Q.If $\vec{a}$ and $\vec{b}$ are two collinear vectors, then which of the following are incorrect: (A) $\vec{b} = \lambda\vec{a}$, for some scalar $\lambda$ (B) $\vec{a} = \pm\vec{b}$ (C) the respective components of $\vec{a}$ and $\vec{b}$ are not proportional (D) both the vectors $\vec{a}$ and $\vec{b}$ have same direction, but different magnitudes.

Showing the 12 most recent of 15 on this concept.

1. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.Let $\vec{a}, \vec{b}$ and $\vec{c}$ are 3 non zero vectors such that no 2 of these are collinear. If vector $\vec{a} + 2\vec{b}$ is collinear with $\vec{c}$ and $\vec{b} + 3\vec{c}$ is collinear with $\vec{a}$ ($\lambda$ being some non-zero scalar) then $\vec{a} + 2\vec{b} + 6\vec{c}$ equals (A) $\lambda \vec{a}$ (B) $\lambda \vec{b}$ (C) $\lambda \vec{c}$ (D) $\vec{0}$

   ›Reveal solutionSolution

   Writing both collinearity conditions as scalar equations and eliminating variables forces both coefficients to zero, giving $\vec a+2\vec b+6\vec c=\vec0$.

   Concept and Intuition

   "$\vec u$ collinear with $\vec v$" means $\vec u=k\vec v$ for some scalar $k$. With two such conditions linking three non-collinear (linearly independent, pairwise) vectors, substituting one into the other and using linear independence (a non-collinear pair can't satisfy a scalar-multiple relation unless the multiplier is zero) pins down all constants.

   Step-by-Step Solution

   1. $\vec a+2\vec b$ collinear with $\vec c$: $\vec a+2\vec b = m\vec c$ for some scalar $m$, i.e. $\vec a = m\vec c-2\vec b$.
   2. $\vec b+3\vec c$ collinear with $\vec a$: $\vec b+3\vec c=n\vec a$, i.e. $\vec b=n\vec a-3\vec c$.
   3. Substitute (2) into (1): $\vec a=m\vec c-2(n\vec a-3\vec c)=m\vec c-2n\vec a+6\vec c=(m+6)\vec c-2n\vec a$. …

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2. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
   Q.The vectors $\bar{a} = 2\bar{i}+3\bar{j}+6\bar{k}$ and $\bar{b}$ are collinear and $|\bar{b}| = 21$, then $\bar{b} =$ (A) $\pm(2\bar{i}+3\bar{j}+6\bar{k})$ (B) $\pm(6\bar{i}+9\bar{j}+18\bar{k})$ (C) $\dfrac{21}{\sqrt{3}}(\bar{i}+\bar{j}+\bar{k})$ (D) $\pm 21(2\bar{i}+3\bar{j}+6\bar{k})$

   ›Reveal solutionSolution

   This tests scaling a unit vector along a given direction to a specified magnitude, allowing for both possible orientations. Answer: $\bar b=\pm(6\bar i+9\bar j+18\bar k)$.

   Concept and Intuition

   Two vectors are collinear if one is a scalar multiple of the other. Given the direction (from $\bar a$) and the desired magnitude of $\bar b$, we first find the unit vector along $\bar a$, then scale it to the required length. Since collinear can mean parallel in either the same or opposite direction, both $+$ and $-$ signs are valid.

   Step-by-Step Solution

   1. Compute $|\bar a| = \sqrt{2^2+3^2+6^2} = \sqrt{4+9+36} = \sqrt{49} = 7$.
   2. The unit vector along $\bar a$ is $\hat a = \dfrac{1}{7}(2\bar i+3\bar j+6\bar k)$.
   3. Since $\bar b$ is collinear with $\bar a$, we can write $\bar b = \lambda\, \hat a$ for some scalar $\lambda = \pm|\bar b|$ (sign accounts for direction).
   4. Given $|\bar b|=21$: $\bar b = \pm 21 \cdot \dfrac{1}{7}(2\bar i+3\bar j+6\bar k) = \pm3(2\bar i+3\bar j+6\bar k)$.
   5. Distribute: $\bar b = \pm(6\bar i+9\bar j+18\bar k)$. …

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3. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.If $\bar{a}=(p,-2,5)$ and $\bar{b}=(1,q,-3)$ are collinear vectors then (A) $p=\dfrac{5}{3}, q=\dfrac{6}{5}$ (B) $p=\dfrac{-5}{3}, q=\dfrac{-6}{5}$ (C) $p=\dfrac{5}{3}, q=\dfrac{-6}{5}$ (D) $p=\dfrac{-5}{3}, q=\dfrac{6}{5}$

   ›Reveal solutionSolution

   Collinear vectors have proportional components; solving the proportion gives (D) $p=-\tfrac53, q=\tfrac65$.

   Concept and Intuition

   Two vectors $\bar a=(a_1,a_2,a_3)$ and $\bar b=(b_1,b_2,b_3)$ are collinear (parallel) if and only if their corresponding components are proportional: $\dfrac{a_1}{b_1}=\dfrac{a_2}{b_2}=\dfrac{a_3}{b_3}$ (equivalently $\bar a = \lambda\bar b$ for some scalar $\lambda$).

   Step-by-Step Solution

   1. Set up the proportionality: $\dfrac{p}{1} = \dfrac{-2}{q} = \dfrac{5}{-3}$.
   2. From the last ratio, the common scalar is $\lambda = \dfrac{5}{-3} = -\dfrac{5}{3}$.
   3. First component: $p = \lambda\times1 = -\dfrac53$.
   4. Second component: $-2 = \lambda q = -\dfrac53 q \Rightarrow q = \dfrac{-2\times(-3)}{5} = \dfrac65$. …

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4. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
   Q.Let $\vec a, \vec b$ and $\vec c$ be three non-zero vectors, no two which are collinear. If $\vec a+2\vec b$ is collinear with $\vec c$ and $\vec b+3\vec c$ is collinear with $\vec a$, then $\vec a+2\vec b=$ (A) $\vec c$ (B) $-4\vec c$ (C) $6\vec c$ (D) $-6\vec c$

   ›Reveal solutionSolution

   Turning both "collinear with" statements into scalar equations and using the fact that non-collinear vectors can't be proportional pins down $\vec a+2\vec b=-6\vec c$.

   Concept and Intuition

   "$\vec u$ is collinear with $\vec v$" simply means $\vec u=k\vec v$ for some scalar $k$. When two of your three given (pairwise non-collinear) vectors combine to something that has to equal a scalar multiple of yet another, and it isn't automatically zero, the only consistent resolution is that the coefficients of the "stray" non-collinear vector must vanish — this is the standard technique for such problems.

   Step-by-Step Solution

   1. $\vec a+2\vec b$ collinear with $\vec c$: $\vec a+2\vec b=\lambda\vec c$ for some scalar $\lambda$. — (1)
   2. $\vec b+3\vec c$ collinear with $\vec a$: $\vec b+3\vec c=\mu\vec a$ for some scalar $\mu$, so $\vec b=\mu\vec a-3\vec c$. — (2)
   3. Substitute (2) into (1): $\vec a+2(\mu\vec a-3\vec c)=\lambda\vec c \Rightarrow (1+2\mu)\vec a - 6\vec c=\lambda\vec c \Rightarrow (1+2\mu)\vec a = (\lambda+6)\vec c$. …

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5. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.$\vec{a}, \vec{b}$ and $\vec{c}$ are three non-zero vectors such that no two of them are collinear. If the vector $\vec{a} + \vec{b}$ is collinear with $\vec{c}$ and $\vec{b} + \vec{c}$ is collinear with $\vec{a}$, then $\vec{a} + \vec{b} + \vec{c} =$ (A) $\vec{a}$ (B) $\vec{b}$ (C) $\vec{c}$ (D) $\vec{0}$

   ›Reveal solutionSolution

   Turning "collinear with" into scalar-multiple equations and using linear independence of $\vec a,\vec b$ pins down both scalars as $-1$, forcing $\vec a+\vec b+\vec c=\vec0$.

   Concept and Intuition

   "$\vec u$ is collinear with $\vec v$" means $\vec u=k\vec v$ for some scalar $k$. Since no two of $\vec a,\vec b,\vec c$ are collinear, any two of them are linearly independent — so if a linear combination of two of them equals the zero vector, both coefficients must vanish.

   Step-by-Step Solution

   1. $\vec a+\vec b$ collinear with $\vec c$ means $\vec a+\vec b=\lambda\vec c$ for some scalar $\lambda$ — (i)
   2. $\vec b+\vec c$ collinear with $\vec a$ means $\vec b+\vec c=\mu\vec a$, so $\vec c=\mu\vec a-\vec b$ — (ii)
   3. Substitute (ii) into (i): $\vec a+\vec b=\lambda(\mu\vec a-\vec b)=\lambda\mu\,\vec a-\lambda\vec b$.
   4. Rearrange: $(1-\lambda\mu)\vec a+(1+\lambda)\vec b=\vec 0$. …

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6. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If the vectors $-3\bar{i} + 4\bar{j} + \lambda\bar{k}$ and $\mu\bar{i} + 8\bar{j} + 6\bar{k}$ are collinear, then $\lambda - \mu =$ (A) $0$ (B) $-3$ (C) $6$ (D) $9$

   ›Reveal solutionSolution

   This tests the collinear-vectors condition (proportional components). Solving gives $\lambda-\mu=9$.

   Concept and Intuition

   Two vectors are collinear (parallel) exactly when one is a scalar multiple of the other, which means their corresponding $i,j,k$ components are all in the same ratio. Matching this ratio for the known pair of components ($j$-components here) pins down the scalar multiple, and then the unknowns follow from the other two components.

   Step-by-Step Solution

   1. The vectors $-3\bar i+4\bar j+\lambda\bar k$ and $\mu\bar i+8\bar j+6\bar k$ are collinear, so:

   $\dfrac{-3}{\mu}=\dfrac{4}{8}=\dfrac{\lambda}{6}$

   2. From the known ratio $\dfrac{4}{8}=\dfrac12$, this common ratio equals $\dfrac12$.
   3. Solve for $\mu$: $\dfrac{-3}{\mu}=\dfrac12 \implies \mu = -6$.
   4. Solve for $\lambda$: $\dfrac{\lambda}{6}=\dfrac12 \implies \lambda=3$. …

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7. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
   Q.The number of values of $m \in R$ for which the vectors $\hat i + 2\hat j + m\hat k$ and $\hat i + m\hat j + 2\hat k$ are collinear is (A) $2$ (B) $3$ (C) $1$ (D) Infinite

   ›Reveal solutionSolution

   Setting the cross product of the two vectors to zero gives three conditions on $m$, and only $m=2$ satisfies all three simultaneously — so exactly one value of $m$ works.

   Concept and Intuition

   Two vectors are collinear (parallel) exactly when their cross product is the zero vector — this gives three scalar equations (one per component), all of which must hold simultaneously for genuine collinearity. It's not enough for just one component equation to be satisfied; a value of the parameter must satisfy every component equation at once.

   Step-by-Step Solution

   1. Let $\vec u = \hat i+2\hat j+m\hat k$ and $\vec v=\hat i+m\hat j+2\hat k$. They are collinear iff $\vec u \times \vec v = \vec 0$.
   2. Compute the cross product: $\vec u\times\vec v = \begin{vmatrix}\hat i&\hat j&\hat k\\1&2&m\\1&m&2\end{vmatrix} = \hat i(2\cdot2-m\cdot m) - \hat j(1\cdot2-m\cdot1) + \hat k(1\cdot m-2\cdot1)$ $= (4-m^2)\hat i - (2-m)\hat j + (m-2)\hat k$.
   3. For collinearity, each component must vanish:
      • $4-m^2=0 \Rightarrow m^2=4 \Rightarrow m=2$ or $m=-2$.
      • $2-m=0 \Rightarrow m=2$.
      • $m-2=0 \Rightarrow m=2$. …

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8. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
   Q.If $\bar{a}$ is collinear with $\bar{b} = 3i+6j+6k$ and $\bar{a}\cdot\bar{b} = 27$ then $|\bar{a}| =$ (A) $1$ (B) $2$ (C) $3$ (D) $4$

   ›Reveal solutionSolution

   Collinearity means $\bar a$ is a scalar multiple of $\bar b$; using the given dot product pins down that scalar and hence $|\bar a|=3$.

   Concept and Intuition

   Two vectors are collinear exactly when one is a scalar multiple of the other: $\bar a = k\bar b$. This single scalar $k$ captures both the direction (same or opposite to $\bar b$) and the relative length. Once we know $k$, both $|\bar a|$ and the dot product with $\bar b$ follow immediately, so a single scalar equation (the given dot product) is enough to solve for $k$.

   Step-by-Step Solution

   1. Since $\bar a$ is collinear with $\bar b=3\hat i+6\hat j+6\hat k$, write $\bar a=k\bar b$.
   2. Compute $|\bar b|=\sqrt{3^2+6^2+6^2}=\sqrt{9+36+36}=\sqrt{81}=9$.
   3. $\bar a\cdot\bar b = k\bar b\cdot\bar b = k|\bar b|^2 = 81k$.
   4. Given $\bar a\cdot\bar b=27$: $81k=27\Rightarrow k=\dfrac13$. …

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9. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.If $2\bar{i}+\bar{j}-\bar{k}$, $\bar{i}-3\bar{j}+5\bar{k}$ and $-3\bar{i}+4\bar{j}+4\bar{k}$ are the position vectors of three points A, B and C respectively, then (A) ABC is a right angled triangle (B) ABC is an isosceles triangle (C) A, B, C are collinear points (D) ABC is a scalene triangle

   ›Reveal solutionSolution

   Computing the three squared side lengths of the triangle formed by $A,B,C$ gives $53,59,66$ — all different, with no Pythagorean relation among them, so the triangle is scalene.

   Concept and Intuition

   Given three position vectors, first check for collinearity (via proportional direction vectors); if not collinear, compute the three side lengths (or their squares) to classify the triangle as scalene/isosceles/equilateral, and check for a right angle via the Pythagorean relation among the squared lengths (or a zero dot product).

   Step-by-Step Solution

   1. Position vectors: $A=(2,1,-1)$, $B=(1,-3,5)$, $C=(-3,4,4)$.
   2. $\vec{AB}=B-A=(-1,-4,6)$, $|\vec{AB}|^2=1+16+36=53$.
   3. $\vec{AC}=C-A=(-5,3,5)$, $|\vec{AC}|^2=25+9+25=59$.
   4. $\vec{BC}=C-B=(-4,7,-1)$, $|\vec{BC}|^2=16+49+1=66$.
   5. Check collinearity: $\vec{AB}=(-1,-4,6)$ and $\vec{AC}=(-5,3,5)$ are not scalar multiples of each other (ratios $-1/-5=0.2$ vs $-4/3\approx-1.33$ disagree), so $A,B,C$ are not collinear — ruling out option (C). …

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10. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
    Q.Assertion (A): If each of the angles $A, B, C$ is not a multiple of $\pi$, then the vectors $\vec{r_1} = (\sec^2 A)\vec{i} + \vec{j} + \vec{k}$, $\vec{r_2} = \vec{i} + (\sec^2 B)\vec{j} + \vec{k}$, $\vec{r_3} = \vec{i} + \vec{j} + (\sec^2 C)\vec{k}$ are coplanar. Reason (R): The three vectors $\vec{a} = a_1\vec{i} + a_2\vec{j} + a_3\vec{k}$, $\vec{b} = b_1\vec{i} + b_2\vec{j} + b_3\vec{k}$ and $\vec{c} = c_1\vec{i} + c_2\vec{j} + c_3\vec{k}$ are coplanar $\Leftrightarrow \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} = 0$ Which one of the following is true? (A) (A) is true, (R) is true and (R) is a correct explanation of (A) (B) (A) is true, (R) is true but (R) is not a correct explanation of (A) (C) (A) is true but (R) is false (D) (A) is false but (R) is true

    ›Reveal solutionSolution

    The Reason is the genuine general coplanarity test and is true; the Assertion, however, fails on a direct counterexample ($A=B=C=\pi/4$), so (A) is false but (R) is true.

    Concept and Intuition

    Assertion-Reason questions require checking BOTH statements independently for truth, and only then checking whether R explains A. Here it pays to just compute the determinant symbolically and test a concrete case rather than trust the assertion's phrasing.

    Step-by-Step Solution

    1. Let $x=\sec^2A,\,y=\sec^2B,\,z=\sec^2C$. The three given vectors are the rows $(x,1,1),(1,y,1),(1,1,z)$.
    2. Expand: $\det=x(yz-1)-1(z-1)+1(1-y)=xyz-x-y-z+2$.
    3. For the Assertion to hold for ALL valid $A,B,C$ (any angles not multiples of $\pi$), this expression would need to vanish identically — but it doesn't.
    4. Test $A=B=C=\pi/4$ (a valid choice, not a multiple of $\pi$): $\sec^2(\pi/4)=2$, so $x=y=z=2$, and $\det=8-2-2-2+2=4\ne0$. The vectors are NOT coplanar here, so the Assertion is FALSE in general.
    5. The Reason statement itself — "three vectors are coplanar iff the determinant of their components is zero" — is a standard, universally true fact of vector algebra, independent of the Assertion. …

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11. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
    Q.If $\bar{a} = (2x+y)\bar{i}+3\bar{j}+9\bar{k}$ and $\bar{b} = 2\bar{i}+\bar{j}-(x-y)\bar{k}$ are two collinear vectors, then $x^3+27y^3 =$ (A) $1241$ (B) $1512$ (C) $1072$ (D) $1729$

    ›Reveal solutionSolution

    Collinearity forces $\bar a=3\bar b$ (matched via the $j$-component); solving the resulting linear system gives $x=1,y=4$, and $x^3+27y^3=1729$ — the famous taxicab number.

    Concept and Intuition

    Two vectors are collinear iff one is a scalar multiple of the other: $\bar a=\lambda\bar b$ for some real $\lambda$. This gives one equation per component ($i,j,k$), and since the $j$-components involve no unknowns, that equation alone fixes $\lambda$ immediately, after which the other two components become simple linear equations in $x,y$.

    Step-by-Step Solution

    1. $\bar a = (2x+y)\bar i+3\bar j+9\bar k$, $\bar b=2\bar i+\bar j-(x-y)\bar k$. Collinear: $\bar a=\lambda\bar b$.
    2. Match $j$-components: $3=\lambda\cdot1 \Rightarrow \lambda=3$.
    3. Match $i$-components: $2x+y=\lambda\cdot2=6$.
    4. Match $k$-components: $9=\lambda\cdot[-(x-y)]=3\cdot[-(x-y)] \Rightarrow -(x-y)=3 \Rightarrow x-y=-3 \Rightarrow y=x+3$. …

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12. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.Let $2\bar{i}-\bar{j}-\bar{k}$, $5\bar{i}+\bar{j}-2\bar{k}$, $-13\bar{i}-11\bar{j}+4\bar{k}$ be the position vectors of three points A, B, C respectively. If $\overline{AB} = \lambda \overline{BC}$ and $\overline{AC} = \mu \overline{CB}$, then $\lambda + \mu =$ (A) $1$ (B) $-1$ (C) $2$ (D) $-2$

    ›Reveal solutionSolution

    This tests recognizing that A, B, C are collinear and computing the scalar ratios between the direction vectors formed. The answer is (B).

    Concept and Intuition

    If three points are collinear, every pair of displacement vectors between them is a scalar multiple of every other — the scalar being determined purely by how far apart the points are along the line. Computing the coordinate vectors directly and comparing component-by-component pins down each scalar exactly.

    Step-by-Step Solution

    1. $A=(2,-1,-1)$, $B=(5,1,-2)$, $C=(-13,-11,4)$.
    2. $\overline{AB}=B-A=(3,2,-1)$. $\overline{BC}=C-B=(-18,-12,6)$.
    3. Check proportionality: $\dfrac{3}{-18}=-\dfrac16$, $\dfrac{2}{-12}=-\dfrac16$, $\dfrac{-1}{6}=-\dfrac16$ — consistent, confirming collinearity and $\lambda=-\dfrac16$ (since $\overline{AB}=\lambda\overline{BC}$).
    4. $\overline{AC}=C-A=(-15,-10,5)$. $\overline{CB}=B-C=(18,12,-6)$. …

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