Q.Show that the points $A(2, 3, -4)$, $B(1, -2, 3)$ and $C(3, 8, -11)$ are collinear.

Showing the 12 most recent of 69 on this concept.

1. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.The four points whose position vectors are given by $2\vec{a} + 3\vec{b} - \vec{c}$, $\vec{a} - 2\vec{b} + 3\vec{c}$, $3\vec{a} + 4\vec{b} - 2\vec{c}$ and $\vec{a} - 6\vec{b} + 6\vec{c}$ are (A) Collinear (B) Coplanar (C) Vertices of a square (D) Vertices of a rectangle

   ›Reveal solutionSolution

   The four points are coplanar because the vectors connecting them are linearly dependent, meaning they all lie in the same plane. The correct option is (B).

   Concept and Intuition

   We are given four points defined by position vectors in terms of three independent vectors $\vec{a}, \vec{b}, \vec{c}$. The key question: are these points collinear, coplanar, or forming a special quadrilateral?

   Collinearity would mean all points lie on a single line — that’s very restrictive. Coplanarity means they all lie in some plane — a much weaker condition. Since we have three basis vectors, any point is in 3D space. Four points in 3D are always coplanar if the vectors from one point to the other three are linearly dependent (i.e., one is a combination of the other two).

   We can test this by picking one point as a reference and checking if the three difference vectors are linearly dependent. If they are, the points are coplanar. If they aren’t, the points are not coplanar (they form a tetrahedron).

   Let’s do exactly that.

   ---

   Step-by-step solution

   1. Label the points

   Let

   $P_1 = 2\vec{a} + 3\vec{b} - \vec{c}, \quad P_2 = \vec{a} - 2\vec{b} + 3\vec{c}, \quad P_3 = 3\vec{a} + 4\vec{b} - 2\vec{c}, \quad P_4 = \vec{a} - 6\vec{b} + 6\vec{c}.$

   2. Choose a reference point

   Take $P_1$ as the reference. Compute the vectors from $P_1$ to the other three points:

   $\vec{v}_2 = P_2 - P_1 = (\vec{a} - 2\vec{b} + 3\vec{c}) - (2\vec{a} + 3\vec{b} - \vec{c}) = -\vec{a} - 5\vec{b} + 4\vec{c}.$

   $\vec{v}_3 = P_3 - P_1 = (3\vec{a} + 4\vec{b} - 2\vec{c}) - (2\vec{a} + 3\vec{b} - \vec{c}) = \vec{a} + \vec{b} - \vec{c}.$

   $\vec{v}_4 = P_4 - P_1 = (\vec{a} - 6\vec{b} + 6\vec{c}) - (2\vec{a} + 3\vec{b} - \vec{c}) = -\vec{a} - 9\vec{b} + 7\vec{c}.$

   3. Check linear dependence

   We ask: can $\vec{v}_4$ be written as a combination of $\vec{v}_2$ and $\vec{v}_3$? That is, do there exist scalars $\alpha, \beta$ such that

   $\vec{v}_4 = \alpha \vec{v}_2 + \beta \vec{v}_3 ?$

   Substitute:

   $-\vec{a} - 9\vec{b} + 7\vec{c} = \alpha(-\vec{a} - 5\vec{b} + 4\vec{c}) + \beta(\vec{a} + \vec{b} - \vec{c}).$

   4. Equate coefficients

   Since $\vec{a}, \vec{b}, \vec{c}$ are independent, we equate coefficients:

   • For $\vec{a}$: $-1 = -\alpha + \beta$
   • For $\vec{b}$: $-9 = -5\alpha + \beta$
   • For $\vec{c}$: $7 = 4\alpha - \beta$

   5. Solve the system

   From the first equation: $\beta = \alpha - 1$.

   Substitute into the second:

   $-9 = -5\alpha + (\alpha - 1) \Rightarrow -9 = -4\alpha - 1 \Rightarrow -8 = -4\alpha \Rightarrow \alpha = 2$.

   Then $\beta = 2 - 1 = 1$.

   Check the third equation: $4(2) - 1 = 8 - 1 = 7$, which matches.

   So $\vec{v}_4 = 2\vec{v}_2 + \vec{v}_3$. The three vectors are linearly dependent. …

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2. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.$\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors. If the three points $\lambda \vec{a} - 2\vec{b} + \vec{c},\ 2\vec{a} + \lambda \vec{b} - 2\vec{c},\ 4\vec{a} + 7\vec{b} - 8\vec{c}$ are collinear, then $\lambda =$ (A) $-1$ (B) $-2$ (C) $2$ (D) $1$

   ›Reveal solutionSolution

   For three points expressed as linear combinations of non‑coplanar vectors to be collinear, the vectors connecting them must be parallel. Setting the cross product of two such vectors to zero yields a quadratic in λ; the only value that works is λ = –2.

   Concept & Intuition

   When points are given in terms of a basis (here the non‑coplanar vectors $\vec{a},\vec{b},\vec{c}$), collinearity means the displacement vectors between any two pairs are scalar multiples of each other. Because the basis is linearly independent, we can equate coefficients after writing one displacement as a scalar times another. This gives a system of equations that determines λ.

   ---

   1. Label the points

   $P = \lambda\vec{a} - 2\vec{b} + \vec{c},\qquad Q = 2\vec{a} + \lambda\vec{b} - 2\vec{c},\qquad R = 4\vec{a} + 7\vec{b} - 8\vec{c}.$

   2. Form two displacement vectors

   $\overrightarrow{PQ} = Q - P = (2-\lambda)\vec{a} + (\lambda+2)\vec{b} + (-2-1)\vec{c} = (2-\lambda)\vec{a} + (\lambda+2)\vec{b} - 3\vec{c}.$

   $\overrightarrow{PR} = R - P = (4-\lambda)\vec{a} + (7+2)\vec{b} + (-8-1)\vec{c} = (4-\lambda)\vec{a} + 9\vec{b} - 9\vec{c}.$

   3. Collinearity condition There exists a scalar $k$ such that $\overrightarrow{PR} = k\,\overrightarrow{PQ}$.

   $(4-\lambda)\vec{a} + 9\vec{b} - 9\vec{c} = k\big[(2-\lambda)\vec{a} + (\lambda+2)\vec{b} - 3\vec{c}\big].$

   4. Equate coefficients (since $\vec{a},\vec{b},\vec{c}$ are linearly independent)

   $\begin{cases} 4-\lambda = k(2-\lambda) & \text{(1)}\\ 9 = k(\lambda+2) & \text{(2)}\\ -9 = k(-3) & \text{(3)} \end{cases}$

   5. Solve from the simplest equation

      From (3): $-9 = -3k \Rightarrow k = 3$.

   6. Substitute $k=3$ into (2)

   $9 = 3(\lambda+2) \;\Rightarrow\; \lambda+2 = 3 \;\Rightarrow\; \lambda = 1.$

   7. Check consistency with (1)

   $4-\lambda = 3(2-\lambda) \;\Rightarrow\; 4-1 = 3(2-1) \;\Rightarrow\; 3 = 3.$

   So $\lambda = 1$ satisfies all three equations. …

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3. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
   Q.If the collinear points A, B and C have position vectors respectively $(1, x, 3)$, $(3, 4, 7)$ and $(y, -2, -5)$, then $x + y =$ (A) $-1$ (B) $1$ (C) $-5$ (D) $5$

   ›Reveal solutionSolution

   For three points to be collinear, the vector formed by any two points must be a scalar multiple of the vector formed by another pair of points. By setting $\vec{AB} = k \vec{BC}$ and equating components, we find $x=2$ and $y=-3$, leading to $x+y = \boxed{-1}$.

   When three points A, B, and C are collinear, it means they lie on the same straight line. In terms of vectors, this implies that the vector connecting any two of these points is parallel to the vector connecting another pair of these points. For instance, vector $\vec{AB}$ must be parallel to vector $\vec{BC}$.

   If two vectors are parallel, one must be a scalar multiple of the other. So, we can write $\vec{AB} = k \vec{BC}$ for some scalar $k$. By calculating the component form of these vectors and equating their corresponding components, we can set up a system of equations to solve for the unknown values $x$, $y$, and the scalar $k$.

   Here's how to solve the problem step-by-step:

   1. Write down the position vectors:

      The position vectors of points A, B, and C are given as:

      $\vec{a} = (1, x, 3)$

      $\vec{b} = (3, 4, 7)$

      $\vec{c} = (y, -2, -5)$

   2. Form vectors $\vec{AB}$ and $\vec{BC}$:

      The vector $\vec{AB}$ is found by subtracting the position vector of A from that of B:

      $\vec{AB} = \vec{b} - \vec{a} = (3-1, 4-x, 7-3) = (2, 4-x, 4)$

      Similarly, the vector $\vec{BC}$ is found by subtracting the position vector of B from that of C:

      $\vec{BC} = \vec{c} - \vec{b} = (y-3, -2-4, -5-7) = (y-3, -6, -12)$

   3. Apply the collinearity condition:

      Since points A, B, and C are collinear, the vector $\vec{AB}$ must be a scalar multiple of $\vec{BC}$. Let this scalar be $k$.

      $\vec{AB} = k \vec{BC}$

      $(2, 4-x, 4) = k(y-3, -6, -12)$

   4. Equate corresponding components:

      This vector equation gives us three scalar equations by equating the $x$, $y$, and $z$ components:

      1. $2 = k(y-3)$
      2. $4-x = k(-6)$
      3. $4 = k(-12)$

   5. Solve for $k$, $x$, and $y$: …

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4. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
   Q.If $\vec{a}, \vec{b}, \vec{c}$ are the non-coplanar vectors and $\vec{a} - 2\vec{b} + 3\vec{c}, -4\vec{a} + 5\vec{b} - 6\vec{c}, x\vec{a} - 9\vec{b} + z\vec{c}$ are collinear points then $2x - z =$ (A) $-10$ (B) $-9$ (C) $0$ (D) $9$

   ›Reveal solutionSolution

   For three vectors representing collinear points, the vectors between them must be parallel — this gives a proportionality condition that lets us solve for $x$ and $z$, yielding $2x - z = -9$.

   The key idea here is that "collinear points" means the position vectors of the three points lie on the same straight line. When vectors are given as linear combinations of a non-coplanar basis $\vec{a}, \vec{b}, \vec{c}$, the condition for collinearity translates into a neat algebraic condition: the differences between consecutive vectors must be scalar multiples of each other.

   Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar, they form a basis — no vector can be expressed as a combination of the other two. This means that when we set up the proportionality of the difference vectors, the coefficients of $\vec{a}, \vec{b}, \vec{c}$ must separately be in the same ratio. That gives us two equations to solve for $x$ and $z$.

   Let’s label the three vectors:

   $\vec{p} = \vec{a} - 2\vec{b} + 3\vec{c}, \quad \vec{q} = -4\vec{a} + 5\vec{b} - 6\vec{c}, \quad \vec{r} = x\vec{a} - 9\vec{b} + z\vec{c}$

   1. Form the difference vectors. For collinearity, $\vec{q} - \vec{p}$ and $\vec{r} - \vec{p}$ must be parallel (or one could be a scalar multiple of the other).

   $\vec{q} - \vec{p} = (-4 - 1)\vec{a} + (5 + 2)\vec{b} + (-6 - 3)\vec{c} = -5\vec{a} + 7\vec{b} - 9\vec{c}$

   $\vec{r} - \vec{p} = (x - 1)\vec{a} + (-9 + 2)\vec{b} + (z - 3)\vec{c} = (x - 1)\vec{a} - 7\vec{b} + (z - 3)\vec{c}$

   2. Apply the parallelism condition. There exists some scalar $k$ such that:

   $\vec{r} - \vec{p} = k (\vec{q} - \vec{p})$

   That is:

   $(x - 1)\vec{a} - 7\vec{b} + (z - 3)\vec{c} = k(-5\vec{a} + 7\vec{b} - 9\vec{c})$

   3. Equate coefficients of the basis vectors. …

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5. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.If $2\vec{i} - \vec{j} + 3\vec{k},\ -12\vec{i} - \vec{j} - 3\vec{k},\ -\vec{i} + 2\vec{j} - 4\vec{k}$ and $\lambda \vec{i} + 2\vec{j} - \vec{k}$ are the position vectors of four coplanar points, then $\lambda =$ (A) $9$ (B) $-2$ (C) $8$ (D) $6$

   ›Reveal solutionSolution

   Set the scalar triple product of the three difference vectors to zero: $18\lambda - 108 = 0$, so $\lambda = 6$.

   Concept. Four points $A, B, C, D$ are coplanar iff $[\vec{AB}\ \vec{AC}\ \vec{AD}] = 0$ (the three difference vectors are linearly dependent).

   Step 1 — difference vectors. With $A(2,-1,3)$, $B(-12,-1,-3)$, $C(-1,2,-4)$, $D(\lambda,2,-1)$:

   $\vec{AB} = (-14,\ 0,\ -6), \quad \vec{AC} = (-3,\ 3,\ -7), \quad \vec{AD} = (\lambda-2,\ 3,\ -4).$

   Step 2 — scalar triple product. …

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6. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.The straight line given by the equation $\vec{r} = (4\vec{i} + 5\vec{j} + \vec{k}) + s(4\vec{i} + 6\vec{j} + 2\vec{k})$ is coplanar with a straight line given below. Choose the correct option (A) $\vec{r} = (\vec{i} - 2\vec{j} + 3\vec{k}) + p(2\vec{i} + 3\vec{j} - 4\vec{k})$ (B) $\vec{r} = (3\vec{i} - 4\vec{j} + 3\vec{k}) + q(-4\vec{i} + 5\vec{j} - 6\vec{k})$ (C) $\vec{r} = (2\vec{i} + 5\vec{j} - 4\vec{k}) + r(\vec{i} + 4\vec{j} - 3\vec{k})$ (D) $\vec{r} = (-4\vec{i} + 4\vec{j} + 4\vec{k}) + t(7\vec{i} + 5\vec{j})$

   ›Reveal solutionSolution

   Two lines are coplanar if the scalar triple product of their direction vectors and the vector joining a point on each line is zero. For the given line, only option (D) satisfies this condition.

   Concept & Intuition

   Two lines in 3D are coplanar if they lie in the same plane. This happens either when they are parallel (direction vectors are scalar multiples) or when they intersect. But there’s a third, more general case: they can be skew (not coplanar) or coplanar without intersecting (parallel but distinct). The universal test: take a point on each line, form the vector between them, and check if the three vectors (the two direction vectors and the connecting vector) are linearly dependent — i.e., their scalar triple product is zero. Geometrically, this means the volume of the parallelepiped they span is zero, so they all lie in a plane.

   Step-by-step solution

   1. Identify the given line

      Line $L_0$: $\vec{r} = (4\vec{i} + 5\vec{j} + \vec{k}) + s(4\vec{i} + 6\vec{j} + 2\vec{k})$

      Point $P_0 = (4,5,1)$, direction $\vec{d}_0 = (4,6,2)$.

      Notice $\vec{d}_0 = 2(2,3,1)$, so its direction is essentially $(2,3,1)$.

   2. Coplanarity condition

      For another line $L$: $\vec{r} = \vec{a} + t\vec{d}$, with point $P$ and direction $\vec{d}$, the lines are coplanar iff

   $(\vec{d}_0 \times \vec{d}) \cdot (\vec{a} - \vec{P}_0) = 0$

   That is, the scalar triple product $[\vec{d}_0, \vec{d}, \vec{P}_0\vec{P}] = 0$.

   3. Test each option

      Option (A):

      $P = (1,-2,3)$, $\vec{d} = (2,3,-4)$

      $\vec{d}_0 \times \vec{d} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 4 & 6 & 2 \\ 2 & 3 & -4 \end{vmatrix} = \vec{i}(6\cdot(-4)-2\cdot3) - \vec{j}(4\cdot(-4)-2\cdot2) + \vec{k}(4\cdot3-6\cdot2)$

      $= \vec{i}(-24-6) - \vec{j}(-16-4) + \vec{k}(12-12) = (-30, 20, 0)$

      $\vec{a} - \vec{P}_0 = (1-4, -2-5, 3-1) = (-3, -7, 2)$

      Dot product: $(-30)(-3) + (20)(-7) + 0\cdot2 = 90 - 140 = -50 \neq 0$ → Not coplanar.

      Option (B):

      $P = (3,-4,3)$, $\vec{d} = (-4,5,-6)$

      $\vec{d}_0 \times \vec{d} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 4 & 6 & 2 \\ -4 & 5 & -6 \end{vmatrix} = \vec{i}(6\cdot(-6)-2\cdot5) - \vec{j}(4\cdot(-6)-2\cdot(-4)) + \vec{k}(4\cdot5-6\cdot(-4))$

      $= \vec{i}(-36-10) - \vec{j}(-24+8) + \vec{k}(20+24) = (-46, 16, 44)$

      $\vec{a} - \vec{P}_0 = (3-4, -4-5, 3-1) = (-1, -9, 2)$

      Dot: $(-46)(-1) + (16)(-9) + (44)(2) = 46 - 144 + 88 = -10 \neq 0$ → Not coplanar.

      Option (C):

      $P = (2,5,-4)$, $\vec{d} = (1,4,-3)$

      Cross: $(4,6,2) \times (1,4,-3)$ …

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7. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.Let $\vec{a}$ and $\vec{b}$ be non-collinear vectors. If the vectors $(\lambda-1)\vec{a} + 2\vec{b}$ and $3\vec{a} + \lambda\vec{b}$ are collinear, then the set of all possible values of $\lambda$ is (A) $\{2, 3\}$ (B) $\{-2, 3\}$ (C) $\{-2, -3\}$ (D) $\{2, -3\}$

   ›Reveal solutionSolution

   For two vectors expressed as linear combinations of non-collinear $\vec{a}$ and $\vec{b}$ to be collinear, their coefficients must be proportional. Solving the proportion gives $\lambda = -2$ or $\lambda = 3$, so the set is $\{-2, 3\}$.

   The key idea here is that when two vectors are collinear (parallel), one is a scalar multiple of the other. Since $\vec{a}$ and $\vec{b}$ are non-collinear, they form a basis — meaning any vector written in terms of them has a unique representation. So if two such combinations are parallel, the coefficients of $\vec{a}$ and $\vec{b}$ must be in the same ratio.

   Let’s work through it.

   1. Set up the collinearity condition. If $(\lambda-1)\vec{a} + 2\vec{b}$ and $3\vec{a} + \lambda\vec{b}$ are collinear, there exists some scalar $k$ such that

   $(\lambda-1)\vec{a} + 2\vec{b} = k(3\vec{a} + \lambda\vec{b}).$

   2. Equate coefficients of $\vec{a}$ and $\vec{b}$. Because $\vec{a}$ and $\vec{b}$ are non-collinear, the representation is unique. So we get two equations:

   $\lambda - 1 = 3k \quad \text{and} \quad 2 = k\lambda.$

   3. Eliminate $k$ to find $\lambda$. From the second equation, $k = \frac{2}{\lambda}$ (provided $\lambda \neq 0$). Substitute into the first: $\lambda - 1 = 3 \cdot \frac{2}{\lambda}.$ …

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8. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.A unit vector $\vec{e} = a\hat{i} + b\hat{j} + c\hat{k}$ is coplanar with the vectors $\hat{i} - 3\hat{j} + 5\hat{k}$ and $3\hat{i} + \hat{j} - 5\hat{k}$. If $\vec{e}$ is perpendicular to the vector $\hat{i} + \hat{j} + \hat{k}$, then $2a^2 + 3b^2 + 4c^2 =$ (A) $1$ (B) $3$ (C) $-1$ (D) $\sqrt{2}$

   ›Reveal solutionSolution

   The key idea is to use the coplanarity condition (scalar triple product = 0) and the perpendicularity condition (dot product = 0) to solve for the components of the unit vector, then compute the required expression. The final result is 3.

   We are given a unit vector $\vec{e} = a\hat{i} + b\hat{j} + c\hat{k}$ that is:

   1. Coplanar with $\vec{u} = \hat{i} - 3\hat{j} + 5\hat{k}$ and $\vec{v} = 3\hat{i} + \hat{j} - 5\hat{k}$.
   2. Perpendicular to $\vec{w} = \hat{i} + \hat{j} + \hat{k}$.
   3. A unit vector: $a^2 + b^2 + c^2 = 1$.

   We need $2a^2 + 3b^2 + 4c^2$.

   Concept and intuition:

   Coplanarity of three vectors means one is a linear combination of the other two, or equivalently, their scalar triple product is zero. Perpendicularity gives a dot product of zero. The unit vector condition gives a third equation. We have three unknowns, so we can solve for $a, b, c$ (up to sign, but squares will be determined). Then we compute the weighted sum of squares.

   ---

   Step-by-step solution:

   1. Coplanarity condition: Vectors $\vec{e}, \vec{u}, \vec{v}$ are coplanar iff their scalar triple product is zero:

   $\vec{e} \cdot (\vec{u} \times \vec{v}) = 0.$

   First compute $\vec{u} \times \vec{v}$:

   $\vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 5 \\ 3 & 1 & -5 \end{vmatrix} = \hat{i}((-3)(-5) - (5)(1)) - \hat{j}((1)(-5) - (5)(3)) + \hat{k}((1)(1) - (-3)(3))$

   $= \hat{i}(15 - 5) - \hat{j}(-5 - 15) + \hat{k}(1 + 9) = 10\hat{i} + 20\hat{j} + 10\hat{k}.$

   So $\vec{u} \times \vec{v} = 10(\hat{i} + 2\hat{j} + \hat{k})$.

   The coplanarity condition becomes:

   $\vec{e} \cdot (10(\hat{i} + 2\hat{j} + \hat{k})) = 0 \quad \Rightarrow \quad a + 2b + c = 0. \qquad(1)$

   2. Perpendicularity condition: $\vec{e} \perp \vec{w}$ means $\vec{e} \cdot \vec{w} = 0$:

   $a + b + c = 0. \qquad(2)$

   3. Solve for relationships: Subtract (2) from (1): $$$$

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9. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.Consider the vectors $\vec{a} = 2\hat{i} + 3\hat{j} - 6\hat{k}$, $\vec{b} = 6\hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{c} = 3\hat{i} - 6\hat{j} - 2\hat{k}$. Assertion (A): The three vectors do not form a triangle Reason (R): The three vectors are non-coplanar The correct option among the following is (A) (A) is true, (R) is true and (R) is the correct explanation for (A) (B) (A) is true, (R) is true but (R) is not the correct explanation for (A) (C) (A) is true but (R) is false (D) (A) is false but (R) is true

   ›Reveal solutionSolution

   The given vectors do not sum to the zero vector, so they cannot form a triangle. Their scalar triple product is non-zero, meaning they are non-coplanar. Since vectors forming a triangle must be coplanar, their non-coplanarity correctly explains why they cannot form a triangle. The correct option is (A).

   To determine if three vectors can form a triangle, we check if their vector sum is the zero vector. If $\vec{a}$, $\vec{b}$, and $\vec{c}$ represent the sides of a triangle taken in order, then their resultant must be $\vec{0}$. This is a fundamental property of closed vector polygons.

   To determine if three vectors are coplanar, we calculate their scalar triple product. If the scalar triple product is zero, the vectors are coplanar; otherwise, they are non-coplanar. A triangle is a planar figure, so its sides must necessarily be coplanar.

   1. Evaluate Assertion (A): The three vectors do not form a triangle.

      For three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ to form a triangle, their vector sum must be $\vec{0}$ when placed head-to-tail. Let's calculate the sum of the given vectors:

      $\vec{a} + \vec{b} + \vec{c} = (2\hat{i} + 3\hat{j} - 6\hat{k}) + (6\hat{i} - 2\hat{j} + 3\hat{k}) + (3\hat{i} - 6\hat{j} - 2\hat{k})$

      Group the components:

      $= (2+6+3)\hat{i} + (3-2-6)\hat{j} + (-6+3-2)\hat{k}$

      $= 11\hat{i} - 5\hat{j} - 5\hat{k}$

      Since $\vec{a} + \vec{b} + \vec{c} \neq \vec{0}$, the three vectors do not form a triangle.

      Therefore, Assertion (A) is true.

   2. Evaluate Reason (R): The three vectors are non-coplanar.

      Three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ are coplanar if their scalar triple product, $[\vec{a} \ \vec{b} \ \vec{c}]$, is equal to zero. If it is non-zero, they are non-coplanar.

      The scalar triple product is given by the determinant of the matrix formed by their components:

   $[\vec{a} \ \vec{b} \ \vec{c}] = \begin{vmatrix} 2 & 3 & -6 \\ 6 & -2 & 3 \\ 3 & -6 & -2 \end{vmatrix}$

   Calculate the determinant:
   $= 2((-2)(-2) - (3)(-6)) - 3((6)(-2) - (3)(3)) + (-6)((6)(-6) - (-2)(3))$ …
   

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10. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
    Q.The position vectors of the points A, B are $\vec{a}, \vec{b}$ respectively. If the position vector of the point C is $\dfrac{\vec{a}}{2} + \dfrac{\vec{b}}{3}$, then (A) C lies inside $\triangle OAB$ (B) C lies outside $\triangle OAB$ but inside $\angle AOB$ (C) C lies outside $\triangle OAB$ but inside $\angle OAB$ (D) C lies outside $\triangle OAB$ but inside $\angle OBA$

    ›Reveal solutionSolution

    The point C is a convex combination of A and B with positive coefficients that sum to less than 1, so it lies strictly inside triangle OAB. The correct option is (A).

    We are given position vectors $\vec{a}, \vec{b}$ for points A and B, with O as the origin. The point C has position vector

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

    We need to decide where C lies relative to triangle OAB and the angles at O, A, and B.

    Concept and intuition:

    Any point inside triangle OAB can be written as a convex combination of O, A, and B:

    $\vec{p} = \alpha \vec{a} + \beta \vec{b} + \gamma \vec{0}, \quad \alpha, \beta, \gamma \ge 0,\ \alpha+\beta+\gamma = 1.$

    Here $\vec{0}$ is the position vector of O. So the condition for a point to be inside (or on the boundary of) triangle OAB is that its position vector is a nonnegative linear combination of $\vec{a}$ and $\vec{b}$ with coefficients summing to at most 1.

    Our $\vec{c}$ has coefficients $\frac12$ and $\frac13$ for $\vec{a}$ and $\vec{b}$. Both are positive, and their sum is

    $\frac12 + \frac13 = \frac{5}{6} < 1.$

    Thus we can write

    $\vec{c} = \frac12 \vec{a} + \frac13 \vec{b} + \left(1 - \frac12 - \frac13\right)\vec{0} = \frac12 \vec{a} + \frac13 \vec{b} + \frac16 \vec{0},$

    which is exactly a convex combination of O, A, B. Hence C lies strictly inside triangle OAB (not on the boundary because all coefficients are positive and none is zero).

    Let’s verify step by step.

    1. Check if C is inside the angle AOB.

       For C to be inside $\angle AOB$, its position vector must be a nonnegative linear combination of $\vec{a}$ and $\vec{b}$ (with no restriction on the sum). Here $\vec{c} = \frac12 \vec{a} + \frac13 \vec{b}$ has both coefficients positive, so C is indeed inside the angle AOB. This eliminates options that claim C is outside the angle.

    2. Check if C is inside triangle OAB.

       As argued, a point inside triangle OAB must be expressible as $\alpha \vec{a} + \beta \vec{b}$ with $\alpha, \beta \ge 0$ and $\alpha + \beta \le 1$. Here $\alpha = \frac12$, $\beta = \frac13$, and $\alpha+\beta = \frac56 < 1$. So C satisfies the condition and lies strictly inside the triangle.

    3. Eliminate other options.

       • Option (B) says C lies outside the triangle but inside $\angle AOB$. We have shown it is inside the triangle, so (B) is false. …

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11. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
    Q.If the algebraic sum of the perpendicular distances from the points $(2,0)$, $(0,2)$ and $(1,1)$ to a variable line is zero, then the variable line always passes through a fixed point. The coordinates of that point are (A) $(0,0)$ (B) $(2,0)$ (C) $(0,2)$ (D) $(1,1)$

    ›Reveal solutionSolution

    The condition that the algebraic sum of signed distances from three points to a line is zero forces the line to pass through the centroid of those points. The centroid of $(2,0)$, $(0,2)$, and $(1,1)$ is $(1,1)$, so the line always passes through $(1,1)$.

    The key idea here is that the signed perpendicular distance from a point $(x_1, y_1)$ to a line $ax + by + c = 0$ is proportional to $ax_1 + by_1 + c$. When you add three such expressions and set the sum to zero, you get a linear condition that the coefficients $a, b, c$ must satisfy — and that condition turns out to be that the line passes through the centroid of the three points.

    Let’s work through it.

    1. Write the general line and the signed distance formula. Let the variable line be $ax + by + c = 0$, where $a$ and $b$ are not both zero. The signed perpendicular distance from a point $(x_1, y_1)$ to this line is

    $\frac{ax_1 + by_1 + c}{\sqrt{a^2 + b^2}}.$

    The denominator is the same for all points, so the condition “algebraic sum of distances is zero” becomes

    $\frac{(ax_1 + by_1 + c) + (ax_2 + by_2 + c) + (ax_3 + by_3 + c)}{\sqrt{a^2 + b^2}} = 0.$

    2. Cancel the denominator and simplify. Since $\sqrt{a^2 + b^2} \neq 0$, we have

    $(ax_1 + by_1 + c) + (ax_2 + by_2 + c) + (ax_3 + by_3 + c) = 0.$

    This simplifies to

    $a(x_1 + x_2 + x_3) + b(y_1 + y_2 + y_3) + 3c = 0.$

    3. Plug in the given points. The points are $(2,0)$, $(0,2)$, and $(1,1)$. So

    $x_1 + x_2 + x_3 = 2 + 0 + 1 = 3, \quad y_1 + y_2 + y_3 = 0 + 2 + 1 = 3.$

    The condition becomes

    $a(3) + b(3) + 3c = 0 \quad \Rightarrow \quad 3a + 3b + 3c = 0 \quad \Rightarrow \quad a + b + c = 0.$

    4. Interpret the condition $a + b + c = 0$. This means that for any line $ax + by + c = 0$ satisfying the given condition, the point $(1,1)$ satisfies the equation because $$$$

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12. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
    Q.Let $\alpha, \beta, \gamma$ be three non-zero real constants and $a, b, c$ be three arbitrary real numbers which satisfy $\alpha a + \beta b + \gamma c = 0$. Then the point of concurrence of the family of lines $ax + by + c = 0$ is (A) $\left( \dfrac{\alpha}{\beta}, \dfrac{\beta}{\gamma} \right)$ (B) $\left( \dfrac{\gamma}{\alpha}, \dfrac{\beta}{\alpha} \right)$ (C) $\left( \dfrac{\alpha}{\gamma}, \dfrac{\gamma}{\beta} \right)$ (D) $\left( \dfrac{\alpha}{\gamma}, \dfrac{\beta}{\gamma} \right)$

    ›Reveal solutionSolution

    The point of concurrence for a family of lines $ax+by+c=0$ constrained by $\alpha a + \beta b + \gamma c = 0$ is found by recognizing that the coefficients of $a, b, c$ in both equations must be proportional. The point is $\boxed{\left( \frac{\alpha}{\gamma}, \frac{\beta}{\gamma} \right)}$.

    Concept and Intuition

    A "point of concurrence" for a family of lines is a single, fixed point through which every line in that family passes. If we denote this point as $(x_0, y_0)$, then for any line $ax + by + c = 0$ belonging to the family, substituting $(x_0, y_0)$ into the line's equation must satisfy it. This means $ax_0 + by_0 + c = 0$ must hold true for all valid sets of coefficients $(a, b, c)$.

    We are given a specific condition that defines which sets of $(a, b, c)$ are "valid": $\alpha a + \beta b + \gamma c = 0$.

    Our goal is to find the unique point $(x_0, y_0)$ such that the equation $ax_0 + by_0 + c = 0$ is satisfied whenever the condition $\alpha a + \beta b + \gamma c = 0$ is satisfied.

    For two linear equations involving the same variables (in this case, $a, b, c$) to represent the same relationship or constraint, their corresponding coefficients must be proportional. This is the core idea we will use to find $(x_0, y_0)$.

    Step-by-step Derivation

    1. Formulate the condition for concurrence: Let $(x_0, y_0)$ be the point of concurrence. By definition, every line $ax + by + c = 0$ in the family must pass through $(x_0, y_0)$. Therefore, for any valid $a, b, c$:

    $ax_0 + by_0 + c = 0 \quad (1)$

    2. State the given constraint: The coefficients $a, b, c$ are not arbitrary; they must satisfy the given condition:

    $\alpha a + \beta b + \gamma c = 0 \quad (2)$

    3. Apply the proportionality principle:

       Equation (1) must hold true for any $a, b, c$ that satisfy equation (2). This implies that equation (1) and equation (2) must be linearly dependent. In other words, the coefficients of $a, b, c$ in both equations must be proportional.

       Comparing the coefficients of $a$, $b$, and $c$ from equation (1) and equation (2):

       From (1): $x_0 \cdot a + y_0 \cdot b + 1 \cdot c = 0$

       From (2): $\alpha \cdot a + \beta \cdot b + \gamma \cdot c = 0$

       For these two equations to be proportional, there must exist a non-zero constant $k$ such that:

    $x_0 = k\alpha \quad (3)$

    $y_0 = k\beta \quad (4)$

    $$1 = k\gamma \quad (5)$$ …
    

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