Q.Given that $\vec{a}\cdot\vec{b}=0$ and $\vec{a}\times\vec{b}=\vec{0}$. What can you conclude about the vectors $\vec{a}$ and $\vec{b}$?

Showing the 12 most recent of 26 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If the unit vector which is perpendicular to the normals drawn to the planes $\bar{r} \cdot (2\bar{i} + \bar{j} - \bar{k}) = 3$ and $\bar{r} \cdot (6\bar{i} + 3\bar{j} + 2\bar{k}) = 4$ is $x\bar{i} + y\bar{j} + z\bar{k}$, then $x + y + z =$ (A) $\pm\dfrac{3}{\sqrt{5}}$ (B) 0 (C) $\pm\dfrac{1}{\sqrt{5}}$ (D) $\pm 1$

   ›Reveal solutionSolution

   A vector perpendicular to two given normals lies along their cross product; normalize it and sum its components. Answer: $x+y+z=\pm\dfrac1{\sqrt5}$.

   Concept and Intuition

   A vector perpendicular to both normals $\bar n_1$ and $\bar n_2$ of two planes must be perpendicular to each of them individually — exactly the defining property of the cross product $\bar n_1\times\bar n_2$. Normalizing this cross product gives the (up to sign) unique unit vector satisfying the requirement; the $\pm$ arises because both directions along that line are legitimate unit vectors.

   Step-by-Step Solution

   1. Read off the normals from the plane equations $\bar r\cdot\bar n=d$: $\bar n_1=(2,1,-1)$ and $\bar n_2=(6,3,2)$.
   2. Compute the cross product: $\bar n_1\times\bar n_2=\begin{vmatrix}\bar i&\bar j&\bar k\\2&1&-1\\6&3&2\end{vmatrix}$.
   3. $\bar i$-component: $(1)(2)-(-1)(3)=2+3=5$.
   4. $\bar j$-component: $-[(2)(2)-(-1)(6)]=-[4+6]=-10$.
   5. $\bar k$-component: $(2)(3)-(1)(6)=6-6=0$.
   6. So $\bar n_1\times\bar n_2=(5,-10,0)$, with magnitude $\sqrt{25+100+0}=\sqrt{125}=5\sqrt5$.
   7. The unit vector is $\pm\dfrac{(5,-10,0)}{5\sqrt5}=\pm\dfrac{(1,-2,0)}{\sqrt5}$, so $x=\pm\dfrac1{\sqrt5}$, $y=\mp\dfrac2{\sqrt5}$, $z=0$ (with matching signs). …

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2. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.If the line with direction ratios $(1, \alpha, \beta)$ is perpendicular to the line with direction ratios $(-1, 2, 1)$ and parallel to the line with direction ratios $(\alpha, 1, \beta)$, then $(\alpha, \beta)$ is (A) $(-1, -1)$ (B) $(1, -1)$ (C) $(-1, 3)$ (D) $(1, 1)$

   ›Reveal solutionSolution

   Use the perpendicularity dot-product condition and the parallel proportionality condition together to pin down $\alpha$ and $\beta$.

   Concept and Intuition

   Two direction vectors are perpendicular iff their dot product is zero, and parallel iff one is a scalar multiple of the other. Combining both conditions on the same vector $(1,\alpha,\beta)$ gives two independent equations.

   Step-by-Step Solution

   1. Perpendicular to $(-1,2,1)$: $1(-1)+\alpha(2)+\beta(1)=0 \Rightarrow 2\alpha+\beta=1$. — (i)
   2. Parallel to $(\alpha,1,\beta)$: $(1,\alpha,\beta)=k(\alpha,1,\beta)$ for some scalar $k$, so $1=k\alpha,\ \alpha=k,\ \beta=k\beta$.
   3. From $\beta=k\beta$: either $\beta=0$ or $k=1$.
   4. If $k=1$: then $\alpha=k=1$, and $1=k\alpha=1\cdot1=1$ ✓ consistent. Using (i): $2(1)+\beta=1\Rightarrow\beta=-1$. …

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3. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
   Q.Let the plane $\pi$ pass through the point $(1, 0, 1)$ and perpendicular to the planes $2x + 3y - z = 2$ and $x - y + 2z = 1$. Let the equation of the plane passing through the point $(11, 7, 5)$ and parallel to the plane $\pi$ be $ax + by - z + d = 0$. Then $\dfrac{a}{b} + \dfrac{b}{d} =$ (A) $3$ (B) $0$ (C) $2$ (D) $-2$

   ›Reveal solutionSolution

   A plane perpendicular to two given planes has normal equal to the cross product of their normals; carrying this through gives $\frac{a}{b}+\frac{b}{d}=-2$.

   Concept and Intuition

   If a plane is perpendicular to two other planes, its normal vector must be perpendicular to both of their normals — so its normal is (parallel to) $\vec n_1\times\vec n_2$. A plane parallel to $\pi$ shares the same normal direction, differing only in the constant term.

   Step-by-Step Solution

   1. Normals: $\vec n_1=(2,3,-1)$, $\vec n_2=(1,-1,2)$.
   2. $\vec n_1\times\vec n_2 = \left(3(2)-(-1)(-1),\ -[2(2)-(-1)(1)],\ 2(-1)-3(1)\right) = (5,-5,-5)$, i.e. direction $(1,-1,-1)$.
   3. Plane $\pi$ through $(1,0,1)$ with this normal: $1(x-1)-1(y-0)-1(z-1)=0 \Rightarrow x-y-z=0$.
   4. The required plane is parallel to $\pi$, so it has the same normal direction. Writing it as $ax+by-z+d=0$ (z-coefficient fixed at $-1$, matching $\pi$'s normal exactly) gives $a=1,\ b=-1$. …

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4. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.If the pair of straight lines $9x^2 + axy + 4y^2 + 6x + by - 3 = 0$ represents two parallel lines then (A) $a = 6, b = 2$ (B) $a = 12, b = 4$ (C) $a = 3, b = 1$ (D) $a = -12, b = 4$

   ›Reveal solutionSolution

   Parallel lines require the quadratic part to be a perfect square ($a=12$), and matching the remaining linear/constant terms fixes $b=4$.

   Concept and Intuition

   A pair of parallel lines has the form $(L+p)(L+q)=0$ where $L$ is a common linear expression (from the perfect-square quadratic part). Expanding and matching coefficients with the given equation pins down the remaining unknowns.

   Step-by-Step Solution

   1. For $9x^2+axy+4y^2$ to represent two parallel lines, it must factor as a perfect square: $9x^2+axy+4y^2=(3x+2y)^2$ requires $a = 2\cdot3\cdot2=12$.
   2. So $L=3x+2y$, and the full equation is $(L+p)(L+q)=L^2+(p+q)L+pq = 9x^2+12xy+4y^2+(p+q)(3x+2y)+pq$. …

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5. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.A plane $\pi$ is passing through the points A(1, -2, 3) and B(6, 4, 5). If the plane $\pi$ is perpendicular the plane $3x - y + z = 2$, then the perpendicular distance from (0, 0, 0) to the plane $\pi$ is (A) $\frac{63}{\sqrt{594}}$ (B) $\frac{32}{\sqrt{594}}$ (C) $\frac{72}{\sqrt{435}}$ (D) $\frac{23}{\sqrt{135}}$

   ›Reveal solutionSolution

   The plane's normal is $\vec{AB}\times\vec n_2$ (perpendicular to both the direction in the plane and the given plane's normal); then use the point-normal form and the distance formula. Answer: $\dfrac{63}{\sqrt{594}}$.

   Concept and Intuition

   A plane through two points $A,B$ has $\vec{AB}$ lying within it, so the plane's normal $\vec n_1$ is perpendicular to $\vec{AB}$. The condition "$\pi$ perpendicular to the plane $3x-y+z=2$" means the two planes' normals are themselves perpendicular, i.e. $\vec n_1\cdot \vec n_2=0$. A vector perpendicular to both $\vec{AB}$ and $\vec n_2$ simultaneously is exactly their cross product, $\vec{AB}\times\vec n_2$.

   Step-by-Step Solution

   1. $\vec{AB}=B-A=(6-1,4-(-2),5-3)=(5,6,2)$.
   2. The reference plane $3x-y+z=2$ has normal $\vec n_2=(3,-1,1)$.
   3. Normal of $\pi$: $\vec n_1=\vec{AB}\times\vec n_2 = \big(6(1)-2(-1),\ -(5(1)-2(3)),\ 5(-1)-6(3)\big) = (8,\,1,\,-23)$.
   4. Plane through $A(1,-2,3)$ with normal $(8,1,-23)$: $8(x-1)+1(y+2)-23(z-3)=0 \Rightarrow 8x+y-23z-8+2+69=0 \Rightarrow 8x+y-23z+63=0$.
   5. Verify $B(6,4,5)$ satisfies it: $48+4-115+63=0$ ✓. …

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6. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.If $L_1$ represents the radical axis of circles $x^2 + y^2 - 4x - 6y + 5 = 0$ and $x^2 + y^2 - 2x - 4y - 1 = 0$, and $L_2$ represents the radical axis of $x^2 + y^2 + 2x + 2y - 7 = 0$ and $x^2 + y^2 + x + y + 9 = 0$, then ______ (A) $L_1$ is parallel to $L_2$ (B) $L_1$ is perpendicular to $L_2$ (C) $L_1$ and $L_2$ intersect at an angle $30°$ (D) $L_1$ and $L_2$ intersect at $(1, 7)$

   ›Reveal solutionSolution

   Tests computing radical axes of pairs of circles (by subtracting their equations) and then comparing slopes.

   Concept and Intuition

   The radical axis of two circles $S_1=0$ and $S_2=0$ (both with unit coefficient of $x^2,y^2$) is simply $S_1-S_2=0$, since this subtraction cancels the quadratic terms and leaves a linear equation — a straight line perpendicular to the line joining the centers.

   Step-by-Step Solution

   1. $L_1$: circles $x^2+y^2-4x-6y+5=0$ and $x^2+y^2-2x-4y-1=0$. Subtract: $(-4x-6y+5)-(-2x-4y-1) = -2x-2y+6=0$, i.e. $x+y-3=0$, or $x+y=3$. Slope $=-1$.
   2. $L_2$: circles $x^2+y^2+2x+2y-7=0$ and $x^2+y^2+x+y+9=0$. Subtract: $(2x+2y-7)-(x+y+9)=x+y-16=0$, i.e. $x+y=16$. Slope $=-1$.
   3. Both $L_1$ and $L_2$ have slope $-1$, so they are parallel (and, since the constants differ, distinct parallel lines, not the same line).

   Common Mistakes …

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7. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
   Q.$A(2,3,4), B(4,5,7), C(2,-6,3), D(4,-4,k)$ are four points. If the line $\overline{AB}$ is parallel to $\overline{CD}$, then $k$ is equal to _______ (A) 2 (B) 4 (C) 5 (D) 6

   ›Reveal solutionSolution

   This tests using proportional direction ratios to express that two lines are parallel. Answer: $k=6$.

   Concept and Intuition

   Two lines (or segments) are parallel exactly when their direction vectors are scalar multiples of each other — i.e., the direction ratios are proportional componentwise.

   Step-by-Step Solution

   1. $\vec{AB} = B-A = (4-2,\,5-3,\,7-4) = (2,2,3)$.
   2. $\vec{CD} = D-C = (4-2,\,-4-(-6),\,k-3) = (2,2,k-3)$.
   3. For $\vec{AB}\parallel\vec{CD}$: $\dfrac{2}{2}=\dfrac{2}{2}=\dfrac{3}{k-3}$.
   4. The first two ratios are both $1$, so we need $\dfrac{3}{k-3}=1 \Rightarrow k-3=3 \Rightarrow k=6$. …

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8. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.If the tangent drawn at A(2,1) to the curve $x=1+\dfrac{1}{y^2}$ meets the curve again at B, then (A) the tangent drawn at B coincides with the tangent drawn at A (B) the angle between the tangents drawn at A and B is neither $0$ nor $\dfrac{\pi}{2}$ (C) the tangent drawn at A and the tangent drawn at B are perpendicular to each other (D) the tangent drawn at A is parallel to the tangent drawn at B

   ›Reveal solutionSolution

   Find the second intersection B of the tangent at A with the curve, compute both slopes, and compare — the answer is (B).

   Concept and Intuition

   For a curve given as $x=g(y)$, it is often easier to differentiate with respect to $y$ first and invert: $\dfrac{dy}{dx}=1/\dfrac{dx}{dy}$. The "tangent meets the curve again at B" is a classic self-intersection problem: substitute the tangent line back into the curve equation; the tangency point at A must appear as a repeated root, and the remaining root gives B.

   Step-by-Step Solution

   1. Curve: $x=1+\dfrac1{y^2}=1+y^{-2}$. Differentiate w.r.t. $y$: $\dfrac{dx}{dy}=-2y^{-3}=-\dfrac{2}{y^3}$.
   2. So $\dfrac{dy}{dx}=-\dfrac{y^3}{2}$. At $A(2,1)$, $y=1$, so slope $m_A=-\dfrac12$.
   3. Tangent at A: $y-1=-\dfrac12(x-2) \Rightarrow x+2y=4$.
   4. Substitute $x=1+\dfrac1{y^2}$ into $x+2y=4$: $1+\dfrac1{y^2}+2y=4 \Rightarrow \dfrac1{y^2}+2y-3=0$. Multiply by $y^2$: $2y^3-3y^2+1=0$.
   5. Factor (knowing $y=1$ must be a root, and it must be a double root since the line is tangent there): $2y^3-3y^2+1=(y-1)^2(2y+1)$. The remaining root is $y=-\dfrac12$.
   6. At $y=-\dfrac12$: $x=1+\dfrac1{(1/4)}=1+4=5$. So $B=(5,-\tfrac12)$.
   7. Slope at B: $m_B=-\dfrac{y^3}{2}=-\dfrac{(-1/8)}{2}=\dfrac{1}{16}$. …

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9. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
   Q.The equation $8x^2-24xy+18y^2-6x+9y-5=0$ represents a ______. (A) pair of perpendicular lines (B) pair of parallel lines (C) pair of coincident lines (D) parabola

   ›Reveal solutionSolution

   When the quadratic part of a conic factors as a perfect square (i.e. $h^2=ab$), the conic represents a pair of parallel (or coincident) straight lines; substitution reduces the whole equation to a quadratic in a single linear combination, revealing two distinct parallel lines here.

   Concept and Intuition

   For $Ax^2+2Hxy+By^2+2Gx+2Fy+C=0$, the nature of the pair of lines (if it degenerates to lines) depends on $H^2-AB$: if it's zero, the two lines are parallel to each other (same quadratic-part direction), and whether they're distinct or coincident depends on the remaining linear/constant terms.

   Step-by-Step Solution

   1. Quadratic part: $8x^2-24xy+18y^2 = 2(4x^2-12xy+9y^2) = 2(2x-3y)^2$ — a perfect square, confirming $H^2=AB$ (parallel-line case; here $A=8,B=18,2H=-24\Rightarrow H=-12$, and indeed $H^2=144=AB=8\times18$).
   2. Let $u=2x-3y$. The full equation becomes $2u^2 - 6x+9y-5=0$. Since $-6x+9y = -3(2x-3y)=-3u$, this is $2u^2-3u-5=0$.
   3. Solve: $u=\dfrac{3\pm\sqrt{9+40}}{4}=\dfrac{3\pm7}{4}$, giving $u=\dfrac52$ or $u=-1$. …

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10. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
    Q.If the line joining the points $A(7, p, 2)$ and $B(q, -2, 5)$ is parallel to the line joining the points $C(2, -3, 5)$ and $D(-6, -15, 11)$, then the value of $p^2 + q^2 =$ (A) $25$ (B) $16$ (C) $9$ (D) $7$

    ›Reveal solutionSolution

    Tests using direction-vector proportionality for parallel lines in 3D to solve for unknown coordinates.

    Concept and Intuition

    Two lines are parallel exactly when their direction vectors are scalar multiples of each other. Writing out the direction vectors of both segments and matching components (using the known, unscaled component to fix the proportionality constant) gives simple linear equations for the unknowns.

    Step-by-Step Solution

    1. Direction of $CD$: $D-C=(-6-2,\,-15-(-3),\,11-5)=(-8,-12,6)$, which simplifies (dividing by 2) to $(-4,-6,3)$.
    2. Direction of $AB$: $B-A=(q-7,\,-2-p,\,5-2)=(q-7,\,-2-p,\,3)$.
    3. For $AB \parallel CD$, $AB$ must be a scalar multiple of $(-4,-6,3)$. Since the $z$-components already match exactly ($3=3$), the scalar is $1$, so $AB=(-4,-6,3)$ directly. …

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11. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
    Q.Condition that 2 curves $y^2 = 4ax$, $xy = c^2$ cut orthogonally is (A) $c^2 = 16a^2$ (B) $c^2 = 32a^2$ (C) $c^4 = 16a^4$ (D) $c^4 = 32a^4$

    ›Reveal solutionSolution

    Setting the product of the two curves' slopes at the intersection point to $-1$ (orthogonality) and combining with both curve equations gives $c^4=32a^4$.

    Concept and Intuition

    Two curves cut orthogonally at a point if their tangent lines there are perpendicular, i.e. the product of their slopes is $-1$. We find each curve's slope as a function of the intersection coordinates, impose orthogonality, then use both original curve equations to eliminate $x$ and $y$ and get a pure relation between $a$ and $c$.

    Step-by-Step Solution

    1. For $y^2=4ax$: differentiate, $2yy'=4a \Rightarrow y'=\dfrac{2a}{y}$.
    2. For $xy=c^2$: differentiate, $y+xy'=0 \Rightarrow y'=-\dfrac{y}{x}$.
    3. Orthogonality at the intersection point $(x,y)$: $\dfrac{2a}{y}\cdot\left(-\dfrac{y}{x}\right)=-1 \Rightarrow -\dfrac{2a}{x}=-1 \Rightarrow x=2a$.
    4. From $y^2=4ax$ with $x=2a$: $y^2=4a(2a)=8a^2$. …

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12. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
    Q.If the curves $2x^2+ky^2=30$ and $3y^2=28x$ cut each other orthogonally, then $k=$ (A) $5$ (B) $3$ (C) $2$ (D) $1$

    ›Reveal solutionSolution

    Orthogonal curves have slopes multiplying to $-1$ at their intersection; substituting the second curve's own equation gives $k=1$.

    Concept and Intuition

    Two curves intersect orthogonally when the tangent lines at their common point are perpendicular, i.e. the product of their slopes there is $-1$. Implicit differentiation gives each curve's slope as a function of $(x,y)$; since the point also lies on both original curves, we can substitute one equation into the other to eliminate variables.

    Step-by-Step Solution

    1. Differentiate $2x^2+ky^2=30$ implicitly: $4x+2ky\,y'=0\Rightarrow y_1'=-\dfrac{2x}{ky}$.
    2. Differentiate $3y^2=28x$ implicitly: $6y\,y'=28\Rightarrow y_2'=\dfrac{14}{3y}$.
    3. Orthogonality at the intersection point: $y_1'\cdot y_2'=-1$. $\left(-\dfrac{2x}{ky}\right)\left(\dfrac{14}{3y}\right)=-1\Rightarrow -\dfrac{28x}{3ky^2}=-1\Rightarrow 28x=3ky^2$.
    4. But the intersection point lies on curve 2, so $3y^2=28x$ there, i.e. $28x=3y^2$ (from the curve's own equation, not yet involving $k$). …

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