Q.Find the values of $p$ so that the lines $\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2}$ and $\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5}$ are at right angles.

Showing the 12 most recent of 25 on this concept.

1. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.If the line joining $A(4,1,2)$ and $B(0,k,1)$ is perpendicular to the line joining $C(-2,1,1)$ and $D(4,2,5)$, then the value of $k =$ ______ (A) $31$ (B) $-29$ (C) $-31$ (D) $29$

   ›Reveal solutionSolution

   Perpendicular lines have direction vectors with zero dot product; setting up $\vec{AB}\cdot\vec{CD}=0$ gives $k=29$.

   Concept and Intuition

   Two lines are perpendicular exactly when their direction vectors have a zero dot product. Here, $\vec{AB}$ is the direction of the line through $A,B$ and $\vec{CD}$ is the direction of the line through $C,D$.

   Step-by-Step Solution

   1. $\vec{AB}=B-A=(0-4,\,k-1,\,1-2)=(-4,\,k-1,\,-1)$.
   2. $\vec{CD}=D-C=(4-(-2),\,2-1,\,5-1)=(6,\,1,\,4)$.
   3. Perpendicularity: $\vec{AB}\cdot\vec{CD}=0$:

   $(-4)(6)+(k-1)(1)+(-1)(4)=0$

   $-24+k-1-4=0 \Rightarrow k-29=0 \Rightarrow k=29.$

   Common Mistakes …

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2. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.The set of real values of λ for which the vectors $\lambda\vec{i}-3\vec{j}+5\vec{k}$ and $2\lambda\vec{i}-\lambda\vec{j}+\vec{k}$ are perpendicular to each other is (A) $\{0, 1\}$ (B) $\{-2\}$ (C) $\{2, -1\}$ (D) $\varphi$

   ›Reveal solutionSolution

   Perpendicular vectors have zero dot product; the resulting quadratic in $\lambda$ has no real roots, so the answer set is empty.

   Concept and Intuition

   Two vectors are perpendicular exactly when their dot product vanishes. Setting up that equation converts a geometry condition into an algebraic one in $\lambda$.

   Step-by-Step Solution

   1. The vectors are $\vec u=(\lambda,-3,5)$ and $\vec v=(2\lambda,-\lambda,1)$.
   2. Perpendicularity: $\vec u\cdot\vec v=0$: $\ \lambda(2\lambda)+(-3)(-\lambda)+5(1)=0$.
   3. Simplify: $2\lambda^2+3\lambda+5=0$.
   4. Discriminant $=3^2-4(2)(5)=9-40=-31<0$. …

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3. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.The locus of a point at which the line joining the points $(-3, 1, 2), (1, -2, 4)$ subtends a right angle, is (A) $x^2 + y^2 + z^2 + 2x + y - 6z - 3 = 0$ (B) $x^2 + y^2 + z^2 + 2x - y - 6z + 3 = 0$ (C) $x^2 + y^2 + z^2 + 2x + y - 6z + 3 = 0$ (D) $x^2 + y^2 + z^2 - 2x + y - 6z + 3 = 0$

   ›Reveal solutionSolution

   This tests the classic 'locus subtending a right angle' problem, which is just the sphere having $AB$ as diameter, expressed via a perpendicularity dot-product condition. Answer: $x^2+y^2+z^2+2x+y-6z+3=0$.

   Concept and Intuition

   If a segment $AB$ subtends a right angle at a variable point $P$, then $PA\perp PB$, i.e. $\vec{PA}\cdot\vec{PB}=0$ for every such $P$ — this is exactly the defining property of a sphere with $AB$ as diameter (angle in a semicircle is a right angle, generalized to 3D). Writing this dot product in coordinates directly gives the sphere's equation.

   Step-by-Step Solution

   1. Let $P=(x,y,z)$. Then $\vec{PA}=A-P=(-3-x,\,1-y,\,2-z)$ and $\vec{PB}=B-P=(1-x,\,-2-y,\,4-z)$.
   2. Right angle at $P$: $\vec{PA}\cdot\vec{PB}=0$.
   3. $(-3-x)(1-x)=x^2+2x-3$ (expand: $-3+3x-x+x^2$).
   4. $(1-y)(-2-y)=y^2+y-2$ (expand: $-2-y+2y+y^2$).
   5. $(2-z)(4-z)=z^2-6z+8$ (expand: $8-2z-4z+z^2$). …

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4. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.Let $\pi$ be the plane passing through the point (3, -3, 1) and perpendicular to the line joining the points (3, 4, -1) and (2, -1, 5). If the equation of the plane containing the points (3, 4, -1), (-1, 2, 5) and perpendicular to the plane $\pi$ is $ax + y + cz - d = 0$ then $3(a + c) =$ (A) $-d$ (B) $2d$ (C) $d$ (D) $-2d$

   ›Reveal solutionSolution

   Find $\pi$'s normal from the given perpendicular line, then build the required plane's normal as a cross product (perpendicular to $\pi$'s normal and lying along the given two points), and match coefficients.

   Concept and Intuition

   A plane through two points and perpendicular to another plane has a normal that must be perpendicular both to the direction joining the two points (since that line lies in the plane) and to the normal of the other plane (since the planes are perpendicular). That normal is exactly the cross product of those two vectors.

   Step-by-Step Solution

   1. Direction of the line joining $(3,4,-1)$ and $(2,-1,5)$: $(2-3,-1-4,5-(-1))=(-1,-5,6)$ — this is normal to $\pi$.
   2. $\pi$ through $(3,-3,1)$: $-1(x-3)-5(y+3)+6(z-1)=0\Rightarrow -x-5y+6z-18=0$, i.e. normal $\vec n_1=(1,5,-6)$ (up to sign).
   3. Direction joining $(3,4,-1)$ and $(-1,2,5)$: $(-4,-2,6)=\vec d$.
   4. Normal of the required plane: $\vec n_2=\vec n_1\times\vec d=(1,5,-6)\times(-4,-2,6)$. Computing: $i:(5\cdot6-(-6)(-2))=18$; $j:-(1\cdot6-(-6)(-4))=18$; $k:(1\cdot(-2)-5(-4))=18$. So $\vec n_2=(18,18,18)\parallel(1,1,1)$. …

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5. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.If the vectors $2\bar{i} + 3\bar{j} + l\bar{k}$, $-3\bar{i} - 2\bar{j} - 4l\bar{k}$ and $\bar{i} - \bar{j} + 3l\bar{k}$ form a right angled triangle for a positive value of $l$, then the length of its hypotenuse is (A) $\sqrt{\dfrac{40}{3}}$ (B) $\sqrt{\dfrac{55}{3}}$ (C) $\sqrt{\dfrac{65}{3}}$ (D) $\sqrt{\dfrac{59}{3}}$

   ›Reveal solutionSolution

   Because the three given vectors sum to zero, they are the side vectors of a closed triangle; finding which pair is mutually perpendicular locates the right angle, and the third side (opposite that angle) is the hypotenuse whose length we compute.

   Concept and Intuition

   If three vectors $u,v,w$ satisfy $u+v+w=\bar0$, they can be laid tip-to-tail to close a triangle — this is exactly the vector-polygon condition. The vertex where two of them (as drawn, not reversed) are mutually perpendicular is the right-angle vertex of the triangle, and the side "opposite" that vertex — i.e. the third vector — is the hypotenuse. So the whole problem reduces to (a) finding which pair dots to zero for some positive $l$, and (b) computing that third vector's magnitude.

   Step-by-Step Solution

   1. Let $u=(2,3,l)$, $v=(-3,-2,-4l)$, $w=(1,-1,3l)$.
   2. Check closure: $u+v+w=(2-3+1,\,3-2-1,\,l-4l+3l)=(0,0,0)$ — confirmed, they form a triangle.
   3. Test each pair's dot product for a value making it zero (this locates the right angle):
      • $u\cdot v = -6-6-4l^2=-12-4l^2$ — never zero for real $l$.
      • $v\cdot w = -3+2-12l^2=-1-12l^2$ — never zero for real $l$.
      • $u\cdot w = 2-3+3l^2=3l^2-1$ — zero when $l^2=\dfrac13$, i.e. $l=\dfrac{1}{\sqrt3}>0$. ✓ (matches "positive value of $l$" in the problem.) …

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6. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
   Q.Find the equation of the plane passing through the point $(2,1,3)$ and perpendicular to the planes $x-2y+2z+3=0$ and $3x-2y+4z-4=0$. (A) $2x-y-2z+3=0$ (B) $x-2y+2z-3=0$ (C) $2x-y+2z-3=0$ (D) $2x+y-2z-3=0$

   ›Reveal solutionSolution

   The normal of a plane perpendicular to two given planes is the cross product of their normals. Answer: $2x-y-2z+3=0$.

   Concept and Intuition

   A plane perpendicular to two other planes must contain both their normal directions in its own plane — equivalently, its normal vector is perpendicular to both given normals, so it is (parallel to) their cross product.

   Step-by-Step Solution

   1. Normals: $\vec{n_1}=(1,-2,2)$ (from $x-2y+2z+3=0$), $\vec{n_2}=(3,-2,4)$ (from $3x-2y+4z-4=0$).
   2. $\vec{n_1}\times\vec{n_2}=\begin{vmatrix}i&j&k\\1&-2&2\\3&-2&4\end{vmatrix}=i(-8+4)-j(4-6)+k(-2+6)=(-4,2,4)$.
   3. Simplify direction: $(-4,2,4)\propto(2,-1,-2)$ (divide by $-2$). …

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7. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.Find the equation of the plane which passes through the points $(0,1,2)$ and $(-1,0,3)$, and is perpendicular to the plane $2x + 3y + z = 5$. (A) $3x - 4y + 18z + 32 = 0$ (B) $3x + 4y - 18z + 32 = 0$ (C) $4x + 3y - z + 1 = 0$ (D) $4x - 3y + z + 1 = 0$

   ›Reveal solutionSolution

   Three linear conditions (two points + perpendicularity to a given plane) on $ax+by+cz+d=0$ pin the plane down to $4x-3y+z+1=0$.

   Concept and Intuition

   A plane through two given points must satisfy each point's coordinates in its equation. "Perpendicular to another plane" means the two planes' normal vectors are perpendicular, i.e. their dot product is zero. Three such linear conditions (two point conditions + one perpendicularity condition) determine the plane's coefficients up to a common scale.

   Step-by-Step Solution

   1. Let the plane be $ax+by+cz+d=0$.
   2. Through $(0,1,2)$: $b+2c+d=0$ … (i)
   3. Through $(-1,0,3)$: $-a+3c+d=0$ … (ii)
   4. Perpendicular to $2x+3y+z=5$ (normal $(2,3,1)$): $2a+3b+c=0$ … (iii)
   5. From (i): $d=-b-2c$. From (ii): $d=a-3c$. Equate: $-b-2c=a-3c \Rightarrow c=a+b$.
   6. Substitute into (iii): $2a+3b+(a+b)=0 \Rightarrow 3a+4b=0 \Rightarrow a=-\dfrac{4b}{3}$.
   7. Choose $b=3$ (clears the fraction): $a=-4$, $c=a+b=-4+3=-1$, $d=-b-2c=-3-2(-1)=-1$. …

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8. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.The plane passing through $(2,1,-3)$ and perpendicular to $3i-j+2k$ contains the points (A) $(1,5,1)$ & $(3,0,-5)$ (B) $\left(\dfrac{1}{3},3,\dfrac{1}{2}\right)$ & $\left(1,5,\dfrac{1}{2}\right)$ (C) $(3,1,-5)$ & $\left(\dfrac{1}{3},3,\dfrac{1}{2}\right)$ (D) $(1,5,3)$ & $(3,0,1)$

   ›Reveal solutionSolution

   This tests writing a plane's equation from a point and normal vector, then checking which pair of points satisfies it. The answer is option (B).

   Concept and Intuition

   A plane through point $P_0=(x_0,y_0,z_0)$ perpendicular to $\vec n=(a,b,c)$ has equation $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$. Once we have this equation, a point "lies in the plane" exactly when it satisfies the equation — we just substitute each candidate point and check.

   Step-by-Step Solution

   1. Normal vector $\vec n=3\hat i-\hat j+2\hat k=(3,-1,2)$, point $(2,1,-3)$.
   2. Plane equation: $3(x-2)-1(y-1)+2(z+3)=0 \Rightarrow 3x-6-y+1+2z+6=0 \Rightarrow 3x-y+2z+1=0$.
   3. Test option (A): $(1,5,1)$: $3(1)-5+2(1)+1=3-5+2+1=1\ne0$ — fails.
   4. Test option (B): $\left(\frac13,3,\frac12\right)$: $3\left(\frac13\right)-3+2\left(\frac12\right)+1=1-3+1+1=0$ — satisfies. $\left(1,5,\frac12\right)$: $3(1)-5+2\left(\frac12\right)+1=3-5+1+1=0$ — satisfies. Both points lie in the plane. …

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9. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.Let $\vec a=2\vec i-3\vec j-5\vec k$ and $\vec b=3\vec i+2\vec j-5\vec k$ be two vectors and $\vec r$ be a vector in the plane of $\vec a$ and $\vec b$. If $\vec r$ is orthogonal to the vector $5\vec i-2\vec j+3\vec k$ and the magnitude of $\vec r$ is $\sqrt{94}$, then $|\vec r \cdot \vec b| =$ (A) $36$ (B) $38$ (C) $42$ (D) $46$

   ›Reveal solutionSolution

   Since $\vec r$ is in the plane of $\vec a,\vec b$ and perpendicular to $\vec n$, it must be parallel to $(\vec a\times\vec b)\times\vec n$; scaling this to the given magnitude $\sqrt{94}$ and dotting with $\vec b$ gives $|\vec r\cdot\vec b|=46$.

   Concept and Intuition

   Two conditions pin down $\vec r$'s direction uniquely (up to sign and scale): (1) $\vec r$ lies in the plane of $\vec a,\vec b$, meaning $\vec r\perp N$ where $N=\vec a\times\vec b$ is the plane's normal; (2) $\vec r\perp\vec n$ (given). A vector perpendicular to both $N$ and $\vec n$ must be parallel to $N\times\vec n$.

   Step-by-Step Solution

   1. $\vec a=(2,-3,-5)$, $\vec b=(3,2,-5)$. Compute $N=\vec a\times\vec b$: $N_i=(-3)(-5)-(-5)(2)=15+10=25$ $N_j=-[(2)(-5)-(-5)(3)]=-[-10+15]=-5$ $N_k=(2)(2)-(-3)(3)=4+9=13$ So $N=(25,-5,13)$.
   2. $\vec n=(5,-2,3)$. Compute $N\times\vec n$: $i$: $(-5)(3)-(13)(-2)=-15+26=11$ $j$: $-[(25)(3)-(13)(5)]=-[75-65]=-10$ $k$: $(25)(-2)-(-5)(5)=-50+25=-25$ So $N\times\vec n=(11,-10,-25)$.
   3. $\vec r=\lambda(11,-10,-25)$ for some scalar $\lambda$. $|N\times\vec n|^2=121+100+625=846$.
   4. $|\vec r|^2=\lambda^2(846)=94\Rightarrow\lambda^2=\dfrac{94}{846}=\dfrac19\Rightarrow\lambda=\pm\dfrac13$. …

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10. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
    Q.If (2, -1, 3) is the foot of the perpendicular drawn from the origin (0, 0, 0) to a plane then the equation of that plane is (A) $2x + y - 3z + 6 = 0$ (B) $2x - y + 3z - 14 = 0$ (C) $2x - y + 3z - 13 = 0$ (D) $2x + y + 3z - 10 = 0$

    ›Reveal solutionSolution

    When the foot of the perpendicular from the origin to a plane is given, that

    point's position vector IS the plane's normal direction, and plugging the point

    back in gives the constant term instantly.

    Concept and Intuition

    The perpendicular from the origin to a plane is, by definition, along the

    plane's normal direction. So if $F=(x_0,y_0,z_0)$ is the foot of that

    perpendicular, the vector $\overrightarrow{OF}=(x_0,y_0,z_0)$ is normal to the plane,

    and the plane's equation is $x_0x+y_0y+z_0z = x_0^2+y_0^2+z_0^2$ (since $F$ itself

    must satisfy the plane equation, and $|OF|^2$ is exactly the dot product of $F$

    with itself).

    Step-by-Step Solution

    1. Normal direction $= (2,-1,3)$ (the given foot of perpendicular).
    2. Plane: $2x - y + 3z = k$ for some constant $k$.
    3. Since $(2,-1,3)$ lies on the plane: $k = 2(2) + (-1)(-1) + 3(3) = 4+1+9 = 14$. …

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11. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
    Q.If $\vec a$ and $\vec b$ are two vectors such that $|\vec a| = 2$, $|\vec b| = 3$ and $\vec a + t\vec b$ and $\vec a - t\vec b$ are perpendicular, where 't' is a positive scalar, then (A) $t = \pm\dfrac{2}{3}$ (B) $t = \dfrac{4}{9}$ (C) $t = \dfrac{2}{3}$ (D) $t = \dfrac{2}{9}$

    ›Reveal solutionSolution

    Perpendicularity of $\vec a+t\vec b$ and $\vec a-t\vec b$ forces $|\vec a|^2=t^2|\vec b|^2$, giving the positive value $t=2/3$.

    Concept and Intuition

    Two vectors are perpendicular exactly when their dot product is zero. Expanding $(\vec a+t\vec b)\cdot(\vec a-t\vec b)$ using the distributive property of the dot product collapses to a simple difference of squared magnitudes, since $\vec a\cdot\vec b$ cancels.

    Step-by-Step Solution

    1. $(\vec a+t\vec b)\cdot(\vec a-t\vec b) = \vec a\cdot\vec a - t\,\vec a\cdot\vec b + t\,\vec b\cdot\vec a - t^2\,\vec b\cdot\vec b = |\vec a|^2 - t^2|\vec b|^2$.
    2. Setting this to zero (perpendicularity): $|\vec a|^2 = t^2|\vec b|^2$.
    3. Substitute $|\vec a|=2$, $|\vec b|=3$: $4 = 9t^2 \Rightarrow t^2=\dfrac49$. …

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12. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
    Q.If the equation of the plane passing through the points $(1,-3,2)$, $(-2,3,1)$ and perpendicular to the plane $x + 2y - 3z = 0$ is $ax + by + cz + d = 0$, then $\dfrac{a+b}{c+d} =$ (A) $\dfrac{3}{11}$ (B) $13$ (C) $\dfrac{13}{11}$ (D) $3$

    ›Reveal solutionSolution

    This tests finding a plane through two points and perpendicular to another plane, using the cross product of the connecting direction vector and the given plane's normal. Answer: $13$.

    Concept and Intuition

    A plane's normal vector must be perpendicular to every direction lying in the plane. Since the plane contains points $P_1(1,-3,2)$ and $P_2(-2,3,1)$, the vector $\vec{P_1P_2}$ lies in the plane, so the required normal $\vec n=(a,b,c)$ satisfies $\vec n\cdot\vec{P_1P_2}=0$. Also, "perpendicular to the plane $x+2y-3z=0$" means the two planes' normals are perpendicular, so $\vec n\cdot(1,2,-3)=0$. A vector perpendicular to both $\vec{P_1P_2}$ and $(1,2,-3)$ is simply their cross product.

    Step-by-Step Solution

    1. Direction vector: $\vec d = P_2-P_1 = (-2-1,\,3-(-3),\,1-2)=(-3,6,-1)$.
    2. Normal of given plane: $\vec n_2=(1,2,-3)$.
    3. Required normal: $\vec n=\vec d\times \vec n_2 = \begin{vmatrix}\vec i&\vec j&\vec k\\-3&6&-1\\1&2&-3\end{vmatrix}$ $= \vec i(6(-3)-(-1)(2)) - \vec j((-3)(-3)-(-1)(1)) + \vec k((-3)(2)-6(1))$ $= \vec i(-18+2) - \vec j(9+1) + \vec k(-6-6) = (-16,-10,-12)$. …

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