Q.Show that the straight lines whose direction cosines are given by $2l + 2m - n = 0$ and $mn + nl + lm = 0$ are at right angles.

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 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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4. 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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5. 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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6. 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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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 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If $(K, 3, 5), (2, -1, 2)$ are direction ratios of two lines and the angle between them is $45^\circ$, then a value of $K$ is (A) $2$ (B) $4$ (C) $6$ (D) $8$

   ›Reveal solutionSolution

   Applying the direction-cosine angle formula and solving the resulting quadratic in $K$ gives $K=4$ (the other root, $52$, isn't among the choices). Answer: (B).

   Concept and Intuition

   The angle between two lines with direction ratios $(l_1,m_1,n_1)$ and $(l_2,m_2,n_2)$ satisfies $\cos\theta=\dfrac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{l_1^2+m_1^2+n_1^2}\sqrt{l_2^2+m_2^2+n_2^2}}$. Setting this equal to $\cos45^\circ$ gives an equation purely in $K$.

   Step-by-Step Solution

   1. $d_1=(K,3,5)$, $d_2=(2,-1,2)$. Dot product: $2K-3+10=2K+7$.
   2. $|d_1|=\sqrt{K^2+9+25}=\sqrt{K^2+34}$, $|d_2|=\sqrt{4+1+4}=3$.
   3. $\cos45^\circ=\frac{2K+7}{3\sqrt{K^2+34}}=\frac{1}{\sqrt2}$
   4. Cross-multiplying: $2(2K+7)=3\sqrt2\sqrt{K^2+34} \Rightarrow 4K+14=3\sqrt2\sqrt{K^2+34}$.
   5. Squaring: $(4K+14)^2=18(K^2+34) \Rightarrow 16K^2+112K+196=18K^2+612$

   $\Rightarrow 2K^2-112K+416=0 \Rightarrow K^2-56K+208=0$

   6. Discriminant $=56^2-4(208)=3136-832=2304=48^2$. $K=\dfrac{56\pm48}{2}=52\text{ or }4$. …

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9. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If the angle between the lines having direction ratios $(3, 1, 2)$ and $(1, -1, 2)$ is $\theta$, then $\cos 2\theta =$ (A) $\dfrac{1}{7}$ (B) $\dfrac{-1}{7}$ (C) $\dfrac{\sqrt{3}}{\sqrt{7}}$ (D) $\dfrac{-3}{\sqrt{7}}$

   ›Reveal solutionSolution

   Computing $\cos\theta$ from the direction-ratio dot-product formula and applying the double-angle identity gives $\cos2\theta=-\dfrac17$.

   Concept and Intuition

   The angle between two lines with direction ratios $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ satisfies

   $\cos\theta=\dfrac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\,\sqrt{a_2^2+b_2^2+c_2^2}}.$

   Once $\cos\theta$ is known, $\cos2\theta=2\cos^2\theta-1$ follows directly from the double angle formula — no need to find $\theta$ itself.

   Step-by-Step Solution

   1. Direction ratios: $(3,1,2)$ and $(1,-1,2)$.
   2. Dot product: $3(1)+1(-1)+2(2)=3-1+4=6$.
   3. Magnitudes: $\sqrt{9+1+4}=\sqrt{14}$, $\sqrt{1+1+4}=\sqrt6$.
   4. $\cos\theta=\dfrac{6}{\sqrt{14}\cdot\sqrt6}=\dfrac{6}{\sqrt{84}}=\dfrac{6}{2\sqrt{21}}=\dfrac{3}{\sqrt{21}}$. …

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10. 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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11. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
    Q.If the angle between the lines joining the origin to the points of intersection of $x + 2y + \lambda = 0$ and $2x^2 - 2xy + 3y^2 + 2x - y - 1 = 0$ is $\frac{\pi}{2}$, then a value of $\lambda$ is (A) $1$ (B) $\frac{1}{2}$ (C) $2$ (D) $\frac{3}{2}$

    ›Reveal solutionSolution

    Homogenize the conic using the given line so the joining lines through the origin become a single degree-2 equation, then apply "coefficient of $x^2$ + coefficient of $y^2$ = 0" for perpendicularity. Answer: $\lambda=1$.

    Concept and Intuition

    The lines joining the origin to the intersection points of a line and a conic can be obtained by making the conic's equation homogeneous of degree 2, using the line's equation (rearranged to equal 1) to convert the linear and constant terms. Once we have a homogeneous pair $Ax^2+Bxy+Cy^2=0$, the two lines are perpendicular exactly when $A+C=0$.

    Step-by-Step Solution

    1. The line is $x+2y+\lambda=0 \Rightarrow x+2y=-\lambda \Rightarrow 1 = -\dfrac{x+2y}{\lambda}$ (assuming $\lambda\ne0$).
    2. Homogenize the conic $2x^2-2xy+3y^2+2x-y-1=0$: multiply the degree-1 terms by $\left(-\dfrac{x+2y}{\lambda}\right)$ once, and the constant term by $\left(-\dfrac{x+2y}{\lambda}\right)^2$:

    $2x^2-2xy+3y^2+(2x-y)\left(-\frac{x+2y}{\lambda}\right)-\left(\frac{x+2y}{\lambda}\right)^2=0$

    3. Expand $(2x-y)(x+2y)=2x^2+4xy-xy-2y^2=2x^2+3xy-2y^2$, and $(x+2y)^2=x^2+4xy+4y^2$. Multiply the whole equation by $\lambda^2$:

    $\lambda^2(2x^2-2xy+3y^2) -\lambda(2x^2+3xy-2y^2) - (x^2+4xy+4y^2)=0$

    4. Collect coefficients: …

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12. 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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