Q.Find the angle between the lines whose direction cosines are given by the equations $l + m + n = 0$, $l^2 + m^2 - n^2 = 0$.

Showing the 12 most recent of 67 on this concept.

1. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
   Q.The angle between the lines whose direction cosines are given by the equations $l^2 + m^2 - n^2 = 0$, $l + m + n = 0$ is ____ (A) $\dfrac{\pi}{6}$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{\pi}{2}$

   ›Reveal solutionSolution

   Eliminate $n$ between the two given relations to find the two actual sets of direction ratios, then compute the angle between them directly.

   Concept and Intuition

   The two given equations jointly define (generically) two lines through the origin whose direction cosines satisfy both. Eliminating one variable reduces the quadratic relation to a simple product-equals-zero form, revealing the two explicit direction-ratio triples.

   Step-by-Step Solution

   1. From $l+m+n=0$: $n=-(l+m)$.
   2. Substitute into $l^2+m^2-n^2=0$: $l^2+m^2-(l+m)^2 = l^2+m^2-l^2-2lm-m^2 = -2lm=0$.
   3. So $lm=0$, meaning $l=0$ or $m=0$.
   4. If $l=0$: $n=-m$, giving direction ratios $(0,1,-1)$.
   5. If $m=0$: $n=-l$, giving direction ratios $(1,0,-1)$. …

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2. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.If the direction cosines of two lines are given by $l + m + n = 0$ and $mn - 2lm - 2nl = 0$, then the acute angle between those lines is (A) $2\pi/5$ (B) $\pi/3$ (C) $\pi/4$ (D) $\pi/60$

   ›Reveal solutionSolution

   Eliminate $n$ using the linear relation, factor the resulting quadratic in $l,m$ to get two sets of direction ratios, then use the cosine formula between two lines.

   Concept and Intuition

   When direction cosines satisfy one linear and one quadratic (or bilinear) relation, substituting the linear relation into the quadratic one reduces it to a single quadratic in the ratio $l:m$, whose two roots give the direction ratios of the two lines being described.

   Step-by-Step Solution

   1. From $l+m+n=0$: $n=-(l+m)$.
   2. Substitute into $mn-2lm-2nl=0$:

   $m(-(l+m)) - 2lm - 2(-(l+m))l = -lm-m^2-2lm+2l^2+2lm = 2l^2-lm-m^2=0$

   3. Solve for $l$ in terms of $m$: $2l^2-lm-m^2=0 \Rightarrow l=\dfrac{m\pm\sqrt{m^2+8m^2}}{4}=\dfrac{m\pm3m}{4}$, giving $l=m$ or $l=-\dfrac{m}{2}$.
   4. Case 1: $l=m=1\Rightarrow n=-(1+1)=-2$. Direction ratios: $(1,1,-2)$.
   5. Case 2: $l=-1,m=2\Rightarrow n=-(-1+2)=-1$. Direction ratios: $(-1,2,-1)\propto(1,-2,1)$. …

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3. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.The acute angle between the lines whose direction cosines satisfy the relations $l^2-5m^2+n^2=0$ and $l+m-n=0$ is (A) $\mathrm{Cos}^{-1}\left(\dfrac{\sqrt{3}}{4}\right)$ (B) $\dfrac{\pi}{3}$ (C) $\mathrm{Cos}^{-1}\left(\dfrac{2}{3}\right)$ (D) $\dfrac{\pi}{6}$

   ›Reveal solutionSolution

   The two relations on direction cosines actually describe a pair of lines; solving them simultaneously extracts both direction ratios, and the angle between them is $\pi/3$.

   Concept and Intuition

   A single homogeneous quadratic relation like $l^2-5m^2+n^2=0$ together with a linear relation like $l+m-n=0$ defines two lines through the origin (the linear relation is a plane, and the quadratic relation restricted to that plane factors into two linear factors — i.e. two direction ratios). Once we have both direction ratio triples, the angle between the lines is just the standard angle-between-vectors formula.

   Step-by-Step Solution

   1. From $l+m-n=0$: $n=l+m$.
   2. Substitute into $l^2-5m^2+n^2=0$: $l^2-5m^2+(l+m)^2=0$.
   3. Expand: $l^2-5m^2+l^2+2lm+m^2=0\Rightarrow 2l^2+2lm-4m^2=0$.
   4. Divide by 2: $l^2+lm-2m^2=0$.
   5. Factor: $(l+2m)(l-m)=0$, so $l=-2m$ or $l=m$.
   6. Case $l=m$: take $m=1\Rightarrow l=1,\ n=l+m=2$. Direction ratios $(1,1,2)$.
   7. Case $l=-2m$: take $m=1\Rightarrow l=-2,\ n=l+m=-1$. Direction ratios $(-2,1,-1)$. …

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4. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.If the direction ratios of two lines are given by $3lm - 4ln + mn = 0$ and $l + 2m + 3n = 0$, then the angle between the lines is ________ (A) $\frac{\pi}{2}$ (B) $\frac{\pi}{3}$ (C) $\frac{\pi}{4}$ (D) $\frac{\pi}{6}$

   ›Reveal solutionSolution

   The linear relation combined with the quadratic relation gives two explicit direction-ratio triples; their dot product is zero. Answer: $\theta=\pi/2$.

   Concept and Intuition

   A pair of homogeneous-degree-2 relation and a linear relation in $(l,m,n)$ together represent two actual lines through a point. Eliminating one variable from the linear relation and substituting into the quadratic relation gives a single-variable quadratic whose two roots correspond to the two lines' direction ratios.

   Step-by-Step Solution

   1. From $l+2m+3n=0$: $l=-2m-3n$.
   2. Substitute into $3lm-4ln+mn=0$: $3(-2m-3n)m - 4(-2m-3n)n + mn = -6m^2-9mn+8mn+12n^2+mn = -6m^2+12n^2=0$.
   3. So $m^2=2n^2 \Rightarrow m=\pm\sqrt2\,n$. Take $n=1$.
   4. Case 1: $m=\sqrt2$, $l=-2\sqrt2-3$. Case 2: $m=-\sqrt2$, $l=2\sqrt2-3$.
   5. Dot product: $l_1l_2+m_1m_2+n_1n_2 = (-2\sqrt2-3)(2\sqrt2-3) + (\sqrt2)(-\sqrt2) + 1\cdot1$. …

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5. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.The angle between the lines whose direction cosines are given by the equations $2l - m + n = 0$ and $lm + 2mn - 10nl = 0$ is $\theta$, then $\cos\theta =$ (A) $\dfrac{20}{3\sqrt{70}}$ (B) $\dfrac{10}{\sqrt{70}}$ (C) $\dfrac{8}{\sqrt{70}}$ (D) $\dfrac{16}{3\sqrt{70}}$

   ›Reveal solutionSolution

   Eliminating $n$ between the two given relations yields a quadratic in $m/l$ whose two roots are the direction ratios of the two lines; their angle has $\cos\theta=8/\sqrt{70}$.

   Concept and Intuition

   When direction cosines satisfy one linear relation and one homogeneous quadratic relation, eliminating one variable between them produces a quadratic whose two roots correspond to the direction ratios of the two lines being jointly described.

   Step-by-Step Solution

   1. From $2l-m+n=0$: $n=m-2l$.
   2. Substitute into $lm+2mn-10nl=0$: $lm+2m(m-2l)-10(m-2l)l=0$.
   3. Expand: $lm+2m^2-4ml-10ml+20l^2=0 \Rightarrow 2m^2+(1-4-10)lm+20l^2=0 \Rightarrow 2m^2-13lm+20l^2=0$.
   4. Divide by $l^2$, let $k=m/l$: $2k^2-13k+20=0\Rightarrow k=\dfrac{13\pm\sqrt{169-160}}{4}=\dfrac{13\pm3}{4}$, giving $k=4$ or $k=2.5$.
   5. Case $k=4$: $m=4l$, $n=m-2l=2l$ — direction ratios $(1,4,2)$ (taking $l=1$).
   6. Case $k=2.5$: taking $l=2$, $m=5$, $n=m-2l=1$ — direction ratios $(2,5,1)$. …

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6. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.If a line L makes angles $\frac{\pi}{3}$ and $\frac{\pi}{4}$ with Y-axis and Z-axis respectively, then the angle between L and another line having direction ratios 1, 1, 1 is (A) $\text{Cos}^{-1}\left(\frac{2}{\sqrt{6}}\right)$ (B) $\text{Cos}^{-1}\left(\frac{\sqrt{2}+1}{3\sqrt{3}}\right)$ (C) $\text{Cos}^{-1}\left(\frac{\sqrt{2}-1}{3}\right)$ (D) $\text{Cos}^{-1}\left(\frac{\sqrt{2}+1}{\sqrt{6}}\right)$

   ›Reveal solutionSolution

   Find the missing direction cosine from $l^2+m^2+n^2=1$, then use $\cos\theta=l l_2+mm_2+nn_2$ against $(1,1,1)$; the answer is $\text{Cos}^{-1}\!\left(\dfrac{\sqrt2+1}{\sqrt6}\right)$.

   Concept and Intuition

   The direction cosines of a line satisfy $l^2+m^2+n^2=1$ where $l=\cos\alpha$, $m=\cos\beta$, $n=\cos\gamma$ are the cosines of the angles the line makes with the $X$, $Y$, $Z$ axes respectively. Once all three are known, the angle between two lines is $\cos\theta=l_1l_2+m_1m_2+n_1n_2$.

   Step-by-Step Solution

   1. Given angle with $Y$-axis is $\dfrac{\pi}{3}$: $m=\cos\dfrac{\pi}{3}=\dfrac12$.
   2. Given angle with $Z$-axis is $\dfrac{\pi}{4}$: $n=\cos\dfrac{\pi}{4}=\dfrac{1}{\sqrt2}$.
   3. From $l^2+m^2+n^2=1$: $l^2=1-\dfrac14-\dfrac12=\dfrac14 \Rightarrow l=\dfrac12$ (taking the positive root).
   4. Direction cosines of the second line with ratios $(1,1,1)$: $\left(\dfrac{1}{\sqrt3},\dfrac{1}{\sqrt3},\dfrac{1}{\sqrt3}\right)$. …

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7. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.The angle between the straight lines $3x + 4y + 9 = 0$ and $x - 7y - 22 = 0$ is ______ (A) $\dfrac{\pi}{4}$ (B) $\dfrac{\pi}{6}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{\pi}{8}$

   ›Reveal solutionSolution

   This tests the standard formula for the angle between two lines given their slopes. The answer is $\dfrac{\pi}{4}$.

   Concept and Intuition

   The angle $\theta$ between two lines with slopes $m_1,m_2$ satisfies $\tan\theta=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|$. We extract each slope from its line equation and substitute.

   Step-by-Step Solution

   1. $3x+4y+9=0 \Rightarrow y=-\dfrac34x-\dfrac94$, so $m_1=-\dfrac34$.
   2. $x-7y-22=0 \Rightarrow y=\dfrac17x-\dfrac{22}{7}$, so $m_2=\dfrac17$.
   3. $\tan\theta=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|=\left|\dfrac{-\frac34-\frac17}{1+\left(-\frac34\right)\left(\frac17\right)}\right|$.
   4. Numerator: $-\dfrac{21}{28}-\dfrac{4}{28}=-\dfrac{25}{28}$. Denominator: $1-\dfrac{3}{28}=\dfrac{25}{28}$. …

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8. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
   Q.Find the angle between the pair of lines represented by the equation $x^2+4xy+y^2=0$. (A) $30°$ (B) $45°$ (C) $60°$ (D) $90°$

   ›Reveal solutionSolution

   A direct application of the standard angle-between-lines formula for a homogeneous pair-of-lines equation.

   Concept and Intuition

   The equation $ax^2+2hxy+by^2=0$ represents two straight lines through the origin, and the angle between them is governed by the formula $\tan\theta=\dfrac{2\sqrt{h^2-ab}}{a+b}$ (valid when $a+b\ne0$), derived from factoring the quadratic into two linear factors $y=m_1x$, $y=m_2x$.

   Step-by-Step Solution

   1. Compare $x^2+4xy+y^2=0$ with $ax^2+2hxy+by^2=0$: $a=1$, $2h=4\Rightarrow h=2$, $b=1$.
   2. $\tan\theta = \dfrac{2\sqrt{h^2-ab}}{a+b} = \dfrac{2\sqrt{4-1}}{1+1} = \dfrac{2\sqrt3}{2}=\sqrt3$. …

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9. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
   Q.The angle between the lines represented by $\cos\theta(\cos\theta+1)x^2-(2\cos\theta+\sin^2\theta)xy+(1-\cos\theta)y^2=0$ is ______. (A) $\dfrac{\pi}{4}$ (B) $\dfrac{\pi}{6}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{\pi}{12}$

   ›Reveal solutionSolution

   A messy-looking trigonometric pair-of-lines equation simplifies beautifully via the Pythagorean identity, giving a constant angle of $45°$ independent of $\theta$.

   Concept and Intuition

   Even though the coefficients depend on $\theta$, applying the angle formula $\tan\phi=\dfrac{2\sqrt{h^2-ab}}{a+b}$ and simplifying using $\sin^2\theta=1-\cos^2\theta$ often produces a $\theta$-independent result — a hallmark of well-designed exam questions.

   Step-by-Step Solution

   1. Identify $A=\cos\theta(\cos\theta+1)$, $2H=-(2\cos\theta+\sin^2\theta)$, $B=1-\cos\theta$.
   2. $A+B = \cos^2\theta+\cos\theta+1-\cos\theta = \cos^2\theta+1$.
   3. $AB = \cos\theta(\cos\theta+1)(1-\cos\theta) = \cos\theta(1-\cos^2\theta) = \cos\theta\sin^2\theta$.
   4. $H^2 = \dfrac{(2\cos\theta+\sin^2\theta)^2}{4}$; expand numerator: $4\cos^2\theta+4\cos\theta\sin^2\theta+\sin^4\theta$.
   5. $H^2-AB = \dfrac{4\cos^2\theta+4\cos\theta\sin^2\theta+\sin^4\theta - 4\cos\theta\sin^2\theta}{4} = \dfrac{4\cos^2\theta+\sin^4\theta}{4}$.
   6. Using $\sin^4\theta=(1-\cos^2\theta)^2=1-2\cos^2\theta+\cos^4\theta$: numerator $=4\cos^2\theta+1-2\cos^2\theta+\cos^4\theta=\cos^4\theta+2\cos^2\theta+1=(\cos^2\theta+1)^2$. …

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10. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
    Q.Suppose $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are the directional cosines of two lines and $\theta$ is the angle between them and $\cos\theta = \pm(l_1 l_2 + m_1 m_2 + n_1 n_2)$. Let $A = (1, -2, 3)$, $B = (3, 1, -3)$ and $C = (-3, 1, 3)$ be the vertices of a triangle ABC. Then $\cos A =$ (A) $-\dfrac{1}{35}$ (B) $\dfrac{1}{7}$ (C) $-\dfrac{1}{7}$ (D) $\dfrac{1}{35}$

    ›Reveal solutionSolution

    The angle at vertex $A$ between sides $AB$ and $AC$ is found from the dot product formula $\cos A=\dfrac{\vec{AB}\cdot\vec{AC}}{|\vec{AB}||\vec{AC}|}=\dfrac{1}{35}$.

    Concept and Intuition

    The angle between two lines through a common vertex of a triangle equals the angle between the vectors from that vertex to the other two vertices. This is exactly the direction-cosine formula for the angle between two lines given in the problem, applied to vectors $\vec{AB}$ and $\vec{AC}$.

    Step-by-Step Solution

    1. $A=(1,-2,3),\,B=(3,1,-3),\,C=(-3,1,3)$.
    2. $\vec{AB}=B-A=(2,3,-6)$; $|\vec{AB}|=\sqrt{4+9+36}=\sqrt{49}=7$.
    3. $\vec{AC}=C-A=(-4,3,0)$; $|\vec{AC}|=\sqrt{16+9+0}=\sqrt{25}=5$.
    4. $\vec{AB}\cdot\vec{AC}=(2)(-4)+(3)(3)+(-6)(0)=-8+9+0=1$.
    5. $\cos A=\dfrac{\vec{AB}\cdot\vec{AC}}{|\vec{AB}||\vec{AC}|}=\dfrac{1}{7\times5}=\dfrac{1}{35}$.

    Common Mistakes …

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11. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
    Q.For $\alpha \in \left[0, \frac{\pi}{2}\right]$, the angle between the lines represented by $[x\cos\theta - y]\left[(\cos\theta + \tan\alpha)x - (1 - \cos\theta.\tan\alpha)y\right] = 0$ is (A) $\alpha$ (B) $\theta$ (C) $\theta + \alpha$ (D) $\theta - \alpha$

    ›Reveal solutionSolution

    Extracting the two slopes and applying the angle-between-lines formula, all the $\theta$-dependence cancels out, leaving the angle equal to $\alpha$.

    Concept and Intuition

    A product of two linear factors set to zero represents a pair of lines through the origin; each factor gives a line's equation directly, from which we read off its slope. The angle between two lines with slopes $m_1,m_2$ is $\tan\phi = \left|\dfrac{m_1-m_2}{1+m_1m_2}\right|$.

    Step-by-Step Solution

    1. First factor $x\cos\theta - y=0 \Rightarrow y=x\cos\theta$, so $m_1=\cos\theta$.
    2. Second factor $(\cos\theta+\tan\alpha)x - (1-\cos\theta\tan\alpha)y=0 \Rightarrow m_2 = \dfrac{\cos\theta+\tan\alpha}{1-\cos\theta\tan\alpha}$.
    3. Compute $m_2-m_1 = \dfrac{(\cos\theta+\tan\alpha) - \cos\theta(1-\cos\theta\tan\alpha)}{1-\cos\theta\tan\alpha} = \dfrac{\tan\alpha(1+\cos^2\theta)}{1-\cos\theta\tan\alpha}$.
    4. Compute $1+m_1m_2 = \dfrac{(1-\cos\theta\tan\alpha)+\cos\theta(\cos\theta+\tan\alpha)}{1-\cos\theta\tan\alpha} = \dfrac{1+\cos^2\theta}{1-\cos\theta\tan\alpha}$. …

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12. AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
    Q.The angle between the lines $ab(x^2 - y^2) + (a^2 - b^2)xy = 0$ is ______ (A) $\dfrac{\pi}{2}$ (B) $\dfrac{\pi}{3}$ (C) $\dfrac{\pi}{4}$ (D) $\dfrac{\pi}{6}$

    ›Reveal solutionSolution

    The coefficients of $x^2$ and $y^2$ in this pair-of-lines equation are exact negatives of each other, which is exactly the condition for the two lines to be perpendicular.

    Concept and Intuition

    For a homogeneous pair of lines $Ax^2+2Hxy+By^2=0$, the lines are perpendicular precisely when the coefficient of $x^2$ plus the coefficient of $y^2$ is zero ($A+B=0$) — this is a standard, easily-checked shortcut, avoiding the need to compute the actual slopes.

    Step-by-Step Solution

    1. Expand the given equation: $ab\,x^2-ab\,y^2+(a^2-b^2)xy=0$, i.e. $ab\,x^2+(a^2-b^2)xy-ab\,y^2=0$.
    2. Compare to $Ax^2+2Hxy+By^2=0$: $A=ab$, $B=-ab$.
    3. Check $A+B=ab+(-ab)=0$.
    4. This satisfies the perpendicularity condition for a homogeneous pair of lines, so the angle between them is $\dfrac{\pi}{2}$, independent of the specific values of $a$ and $b$. …

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