Q.If the lines $\frac{x-1}{-3} = \frac{y-2}{2k} = \frac{z-3}{2}$ and $\frac{x-1}{3k} = \frac{y-1}{1} = \frac{z-6}{-5}$ are perpendicular, find the value of $k$.

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 2024Set eng-2024-05-19-AN1 markMCQ
   Q.Let ABC be an equilateral triangle of side $a$. M and N are two points on the sides AB and AC respectively such that $\overline{AN} = K\overline{AC}$ and $\overline{AB} = 3\overline{AM}$. If the vectors $\overline{BN}$ and $\overline{CM}$ are perpendicular, then $K =$ (A) $\dfrac{1}{5}$ (B) $\dfrac{2}{5}$ (C) $-\dfrac{1}{5}$ (D) $-\dfrac{2}{5}$

   ›Reveal solutionSolution

   Express $\overline{BN}$ and $\overline{CM}$ in terms of the two sides from $A$, use the $60^\circ$ dot product of an equilateral triangle, and set the perpendicularity condition to zero to solve for $K=\dfrac15$.

   Concept and Intuition

   Placing the vertex $A$ at the origin turns every other point into a simple scalar multiple of the two side vectors $\overline{AB}$ and $\overline{AC}$. Perpendicularity of two vectors becomes an algebraic condition: their dot product is zero. For an equilateral triangle, $\overline{AB}.\overline{AC}=a^2\cos 60^\circ=\dfrac{a^2}{2}$.

   Step-by-Step Solution

   1. Let $A$ be the origin, $\bar c=\overline{AB}$, $\bar b=\overline{AC}$, with $|\bar b|=|\bar c|=a$ and $\bar b.\bar c=\dfrac{a^2}{2}$.
   2. Since $\overline{AB}=3\overline{AM}$, $M=\dfrac{\bar c}{3}$. Since $\overline{AN}=K\overline{AC}$, $N=K\bar b$.
   3. $\overline{BN}=N-B=K\bar b-\bar c$, and $\overline{CM}=M-C=\dfrac{\bar c}{3}-\bar b$.
   4. Perpendicularity: $\overline{BN}.\overline{CM}=0$: $(K\bar b-\bar c).\left(\dfrac{\bar c}{3}-\bar b\right)= \dfrac{K}{3}(\bar b.\bar c) - K|\bar b|^2 - \dfrac13|\bar c|^2 + \bar b.\bar c = 0$.
   5. Substitute $|\bar b|^2=|\bar c|^2=a^2$, $\bar b.\bar c=a^2/2$: …

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3. 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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4. 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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5. 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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6. 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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7. 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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8. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
   Q.Let $u = 2i + 3j + k$, $v = -3i + 2j$ and $w = i - j + 4k$. Then which of the following statement is true? (A) $u$ is perpendicular to $v$ but not $w$ (B) $v$ is perpendicular to $w$ but not $u$ (C) $w$ is perpendicular to $u$ but not $v$ (D) $u$ is perpendicular to both $v$ and $w$

   ›Reveal solutionSolution

   Direct dot products show $u\cdot v=0$ (perpendicular) and $u\cdot w=3\ne0$ (not perpendicular), matching option (A).

   Concept and Intuition

   Two vectors are perpendicular exactly when their dot product is zero. Checking each pair's dot product directly settles every option — no need for angle or cross-product computation.

   Step-by-Step Solution

   1. $u\cdot v = (2)(-3)+(3)(2)+(1)(0) = -6+6+0 = 0$ — so $u\perp v$.
   2. $u\cdot w = (2)(1)+(3)(-1)+(1)(4) = 2-3+4 = 3 \ne 0$ — so $u$ is not perpendicular to $w$.
   3. For completeness, $v\cdot w=(-3)(1)+(2)(-1)+(0)(4)=-3-2+0=-5\ne0$ — $v$ is also not perpendicular to $w$. …

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9. 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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10. 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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11. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
    Q.If $\bar{a}=2\bar{i}+3\bar{j}, \bar{b}=3\bar{j}+4\bar{k}$ and $\bar{c}=5\bar{i}+4\bar{k}$ are three vectors, then a vector which is perpendicular to $\bar{a}$ and $\bar{b}\times\bar{c}$ is (A) $45\bar{i}-30\bar{j}+15\bar{k}$ (B) $3\bar{i}-2\bar{j}+\bar{k}$ (C) $-30\bar{i}+20\bar{j}+4\bar{k}$ (D) $-45\bar{i}+30\bar{j}+4\bar{k}$

    ›Reveal solutionSolution

    This tests the vector-triple-product idea: a vector perpendicular to both $\bar a$ and $\bar b\times\bar c$ is simply $\bar a\times(\bar b\times\bar c)$.

    Concept and Intuition

    The cross product of any two vectors is perpendicular to both of them. So if we want a single vector perpendicular to $\bar a$ AND to $\bar b\times\bar c$, the natural candidate is $\bar a\times(\bar b\times\bar c)$ — it is perpendicular to $\bar a$ by definition of cross product, and perpendicular to $\bar b\times\bar c$ for the same reason. No need to invoke the full triple-product expansion formula; we just compute it directly.

    Step-by-Step Solution

    1. Given $\bar a=2\bar i+3\bar j+0\bar k$, $\bar b=0\bar i+3\bar j+4\bar k$, $\bar c=5\bar i+0\bar j+4\bar k$.
    2. Compute $\bar b\times\bar c=\begin{vmatrix}\bar i&\bar j&\bar k\\0&3&4\\5&0&4\end{vmatrix}$ $=\bar i(3\cdot4-4\cdot0)-\bar j(0\cdot4-4\cdot5)+\bar k(0\cdot0-3\cdot5)=12\bar i+20\bar j-15\bar k$.
    3. Compute $\bar a\times(\bar b\times\bar c)=\begin{vmatrix}\bar i&\bar j&\bar k\\2&3&0\\12&20&-15\end{vmatrix}$ …

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12. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
    Q.The vector of magnitude 2 lying in the plane of $\bar{a} = 2\bar{i} - \bar{j} + \bar{k}$ and $\bar{b} = \bar{i} + 3\bar{j} - 5\bar{k}$ and perpendicular to the vector $\bar{c} = \bar{i} + \bar{j} + \bar{k}$ is (A) $\dfrac{\sqrt{2}}{\sqrt{61}}(4\bar{i}+5\bar{j}-9\bar{k})$ (B) $\dfrac{\sqrt{2}}{\sqrt{9}}(2\bar{i}+3\bar{j}-5\bar{k})$ (C) $\dfrac{\sqrt{2}}{\sqrt{31}}(\bar{i}+5\bar{j}-6\bar{k})$ (D) $\dfrac{\sqrt{2}}{\sqrt{13}}(-\bar{i}-3\bar{j}+4\bar{k})$

    ›Reveal solutionSolution

    This tests writing a vector "in the plane of $\bar a,\bar b$" as a linear combination $\alpha\bar a+\beta\bar b$, using perpendicularity to $\bar c$ to pin the ratio $\alpha:\beta$, and finally scaling the resulting direction to the required magnitude.

    Concept and Intuition

    Every vector lying in the plane spanned by $\bar a$ and $\bar b$ is some linear combination $\alpha\bar a+\beta\bar b$ — that's what "lying in the plane" means. The extra condition (perpendicular to $\bar c$) gives one linear equation in $\alpha,\beta$, which fixes their ratio (the direction is determined up to an overall scale). The magnitude condition then fixes that scale.

    Step-by-Step Solution

    1. Let $\bar d=\alpha\bar a+\beta\bar b$ for some scalars $\alpha,\beta$ (this covers every vector in the plane of $\bar a,\bar b$).
    2. Require $\bar d\cdot\bar c=0$: $\alpha(\bar a\cdot\bar c)+\beta(\bar b\cdot\bar c)=0$.
    3. $\bar a\cdot\bar c=(2)(1)+(-1)(1)+(1)(1)=2-1+1=2$. $\bar b\cdot\bar c=(1)(1)+(3)(1)+(-5)(1)=1+3-5=-1$.
    4. So $2\alpha-\beta=0\Rightarrow\beta=2\alpha$. Taking $\alpha=1,\beta=2$: direction $=\bar a+2\bar b=(2+2,\,-1+6,\,1-10)=(4,5,-9)$.
    5. Magnitude of this direction: $\sqrt{4^2+5^2+(-9)^2}=\sqrt{16+25+81}=\sqrt{122}$. …

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