Q.If $x \begin{bmatrix} 2 \\ 3 \end{bmatrix} + y \begin{bmatrix} -1 \\ 1 \end{bmatrix} = \begin{bmatrix} 10 \\ 5 \end{bmatrix}$, find the values of $x$ and $y$.

Showing the 12 most recent of 17 on this concept.

1. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
   Q.$\bar{a}, \bar{b}, \bar{c}$ are non-coplanar vectors. If $\bar{a}+3\bar{b}+4\bar{c} = x(\bar{a}-2\bar{b}+3\bar{c})+y(\bar{a}+5\bar{b}-2\bar{c})+z(6\bar{a}+14\bar{b}+4\bar{c})$ then $x+y+z =$ (A) $-5$ (B) $-4$ (C) $4$ (D) $5$

   ›Reveal solutionSolution

   Non-coplanar vectors are linearly independent, so matching coefficients turns a vector equation into a $3\times3$ linear system; solving gives $x+y+z=-4$.

   Concept and Intuition

   Because $\bar a,\bar b,\bar c$ are non-coplanar, no nontrivial combination of them is zero — they behave exactly like an independent basis $\{\bar a,\bar b,\bar c\}$. So if two linear combinations of them are equal as vectors, the coefficients of $\bar a$, of $\bar b$, and of $\bar c$ must separately match on both sides. This converts a single vector equation into three independent scalar equations in $x,y,z$.

   Step-by-Step Solution

   1. Expand the right side: $x(\bar a-2\bar b+3\bar c)+y(\bar a+5\bar b-2\bar c)+z(6\bar a+14\bar b+4\bar c) = (x+y+6z)\bar a+(-2x+5y+14z)\bar b+(3x-2y+4z)\bar c$.
   2. Match with the left side $\bar a+3\bar b+4\bar c$: $x+y+6z=1$ …(i) $-2x+5y+14z=3$ …(ii) $3x-2y+4z=4$ …(iii)
   3. From (i): $x=1-y-6z$. Substitute into (ii): $-2(1-y-6z)+5y+14z=3\Rightarrow 7y+26z=5$ …(ii′).
   4. Substitute into (iii): $3(1-y-6z)-2y+4z=4\Rightarrow -5y-14z=1\Rightarrow 5y+14z=-1$ …(iii′). …

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2. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If the two systems of equations $x + y - 2z = 0,\ 4x + 4y - 8z = 0,\ 3x + 3y - 6z = 0$ and $x + y + z = 3,\ 2x + 2y - z = \lambda,\ x + y - \mu z = 1$ have the same set of solutions, then $\lambda + \mu =$ (A) $0$ (B) $-5$ (C) $2$ (D) $4$

   ›Reveal solutionSolution

   The first system reduces to $x+y=2z$; forcing the second system to hold on that solution set gives $\lambda=3,\ \mu=1$, so $\lambda+\mu=4$.

   First system. The three equations $x+y-2z=0,\ 4x+4y-8z=0,\ 3x+3y-6z=0$ are all proportional, so the system collapses to the single relation

   $x+y-2z=0 \quad\Longrightarrow\quad x+y=2z.$

   Its solution set is this plane.

   Match with the second system. For both systems to share the same solutions, every solution must also satisfy $x+y+z=3,\ 2x+2y-z=\lambda,\ x+y-\mu z=1$. Substitute $x+y=2z$: …

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3. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.$\bar{a} = \alpha\bar{i} + \beta\bar{j} + 3\bar{k}$, $\bar{b} = \bar{j} + 2\bar{k}$, $\bar{c} = 3\bar{i} + 2\bar{j} + \bar{k}$ are linearly dependent vectors and magnitude of $\bar{a}$ is $\sqrt{14}$. If $\alpha, \beta$ are integers then $\alpha + \beta =$ (A) $3$ (B) $-3$ (C) $5$ (D) $-5$

   ›Reveal solutionSolution

   This tests linear dependence of vectors (zero scalar triple product) combined with a magnitude condition; the answer is $\alpha+\beta=3$.

   Concept and Intuition

   Three vectors in 3D are linearly dependent exactly when their scalar triple product (the determinant of their components) is zero — this is the vector-algebra analogue of coplanarity.

   Step-by-Step Solution

   1. Set up the determinant condition: $\begin{vmatrix}\alpha&\beta&3\\0&1&2\\3&2&1\end{vmatrix}=0$.
   2. Expand: $\alpha(1\cdot1-2\cdot2)-\beta(0\cdot1-2\cdot3)+3(0\cdot2-1\cdot3)=-3\alpha+6\beta-9=0$.
   3. Simplify: $\alpha=2\beta-3$.
   4. Magnitude condition: $\alpha^2+\beta^2+9=14\Rightarrow\alpha^2+\beta^2=5$.
   5. Substitute: $(2\beta-3)^2+\beta^2=5\Rightarrow4\beta^2-12\beta+9+\beta^2=5\Rightarrow5\beta^2-12\beta+4=0$. …

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4. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.Let $\bar{a}=2\bar{i}+3\bar{j}+\bar{k}$, $\bar{b}=4\bar{i}+\bar{j}$, $\bar{c}=\bar{i}-3\bar{j}-7\bar{k}$. If $\bar{r}=x\bar{i}+y\bar{j}+z\bar{k}$, $\bar{r}.\bar{a}=9$, $\bar{r}.\bar{b}=7$, $\bar{r}.\bar{c}=6$ then $(x,y,z) =$ (A) $(1, -3, 2)$ (B) $(-1, 3, -2)$ (C) $(1, 3, 2)$ (D) $(1, 3, -2)$

   ›Reveal solutionSolution

   Translating each dot-product condition into a linear equation in $x,y,z$ and solving the resulting system gives (D) $(1,3,-2)$.

   Concept and Intuition

   $\bar r\cdot\bar a=9$ etc. are just linear equations in the unknown components $x,y,z$ of $\bar r$, obtained by expanding the dot products component-wise. This reduces a vector problem to ordinary simultaneous linear equations.

   Step-by-Step Solution

   1. $\bar r\cdot\bar a = 2x+3y+z=9$ …(1)
   2. $\bar r\cdot\bar b = 4x+y=7$ …(2)
   3. $\bar r\cdot\bar c = x-3y-7z=6$ …(3)
   4. From (2): $y=7-4x$.
   5. Substitute into (1): $2x+3(7-4x)+z=9 \Rightarrow 2x+21-12x+z=9 \Rightarrow z=10x-12$. …

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5. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If the values $x=\alpha, y=\beta, z=\gamma$ satisfy all the 3 equations $x+2y+3z=4$, $3x+y+z=3$ and $x+3y+3z=2$, then $3\alpha+\gamma =$ (A) $\beta$ (B) $2\beta$ (C) $1-2\beta$ (D) $2\beta+1$

   ›Reveal solutionSolution

   Solving the linear system gives $\alpha=\tfrac78,\ \beta=-2,\ \gamma=\tfrac{19}{8}$, so $3\alpha+\gamma=5=1-2\beta$ — option (C).

   Concept and Intuition

   When a question asks for a combination of the unknowns in terms of one of them, the cleanest path is usually still to just solve the (consistent, unique) linear system directly and then match the numeric answer against each option evaluated at the found value of $\beta$ — this avoids needing to spot a clever linear-combination trick.

   Step-by-Step Solution

   1. The system is: (1) $x+2y+3z=4$; (2) $3x+y+z=3$; (3) $x+3y+3z=2$.
   2. From (1): $x=4-2y-3z$.
   3. Substitute into (3): $(4-2y-3z)+3y+3z=2 \Rightarrow 4+y=2 \Rightarrow y=-2$. So $\beta=-2$.
   4. Substitute $x=4-2y-3z$ into (2): $3(4-2y-3z)+y+z=3 \Rightarrow 12-6y-9z+y+z=3 \Rightarrow -5y-8z=-9 \Rightarrow 5y+8z=9$.
   5. With $y=-2$: $5(-2)+8z=9 \Rightarrow -10+8z=9 \Rightarrow 8z=19 \Rightarrow z=\dfrac{19}{8}$. So $\gamma=\dfrac{19}{8}$.
   6. Then $x=4-2(-2)-3\cdot\dfrac{19}{8} = 4+4-\dfrac{57}{8} = 8-\dfrac{57}{8}=\dfrac{64-57}{8}=\dfrac78$. So $\alpha=\dfrac78$. …

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6. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.If $\vec{a} = \vec{i}+2\vec{j}+3\vec{k}$, $\vec{b} = 2\vec{i}+3\vec{j}+\vec{k}$, $\vec{c} = 8\vec{i}+13\vec{j}+9\vec{k}$ and $x\vec{a}+y\vec{b}+z\vec{c}=0$, then $\frac{xy}{z^2} =$ (A) -1 (B) -6 (C) 6 (D) 1

   ›Reveal solutionSolution

   A homogeneous vector equation in three unknowns reduces to solving a linear system for the ratio $x:y:z$; substituting back gives $xy/z^2=6$.

   Concept and Intuition

   When $x\vec a+y\vec b+z\vec c=\vec 0$ for vectors in 3D, each Cartesian component gives one linear equation, so we get a homogeneous $3\times3$ system. Since the vectors are linearly dependent (the problem guarantees a nontrivial solution), the system has a one-parameter family of solutions — we only need the ratios $x:y:z$.

   Step-by-Step Solution

   1. $\vec a=(1,2,3)$, $\vec b=(2,3,1)$, $\vec c=(8,13,9)$.
   2. Component equations: $x+2y+8z=0$; $2x+3y+13z=0$; $3x+y+9z=0$.
   3. From the first equation: $x=-2y-8z$.
   4. Substitute into the second: $2(-2y-8z)+3y+13z=0 \Rightarrow -4y-16z+3y+13z=0 \Rightarrow -y-3z=0 \Rightarrow y=-3z$.
   5. Then $x=-2(-3z)-8z=6z-8z=-2z$. …

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7. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.The equation of straight line passing through the point of intersection of the lines represented by $x^2+4xy+3y^2-4x-10y+3=0$ and the point $(2,2)$ is (A) $2x+3y-10=0$ (B) $3x+2y-10=0$ (C) $2x+y-6=0$ (D) $x+2y-6=0$

   ›Reveal solutionSolution

   Factor the pair-of-lines conic to find its two component lines, get their intersection point, then find the line joining that point to $(2,2)$. Answer: $3x+2y-10=0$.

   Concept and Intuition

   A second-degree equation representing two straight lines can be factored as $(x+y+a)(x+3y+b)=0$ (matching the $x^2+4xy+3y^2$ part, which factors as $(x+y)(x+3y)$), and the constants $a,b$ are found by matching the linear and constant terms.

   Step-by-Step Solution

   1. $x^2+4xy+3y^2 = (x+y)(x+3y)$ (check: $x^2+3xy+xy+3y^2=x^2+4xy+3y^2$ ✓).
   2. Write the full conic as $(x+y+a)(x+3y+b)=0$. Expanding: $x^2+4xy+3y^2+(a+b)x+(3a+b)y+ab$.
   3. Match to $-4x-10y+3$: $a+b=-4$, $3a+b=-10$, $ab=3$.
   4. Subtract: $(3a+b)-(a+b) = -10-(-4) \Rightarrow 2a=-6 \Rightarrow a=-3$. Then $b=-4-(-3)=-1$. Check: $ab=(-3)(-1)=3$ ✓. …

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8. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.The vector equation of a plane passing through the line of intersection of the planes $\bar r.(\bar i-2\bar k)=3$, $\bar r.(2\bar j+\bar k)=5$ and the point $\bar i+2\bar j+3\bar k$ is (A) $\bar r.(\bar i+4\bar j)=13$ (B) $\bar r.(\bar i+6\bar j+\bar k)=18$ (C) $\bar r.(\bar i+2\bar j-\bar k)=8$ (D) $\bar r.(\bar i+8\bar j+2\bar k)=23$

   ›Reveal solutionSolution

   Use the one-parameter family of planes through the line of intersection of two given planes, then pin down the parameter using the extra point. Answer: $\bar r\cdot(\bar i+8\bar j+2\bar k)=23$.

   Concept and Intuition

   Any plane containing the intersection line of two planes $P_1=d_1$ and $P_2=d_2$ can be written as $P_1+\lambda P_2 = d_1+\lambda d_2$ for some scalar $\lambda$ — this is the standard "family of planes" trick, and a single extra point pins down $\lambda$ uniquely.

   Step-by-Step Solution

   1. Given planes: $\bar r\cdot(\bar i-2\bar k)=3$ and $\bar r\cdot(2\bar j+\bar k)=5$.
   2. Family: $\bar r\cdot\left[(\bar i-2\bar k)+\lambda(2\bar j+\bar k)\right]=3+5\lambda$, i.e. $\bar r\cdot\left[\bar i+2\lambda\bar j+(\lambda-2)\bar k\right]=3+5\lambda$.
   3. This plane must pass through $(1,2,3)$ (from $\bar i+2\bar j+3\bar k$):

   $1(1)+2\lambda(2)+(\lambda-2)(3)=3+5\lambda$

   $1+4\lambda+3\lambda-6=3+5\lambda\Rightarrow7\lambda-5=3+5\lambda\Rightarrow2\lambda=8\Rightarrow\lambda=4.$

   4. Substitute $\lambda=4$: coefficient vector becomes $\bar i+8\bar j+(4-2)\bar k=\bar i+8\bar j+2\bar k$, and RHS $=3+5(4)=23$. …

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9. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
   Q.$X$ intercept of the plane containing the line of intersection of the planes $x - 2y + z + 2 = 0$ and $3x - y - z + 1 = 0$ and also passing through $(1,1,1)$ is ________ (A) $\dfrac{1}{3}$ (B) 2 (C) $\dfrac{1}{2}$ (D) $\dfrac{1}{4}$

   ›Reveal solutionSolution

   Using the standard family-of-planes trick through the given point pins down $\lambda$, and the resulting plane has $x$-intercept $1/2$.

   Concept and Intuition

   Any plane containing the line of intersection of two given planes can be written as their linear combination $(P_1)+\lambda(P_2)=0$; requiring it to pass through an extra point fixes $\lambda$ uniquely.

   Step-by-Step Solution

   1. Family: $(x-2y+z+2)+\lambda(3x-y-z+1)=0$.
   2. Substitute $(1,1,1)$: $(1-2+1+2)+\lambda(3-1-1+1)=2+2\lambda=0\Rightarrow\lambda=-1$. …

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10. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
    Q.If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{c} = x\hat{i} + (x-2)\hat{j} - \hat{k}$ and if the vector $\vec{c}$ lies in the plane of vectors $\vec{a}$ and $\vec{b}$ then 'x' equals (A) $0$ (B) $1$ (C) $2$ (D) $-2$

    ›Reveal solutionSolution

    A vector lies in the plane spanned by two other vectors exactly when their scalar triple product is zero; solving that determinant equation gives $x=-2$.

    Concept and Intuition

    Three vectors are coplanar (i.e. one lies in the plane of the other two, assuming those two are independent) exactly when their scalar triple product $[\vec a\ \vec b\ \vec c] = \vec a\cdot(\vec b\times\vec c)$ vanishes, since that product measures the (signed) volume of the parallelepiped they'd form — zero volume means they're flat/coplanar.

    Step-by-Step Solution

    1. Set up the determinant $[\vec a\ \vec b\ \vec c] = \begin{vmatrix}1&1&1\\1&-1&2\\x&x-2&-1\end{vmatrix}$.
    2. Expand along the first row: $1\big[(-1)(-1)-2(x-2)\big] - 1\big[1(-1)-2x\big] + 1\big[1(x-2)-(-1)x\big]$.
    3. First bracket: $1-2(x-2) = 1-2x+4 = 5-2x$.
    4. Second bracket: $-1-2x$, so $-1\times(-1-2x) = 1+2x$.
    5. Third bracket: $(x-2)+x = 2x-2$. …

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11. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
    Q.D, E, F are respectively the points on the sides BC, CA and AB of a $\triangle ABC$ dividing them in the ratio 2:3, 1:2, 3:1 internally. The lines $\overline{BE}$ and $\overline{CF}$ intersect on the line $\overline{AD}$ at P. If $\overline{AP} = x_1.\overline{AB} + y_1.\overline{AC}$ then $x_1+y_1 =$ (A) $5/6$ (B) $1$ (C) $3/2$ (D) $2$

    ›Reveal solutionSolution

    Setting up position vectors from $A$ for $D$, $E$, $F$ and intersecting cevian $AD$ with cevian $CF$ gives $P=\frac12\vec{AB}+\frac13\vec{AC}$, so $x_1+y_1=\frac56$.

    Concept and Intuition

    This is a section-formula/vector-geometry problem: express each division point as a weighted combination of the triangle's vertices (using $A$ as the origin for position vectors), then find where two cevians cross by equating their parametrised forms.

    Step-by-Step Solution

    1. Let $\vec{AB}=b$, $\vec{AC}=c$ (so $A$ is the origin, $B$ has position vector $b$, $C$ has position vector $c$).
    2. $D$ divides $BC$ with $BD:DC=2:3$, so $D=\dfrac{3B+2C}{5}=\dfrac35b+\dfrac25c$.
    3. $E$ divides $CA$ with $CE:EA=1:2$, so $E=\dfrac{2C+1\cdot A}{3}=\dfrac23c$ (since $A=0$).
    4. $F$ divides $AB$ with $AF:FB=3:1$, so $F=\dfrac{1\cdot A+3B}{4}=\dfrac34b$.
    5. Line $AD$: any point on it is $P=tD=t\left(\dfrac35b+\dfrac25c\right)$ for parameter $t$.
    6. Line $CF$: parametrise as $P=c+s(F-c)=s\cdot\dfrac34b+(1-s)c$.
    7. Equate $b$-coefficients: $\dfrac{3t}{5}=\dfrac{3s}{4}\Rightarrow s=\dfrac{4t}{5}$.
    8. Equate $c$-coefficients: $\dfrac{2t}{5}=1-s=1-\dfrac{4t}{5}\Rightarrow \dfrac{6t}{5}=1\Rightarrow t=\dfrac56$.
    9. So $P=\dfrac56\left(\dfrac35b+\dfrac25c\right)=\dfrac12 b+\dfrac13c$. …

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12. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
    Q.Let $\bar{a}, \bar{b}$ and $\bar{c}$ be any three non coplanar vectors. If m, n are scalars such that $\bar{a}+\bar{b} = m\bar{d}-\bar{c}$ and $\bar{b}+\bar{c} = n\bar{a}-\bar{d}$, then $3\bar{a}+2\bar{b}+2\bar{c}+\bar{d} =$ (A) $\bar{a}-\bar{d}$ (B) $\bar{a}+\bar{d}$ (C) $\bar{0}$ (D) $\bar{b}+\bar{c}+2\bar{d}$

    ›Reveal solutionSolution

    The two given vector equations are simultaneously consistent for arbitrary non-coplanar $\bar a,\bar b,\bar c$ only when $m=n=-1$, forcing $\bar a+\bar b+\bar c+\bar d=\bar 0$; substituting this into $3\bar a+2\bar b+2\bar c+\bar d$ gives $\bar a-\bar d$.

    Concept and Intuition

    When a system of vector equations must hold for a general (non-coplanar/independent) set of vectors, matching coefficients on each side forces the scalar unknowns to specific values rather than leaving them free — here it collapses both given relations into one clean identity.

    Step-by-Step Solution

    1. Rewrite the given equations: $\bar a+\bar b=m\bar d-\bar c\ \Rightarrow\ \bar a+\bar b+\bar c=m\bar d$ … (i), and $\bar b+\bar c=n\bar a-\bar d\ \Rightarrow\ \bar b+\bar c+\bar d=n\bar a$ … (ii).
    2. Adding (i) and (ii): $\bar a+2\bar b+2\bar c+\bar d = m\bar d+n\bar a$, i.e. $(1-n)\bar a+2\bar b+2\bar c+(1-m)\bar d=0$.
    3. From (i), $\bar b+\bar c=m\bar d-\bar a$; substituting into (ii): $(m\bar d-\bar a)+\bar d=n\bar a\Rightarrow(m+1)\bar d=(n+1)\bar a$.
    4. Since $\bar a,\bar b,\bar c$ are non-coplanar (linearly independent, spanning all of 3D space) and this relation must hold as a genuine vector identity for the given (general) $\bar a,\bar d$, the only way $(n+1)\bar a=(m+1)\bar d$ can hold in general (without a special, unstated relation forcing $\bar a\parallel\bar d$) is if both coefficients vanish: $m=-1$ and $n=-1$. …

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