Q.$\vec{AB} = 3\hat{i} - \hat{j} + \hat{k}$ and $\vec{CD} = -3\hat{i} + 2\hat{j} + 4\hat{k}$ are two vectors. The position vectors of the points $A$ and $C$ are $6\hat{i} + 7\hat{j} + 4\hat{k}$ and $-9\hat{j} + 2\hat{k}$, respectively. Find the position vector of a point $P$ on the line $AB$ and a point $Q$ on the line $CD$ such that $\vec{PQ}$ is perpendicular to $\vec{AB}$ and $\vec{CD}$ both.

Showing the 12 most recent of 16 on this concept.

1. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.If the equation of the plane passing through the point (2, -1, 3) and perpendicular to each of the planes $3x - 2y + z = 8$ and $x + y + z = 6$ is $lx + my + nz = 1$, then $4m + 2n - 3l =$ (A) $0$ (B) $\dfrac{-20}{11}$ (C) $1$ (D) $3$

   ›Reveal solutionSolution

   The plane's normal is the cross product of the two given planes' normals; writing the plane in the form $lx+my+nz=1$ and substituting gives $4m+2n-3l=1$.

   Concept and Intuition

   A plane perpendicular to two other planes must have a normal vector perpendicular to both of their normals — that is exactly the cross product of the two given normal vectors. Once we have the plane's Cartesian equation through the given point, we simply normalize it into the form $lx+my+nz=1$ (by dividing through by the constant term) to read off $l,m,n$.

   Step-by-Step Solution

   1. Normals of given planes: $\vec{n_1}=(3,-2,1)$ (from $3x-2y+z=8$), $\vec{n_2}=(1,1,1)$ (from $x+y+z=6$).
   2. Required normal $\vec n=\vec n_1\times\vec n_2$:

   $\vec n=\begin{vmatrix}\hat i&\hat j&\hat k\\3&-2&1\\1&1&1\end{vmatrix}=\hat i(-2-1)-\hat j(3-1)+\hat k(3+2)=(-3,-2,5).$

   3. Plane through $(2,-1,3)$ with this normal:

   $-3(x-2)-2(y+1)+5(z-3)=0\Rightarrow -3x+6-2y-2+5z-15=0\Rightarrow -3x-2y+5z-11=0.$

   4. Rewrite as $lx+my+nz=1$: from $-3x-2y+5z=11$, divide by $11$: $\frac{-3}{11}x+\frac{-2}{11}y+\frac{5}{11}z=1\ \Rightarrow\ l=\frac{-3}{11},\ m=\frac{-2}{11},\ n=\frac{5}{11}.$ …

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2. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If two sides of a triangle are represented by $3x^2 - 5xy + 2y^2 = 0$ and its orthocentre is $(2,1)$, then the equation of the third side is (A) $2x + y - 4 = 0$ (B) $6x + 3y - 13 = 0$ (C) $8x + 4y - 17 = 0$ (D) $10x + 5y - 21 = 0$

   ›Reveal solutionSolution

   The homogeneous pair $3x^2-5xy+2y^2=0$ gives two sides through the origin (vertex $O$); using the orthocentre to locate the other two vertices on those sides pins down the third side as $10x+5y-21=0$.

   Concept and Intuition

   A homogeneous second-degree equation $ax^2+2hxy+by^2=0$ always represents a pair of lines through the origin — here they are two sides of the triangle, meeting at the vertex $O=(0,0)$. The orthocentre lies on every altitude, and each altitude is perpendicular to the side it is dropped onto. So if $B$ lies on one side, the altitude from $B$ is perpendicular to the opposite side (through $O$ and $C$), and it passes through both $B$ and the orthocentre $H$ — this pins down $B$ exactly, and symmetrically for $C$.

   Step-by-Step Solution

   1. Factor $3x^2-5xy+2y^2=(3x-2y)(x-y)=0$, so the two sides through $O$ are $3x-2y=0$ (direction $(2,3)$) and $x-y=0$ (direction $(1,1)$).
   2. Let $B=(b,b)$ on $x-y=0$, and $C=(2c,3c)$ on $3x-2y=0$ (both parametrised along their line's direction).
   3. Altitude from $B$ is perpendicular to side $OC$ (direction $(2,3)$) and passes through $B$ and $H(2,1)$: vector $BH=(2-b,\,1-b)$ must satisfy $2(2-b)+3(1-b)=0\Rightarrow 7-5b=0\Rightarrow b=\tfrac75$. So $B=\left(\tfrac75,\tfrac75\right)$. …

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3. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.One line of the pair of lines $x^2 + xy - 2y^2 = 0$ is perpendicular to one line of the pair of lines $3y^2 - 5xy - 2x^2 = 0$. If the combined equation of the two lines other than those two perpendicular lines is $ax^2 + 2hxy + by^2 = 0$, then $a + 2h + b =$ (A) $-1$ (B) $1$ (C) $0$ (D) $-5$

   ›Reveal solutionSolution

   Factor both pairs of lines, find the perpendicular pair by matching slopes whose product is $-1$, then combine the two remaining ("other") lines. Answer: $a+2h+b=0$.

   Concept and Intuition

   A homogeneous second-degree equation $px^2+qxy+ry^2=0$ always factors into two lines through the origin. Once each pair is factored into individual lines with known slopes, we identify which specific line from each pair is perpendicular to a line from the other pair, then combine the two lines that are not part of that perpendicular pair.

   Step-by-Step Solution

   1. Factor the first pair: $x^2+xy-2y^2 = (x+2y)(x-y)$.

      Check: $(x+2y)(x-y) = x^2 -xy+2xy-2y^2 = x^2+xy-2y^2$ ✓.

      Lines: $x-y=0$ (slope $1$) and $x+2y=0$ (slope $-\tfrac12$).

   2. Factor the second pair: treat $3y^2-5xy-2x^2=0$ as a quadratic in $y$:

   $y = \frac{5x\pm\sqrt{25x^2+24x^2}}{6}=\frac{5x\pm7x}{6}$

   So $y=2x$ (slope $2$) or $y=-\dfrac x3$ (slope $-\tfrac13$), i.e. lines $2x-y=0$ and $x+3y=0$.

   3. Find the perpendicular pair (one line from each set): check products of slopes:

   • $1\times2=2$, $1\times(-\tfrac13)=-\tfrac13$, $(-\tfrac12)\times2=-1$ ✓ (perpendicular!), $(-\tfrac12)\times(-\tfrac13)=\tfrac16$. …

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4. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If the equation of the pair of lines passing through (1, 1) and perpendicular to the pair of lines $2x^2 + xy - y^2 - x + 2y - 1 = 0$ is $ax^2 + 2hxy + by^2 + 2gx + 3y = 0$, then $\dfrac{b}{a} =$ (A) $g/h$ (B) $2(g+h)$ (C) $2(g-h)$ (D) $gh$

   ›Reveal solutionSolution

   Factor the given pair of lines, replace each by its perpendicular through $(1,1)$, form the new pair, match coefficients against the template, then check which option equals $b/a$. Answer: (B) $2(g+h)$.

   Concept and Intuition

   A homogeneous-plus-linear second-degree equation $ax^2+2hxy+by^2+\dots=0$ represents a pair of straight lines when it factors into two linear factors. To find lines perpendicular to a given pair, we replace each line's slope $m$ by $-1/m$ and force the new pair through the given point.

   Step-by-Step Solution

   1. Factor $2x^2+xy-y^2-x+2y-1=0$. The quadratic part $2x^2+xy-y^2=(2x-y)(x+y)$, so try $(2x-y+c_1)(x+y+c_2)=0$.
   2. Expanding: $2x^2+xy-y^2+(2c_2+c_1)x+(c_1-c_2)y+c_1c_2$. Matching: $2c_2+c_1=-1$, $c_1-c_2=2$, $c_1c_2=-1$. Solving gives $c_1=1,\ c_2=-1$.
   3. So the pair is $(2x-y+1)(x+y-1)=0$: lines $2x-y+1=0$ (slope $2$) and $x+y-1=0$ (slope $-1$).
   4. Line through $(1,1)$ perpendicular to slope $2$: slope $-\tfrac12$, giving $y-1=-\tfrac12(x-1)\Rightarrow x+2y-3=0$.
   5. Line through $(1,1)$ perpendicular to slope $-1$: slope $1$, giving $y-1=(x-1)\Rightarrow x-y=0$.
   6. Combined equation: $(x+2y-3)(x-y)=0$. Expand: $x^2-xy+2xy-2y^2-3x+3y=x^2+xy-2y^2-3x+3y=0$. …

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5. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.If the direction ratios of two lines $L_1$ and $L_2$ are given by (1, −2, 2) and (−2, 3, −6) respectively, then the direction ratios of the line which is perpendicular to the lines $L_1$ and $L_2$ are (A) $(1,-2,3)$ (B) $(-2,3,5)$ (C) $(6,2,-1)$ (D) $(2,-1,3)$

   ›Reveal solutionSolution

   The direction of a line perpendicular to two given lines is their cross product. Answer: $(6,2,-1)$.

   Concept and Intuition

   If two lines have direction vectors $\vec{u}$ and $\vec{v}$, any vector perpendicular to both lines lies along $\vec{u}\times\vec{v}$, since the cross product is by definition perpendicular to each of its factors.

   Step-by-Step Solution

   1. $\vec{u}=(1,-2,2)$, $\vec{v}=(-2,3,-6)$.
   2. $\vec{u}\times\vec{v} = \left(\ (-2)(-6)-(2)(3),\ \ (2)(-2)-(1)(-6),\ \ (1)(3)-(-2)(-2)\ \right)$
   3. $= (12-6,\ -4+6,\ 3-4) = (6,2,-1)$.

   Common Mistakes …

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6. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.Equation of the plane passing through the origin and perpendicular to the planes $x + 2y - z = 1$ and $3x - 4y + z = 5$ is (A) $x + 2y - 5z = 0$ (B) $x - 2y + 5z = 0$ (C) $x + 2y + 5z = 0$ (D) $3x + y - 5z = 0$

   ›Reveal solutionSolution

   The normal of a plane perpendicular to two other planes is the cross product of their normals; passing through the origin fixes the constant. Answer: $x+2y+5z=0$.

   Concept and Intuition

   If a plane is perpendicular to two given planes, its normal vector must be perpendicular to both of their normals (because the required plane contains directions along both given planes' lines of intersection with it, or more directly: being perpendicular to a plane with normal $\vec n$ means the required plane's normal is perpendicular to $\vec n$... actually the standard fact used here is: the normal of the new plane is perpendicular to both given normals, so it is their cross product).

   Step-by-Step Solution

   1. Normal of $x+2y-z=1$: $\vec n_1=(1,2,-1)$.
   2. Normal of $3x-4y+z=5$: $\vec n_2=(3,-4,1)$.
   3. Required plane's normal $\vec n=\vec n_1\times\vec n_2$: $\vec n=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\1&2&-1\\3&-4&1\end{vmatrix}=\mathbf i(2\cdot1-(-1)(-4))-\mathbf j(1\cdot1-(-1)(3))+\mathbf k(1\cdot(-4)-2\cdot3)$ …

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7. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.If the equation of the pair of straight lines passing through the point (1, 1) and perpendicular to the pair of lines $3x^2+11xy-4y^2=0$ is $ax^2+2hxy+by^2+2gx+2fy+12=0$, then $2(a-h+b-g+f-12)=$ (A) $0$ (B) $-7$ (C) $-19$ (D) $13$

   ›Reveal solutionSolution

   Factor the given pair of lines, form the perpendicular pair through $(1,1)$, multiply the two new lines together, match coefficients, and evaluate the expression.

   Concept and Intuition

   A homogeneous pair of lines factors into two linear factors when possible; perpendicular lines have slopes that are negative reciprocals of each other. Once the two new individual lines are known, their combined (product) equation is the pair-of-lines equation through the required point.

   Step-by-Step Solution

   1. Factor: $3x^2+11xy-4y^2=(3x-y)(x+4y)$ (check: $3x\cdot x=3x^2$, cross terms $12xy-xy=11xy$, $-y\cdot4y=-4y^2$ ✓). So original lines have slopes $3$ and $-1/4$.
   2. Perpendicular slopes: $-1/3$ and $4$.
   3. Line 1 through $(1,1)$, slope $-1/3$: $y-1=-\tfrac13(x-1)\Rightarrow x+3y-4=0$.
   4. Line 2 through $(1,1)$, slope $4$: $y-1=4(x-1)\Rightarrow 4x-y-3=0$.
   5. Multiply: $(x+3y-4)(4x-y-3)=4x^2-xy-3x+12xy-3y^2-9y-16x+4y+12$ $=4x^2+11xy-3y^2-19x-5y+12=0$. …

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8. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.If $(\alpha, \beta)$ is the orthocentre of the triangle with the vertices $(2,2), (5,1), (4,4)$, then $\alpha + \beta =$ (A) $6$ (B) $5$ (C) $\dfrac{5}{2}$ (D) $\dfrac{7}{2}$

   ›Reveal solutionSolution

   Intersect two altitudes (each perpendicular to a side, through the opposite vertex) to get the orthocentre $(3.5,2.5)$, so $\alpha+\beta=6$.

   Concept and Intuition

   The orthocentre is the common intersection point of the three altitudes of a triangle. We only need two altitudes: each is the line through a vertex, perpendicular to the opposite side (using the negative-reciprocal slope rule).

   Step-by-Step Solution

   1. Vertices: $A(2,2),\,B(5,1),\,C(4,4)$.
   2. Slope of $BC = \dfrac{4-1}{4-5} = -3$. Altitude from $A$ is perpendicular to $BC$: slope $=\dfrac{1}{3}$. Line through $A(2,2)$: $y-2=\dfrac13(x-2) \implies x=3y-4$.
   3. Slope of $AC = \dfrac{4-2}{4-2}=1$. Altitude from $B$ is perpendicular to $AC$: slope $=-1$. Line through $B(5,1)$: $y-1=-(x-5) \implies x+y=6$.
   4. Solve $x=3y-4$ and $x+y=6$: $(3y-4)+y=6 \implies 4y=10 \implies y=2.5,\ x=3.5$. …

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9. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.If $l, m, n$ are the direction cosines of a line that is perpendicular to the lines having the direction ratios 1, 2, -1 and 1, -2, 1 then $(l+m+n)^2=$ (A) $\frac{1}{20}$ (B) $\frac{9}{5}$ (C) $\frac{1}{5}$ (D) $\frac{3}{20}$

   ›Reveal solutionSolution

   The direction of a line perpendicular to two given lines is their cross product; normalizing gives $(l+m+n)^2=\dfrac95$.

   Concept and Intuition

   A vector perpendicular to two given direction vectors is found via their cross product. Once we have direction ratios, dividing by their magnitude gives the direction cosines, whose squares sum to 1.

   Step-by-Step Solution

   1. Direction ratios given: $\vec a=(1,2,-1)$, $\vec b=(1,-2,1)$.
   2. $\vec a\times\vec b = \big(2\cdot1-(-1)(-2),\,-[1\cdot1-(-1)(1)],\,1\cdot(-2)-2\cdot1\big) = (2-2,\,-(1+1),\,-2-2) = (0,-2,-4)$.
   3. Simplify: $(0,-2,-4)\propto(0,1,2)$ (dividing by $-2$).
   4. Magnitude $=\sqrt{0^2+1^2+2^2}=\sqrt5$, so direction cosines are $\left(0,\dfrac{1}{\sqrt5},\dfrac{2}{\sqrt5}\right)$. …

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10. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
    Q.The lines $p(p^2+1)x - y + q = 0$ and $(p^2+1)^2x + (p^2+1)y + 2q = 0$ are perpendicular to a line L for (A) exactly one value of p (B) exactly two values of p (C) more than two values of p (D) no value of p

    ›Reveal solutionSolution

    Two lines each perpendicular to a common line $L$ must be parallel to each other; equating their slopes gives a unique solution $p=-1$.

    Concept and Intuition

    If line 1 $\perp L$ and line 2 $\perp L$, then lines 1 and 2 share the same direction relative to $L$ — they must be parallel to each other (same slope), since there's only one direction perpendicular to a given line's direction (in the plane). So the condition "both perpendicular to some line $L$" reduces to "the two given lines are parallel to each other."

    Step-by-Step Solution

    1. Line 1: $p(p^2+1)x-y+q=0$. Writing as $y=p(p^2+1)x+q$, its slope is $m_1=p(p^2+1)$.
    2. Line 2: $(p^2+1)^2x+(p^2+1)y+2q=0$. Dividing by $(p^2+1)$ (nonzero for all real $p$): $(p^2+1)x+y+\dfrac{2q}{p^2+1}=0$, so slope is $m_2=-(p^2+1)$.
    3. For both to be perpendicular to a common line $L$, we need $m_1=m_2$ (parallel to each other): $p(p^2+1)=-(p^2+1)$. …

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11. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
    Q.If the lines given by $(x^2+y^2)\sin^2\alpha = (x\cos\alpha - y\sin\alpha)^2$ are perpendicular to each other, then $\sin^2\alpha + \tan^2\alpha =$ (A) $\dfrac{15}{4}$ (B) $0$ (C) $\dfrac{3}{2}$ (D) $\dfrac{7}{12}$

    ›Reveal solutionSolution

    The given homogeneous equation factors into $x=0$ and a second line of slope $\cot2\alpha$; forcing perpendicularity pins $\alpha=45°$, giving $\sin^2\alpha+\tan^2\alpha=\dfrac32$.

    Concept and Intuition

    An equation of the form (something)$\,=0$ that is homogeneous of degree 2 in $x,y$ represents a pair of straight lines through the origin. Instead of using the generic "$a+b=0$ for perpendicularity" test, it's often faster to actually factor the expression, since here one factor drops out very simply.

    Step-by-Step Solution

    1. Expand $(x\cos\alpha-y\sin\alpha)^2 = x^2\cos^2\alpha - 2xy\sin\alpha\cos\alpha + y^2\sin^2\alpha$.
    2. Given equation: $x^2\sin^2\alpha+y^2\sin^2\alpha - x^2\cos^2\alpha+2xy\sin\alpha\cos\alpha - y^2\sin^2\alpha = 0$.
    3. The $y^2\sin^2\alpha$ terms cancel, leaving $x^2(\sin^2\alpha-\cos^2\alpha)+2xy\sin\alpha\cos\alpha=0$.
    4. Factor out $x$: $x\big[x(\sin^2\alpha-\cos^2\alpha)+2y\sin\alpha\cos\alpha\big]=0$.
    5. So the two lines are $x=0$ and $x(\sin^2\alpha-\cos^2\alpha)+2y\sin\alpha\cos\alpha=0$, i.e. $-x\cos2\alpha+y\sin2\alpha=0$, slope $m=\cot2\alpha$. …

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12. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
    Q.The equation of the plane passing through the point (1,2,2) and perpendicular to the planes $x-y+2z=3$ and $2x-2y+z+12=0$, is (A) $x-2y+2z-1=0$ (B) $2x-3y+4z-4=0$ (C) $x+y+z-5=0$ (D) $x+y-3=0$

    ›Reveal solutionSolution

    The normal to a plane perpendicular to two given planes is the cross product of their normals; then use the point-normal form.

    Concept and Intuition

    If a plane is perpendicular to two other planes, its normal vector must be perpendicular to both of their normals — i.e., it is (a scalar multiple of) the cross product of the two given normals.

    Step-by-Step Solution

    1. Normals of the given planes: $\vec n_1=(1,-1,2)$ (from $x-y+2z=3$), $\vec n_2=(2,-2,1)$ (from $2x-2y+z+12=0$).
    2. Required normal $\vec n = \vec n_1\times\vec n_2 = \big((-1)(1)-(2)(-2),\ -[(1)(1)-(2)(2)],\ (1)(-2)-(-1)(2)\big) = (3,3,0)$.
    3. Simplify direction to $(1,1,0)$. …

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