c}$, which is exactly what we need.

   Here is a step-by-step solution:

   1. Understand the given information:

      We are given that $\vec{a}, \vec{b}, \vec{c}$ are unit vectors. This means their magnitudes are 1:

      $|\vec{a}| = 1$

      $|\vec{b}| = 1$

      $|\vec{c}| = 1$

      We are also provided with the vector equation:

      $2\vec{a} + 3\vec{b} + 4\vec{c} = \vec{0}$

   2. Isolate a term to simplify the equation:

      To establish a relationship between $\vec{b}$ and $\vec{c}$ that involves their dot product, we can move the term containing $\vec{a}$ to one side of the equation. This allows us to eliminate $\vec{a}$ when we take the dot product of the equation with itself.

      $2\vec{a} = -(3\vec{b} + 4\vec{c})$

   3. Square both sides (take the dot product with itself):

      Taking the dot product of each side with itself is a standard technique to introduce magnitudes and dot products of vectors.

      $(2\vec{a}) \cdot (2\vec{a}) = (-(3\vec{b} + 4\vec{c})) \cdot (-(3\vec{b} + 4\vec{c}))$

      Using the property $\vec{x} \cdot \vec{x} = |\vec{x}|^2$, the left side becomes $4|\vec{a}|^2$.

      The right side simplifies to $(3\vec{b} + 4\vec{c}) \cdot (3\vec{b} + 4\vec{c})$.

      So, we have:

      $4|\vec{a}|^2 = (3\vec{b} + 4\vec{c}) \cdot (3\vec{b} + 4\vec{c})$

   4. Expand the dot product and substitute magnitudes:

      Expand the dot product on the right side using the distributive property:

      $(3\vec{b} + 4\vec{c}) \cdot (3\vec{b} + 4\vec{c}) = (3\vec{b}) \cdot (3\vec{b}) + (3\vec{b}) \cdot (4\vec{c}) + (4\vec{c}) \cdot (3\vec{b}) + (4\vec{c}) \cdot (4\vec{c})$

      $= 9(\vec{b} \cdot \vec{b}) + 12(\vec{b} \cdot \vec{c}) + 12(\vec{c} \cdot \vec{b}) + 16(\vec{c} \cdot \vec{c})$

      Since $\vec{b} \cdot \vec{b} = |\vec{b}|^2$, $\vec{c} \cdot \vec{c} = |\vec{c}|^2$, and $\vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{b}$, this simplifies to:

      $= 9|\vec{b}|^2 + 16|\vec{c}|^2 + 24(\vec{b} \cdot \vec{c})$

      Now, substitute this back into the equation from step 3:

      $4|\vec{a}|^2 = 9|\vec{b}|^2 + 16|\vec{c}|^2 + 24(\vec{b} \cdot \vec{c})$

      Since $\vec{a}, \vec{b}, \vec{c}$ are unit vectors, we substitute $|\vec{a}|=1, |\vec{b}|=1, |\vec{c}|=1$:

      $4(1)^2 = 9(1)^2 + 16(1)^2 + 24(\vec{b} \cdot \vec{c})$

      $4 = 9 + 16 + 24(\vec{b} \cdot \vec{c})$

      $4 = 25 + 24(\vec{b} \cdot \vec{c})$ …

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6. 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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7. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.A, B, C, D are any four points. If E and F are mid points of AC and BD respectively, then $\overline{AB} + \overline{CB} + \overline{CD} + \overline{AD} =$ (A) $\overline{EF}$ (B) $2\overline{EF}$ (C) $3\overline{EF}$ (D) $4\overline{EF}$

   ›Reveal solutionSolution

   The sum of the four vectors equals four times the vector joining the midpoints of the diagonals, so the answer is 4 EF.

   ---

   Concept and Intuition

   We are dealing with vectors in free space — points A, B, C, D are arbitrary, and we want a neat expression for

   $\overrightarrow{AB} + \overrightarrow{CB} + \overrightarrow{CD} + \overrightarrow{AD}$

   in terms of the segment joining the midpoints of the diagonals AC and BD.

   The key trick: midpoint vectors let us rewrite each side as a difference of position vectors. Then, by grouping terms cleverly, the sum collapses into a multiple of $\overrightarrow{EF}$.

   ---

   Step‑by‑Step Solution

   1. Assign position vectors Let the position vectors of A, B, C, D be $\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}$ respectively (with respect to some origin). Then:

   $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}, \quad \overrightarrow{CB} = \mathbf{b} - \mathbf{c}, \quad \overrightarrow{CD} = \mathbf{d} - \mathbf{c}, \quad \overrightarrow{AD} = \mathbf{d} - \mathbf{a}.$

   2. Write the sum

   $S = (\mathbf{b} - \mathbf{a}) + (\mathbf{b} - \mathbf{c}) + (\mathbf{d} - \mathbf{c}) + (\mathbf{d} - \mathbf{a}).$

   3. Collect like terms

      • Terms with $\mathbf{b}$: $\mathbf{b} + \mathbf{b} = 2\mathbf{b}$
      • Terms with $\mathbf{d}$: $\mathbf{d} + \mathbf{d} = 2\mathbf{d}$
      • Terms with $-\mathbf{a}$: $-\mathbf{a} - \mathbf{a} = -2\mathbf{a}$
      • Terms with $-\mathbf{c}$: $-\mathbf{c} - \mathbf{c} = -2\mathbf{c}$

      So

   $S = 2\mathbf{b} + 2\mathbf{d} - 2\mathbf{a} - 2\mathbf{c} = 2\bigl[(\mathbf{b} + \mathbf{d}) - (\mathbf{a} + \mathbf{c})\bigr].$

   4. Introduce midpoints E is the midpoint of AC, so $\mathbf{e} = \frac{\mathbf{a} + \mathbf{c}}{2}.$ …

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8. TG EAPCET 2023Set eng-2023-05-13-FN1 markMCQ
   Q.Let $\vec{a} = \lambda \hat{i} + 3\hat{j} + 4\hat{k}$, $\vec{b} = 3\hat{i} - \hat{j} + \lambda \hat{k}$ and $\vec{c} = \lambda \hat{i} + \hat{j} - 3\hat{k}$ be three vectors for some integer $\lambda$. If the volume of the parallelepiped with $\vec{a}, \vec{b}, \vec{c}$ as coterminous edges is $61$ cubic units, then the number of possible values of $\lambda$ is (A) 4 (B) 3 (C) 2 (D) 1

   ›Reveal solutionSolution

   The volume of a parallelepiped is the absolute value of the scalar triple product. Setting $|[\vec{a},\vec{b},\vec{c}]| = 61$ gives a cubic equation in $\lambda$; counting its integer solutions yields the answer.

   Concept & Intuition

   The volume of a parallelepiped formed by three vectors is the absolute value of their scalar triple product:

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

   This is also the absolute value of the determinant of the matrix whose rows (or columns) are the vectors.

   We are told $V = 61$, so we compute the determinant, set its absolute value equal to 61, and solve for the integer $\lambda$. The number of integer solutions is what we count.

   Step-by-step solution

   1. Write the vectors in component form

   $\vec{a} = (\lambda,\;3,\;4),\quad \vec{b} = (3,\;-1,\;\lambda),\quad \vec{c} = (\lambda,\;1,\;-3).$

   2. Compute the scalar triple product The scalar triple product $[\vec{a},\vec{b},\vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c})$ equals the determinant:

   $\Delta = \begin{vmatrix} \lambda & 3 & 4 \\ 3 & -1 & \lambda \\ \lambda & 1 & -3 \end{vmatrix}.$

   Expand along the first row:

   $\Delta = \lambda \begin{vmatrix} -1 & \lambda \\ 1 & -3 \end{vmatrix} - 3 \begin{vmatrix} 3 & \lambda \\ \lambda & -3 \end{vmatrix} + 4 \begin{vmatrix} 3 & -1 \\ \lambda & 1 \end{vmatrix}.$

   Compute each minor:

   $\begin{vmatrix} -1 & \lambda \\ 1 & -3 \end{vmatrix} = (-1)(-3) - (\lambda)(1) = 3 - \lambda,$

   $\begin{vmatrix} 3 & \lambda \\ \lambda & -3 \end{vmatrix} = (3)(-3) - (\lambda)(\lambda) = -9 - \lambda^2,$

   $\begin{vmatrix} 3 & -1 \\ \lambda & 1 \end{vmatrix} = (3)(1) - (-1)(\lambda) = 3 + \lambda.$

   Substitute back:

   $\Delta = \lambda(3 - \lambda) - 3(-9 - \lambda^2) + 4(3 + \lambda).$

   3. Simplify the expression

   $\Delta = 3\lambda - \lambda^2 + 27 + 3\lambda^2 + 12 + 4\lambda.$

   Combine like terms:

   $\Delta = (-\lambda^2 + 3\lambda^2) + (3\lambda + 4\lambda) + (27 + 12) = 2\lambda^2 + 7\lambda + 39.$

   4. Set the absolute value equal to 61 Volume is $|\Delta| = 61$, so:

   $|2\lambda^2 + 7\lambda + 39| = 61.$

   This gives two cases:

   $2\lambda^2 + 7\lambda + 39 = 61 \quad \text{or} \quad 2\lambda^2 + 7\lambda + 39 = -61.$

   5. Solve each equation
      • First equation: $$$$

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9. 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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10. 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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11. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
    Q.Let $\vec{a}$ be a vector in the plane containing vectors $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{c} = 2\hat{i} - \hat{j} + \hat{k}$. If $\vec{a}$ is perpendicular to $\hat{i} + \hat{j} + 3\hat{k}$ and its projection on $\vec{b}$ is $3\sqrt{6}$, then $|\vec{a}|^2 =$ (A) 186 (B) 36 (C) 128 (D) 264

    ›Reveal solutionSolution

    Writing $\vec a=\alpha\vec b+\beta\vec c$, the perpendicular and projection conditions give $\alpha=4,\ \beta=-6$, so $\vec a=(-8,14,-2)$ and $|\vec a|^2=264$ — option (D).

    Plane condition. Since $\vec a$ lies in the plane of $\vec b=(1,2,1)$ and $\vec c=(2,-1,1)$,

    $\vec a=\alpha\vec b+\beta\vec c=(\alpha+2\beta,\ 2\alpha-\beta,\ \alpha+\beta).$

    Perpendicular to $\vec d=(1,1,3)$: $\vec a\cdot\vec d=0$ gives

    $(\alpha+2\beta)+(2\alpha-\beta)+3(\alpha+\beta)=6\alpha+4\beta=0\ \Rightarrow\ 3\alpha+2\beta=0.$

    Projection on $\vec b$: $\dfrac{\vec a\cdot\vec b}{|\vec b|}=3\sqrt6$ with $|\vec b|=\sqrt6$, so $\vec a\cdot\vec b=18$. Using $\vec b\cdot\vec b=6$ and $\vec c\cdot\vec b=1$: …

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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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