Q.Find the equations of the two lines through the origin which intersect the line $\dfrac{x-3}{2} = \dfrac{y-3}{1} = \dfrac{z}{1}$ at angles of $\dfrac{\pi}{3}$ each.

Showing the 12 most recent of 67 on this concept.

1. 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 $(1,1,1)$: $\left(\dfrac{1}{\sqrt3},\dfrac{1}{\sqrt3},\dfrac{1}{\sqrt3}\right)$. …

   🔒 Start your 14-day free trial to see the full solution →
2. 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 $(-2,1,-1)$. …

   🔒 Start your 14-day free trial to see the full solution →
3. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If the line L passing through origin and making an angle of $60°$ with the line $2x + 3y - 4 = 0$ intersects the line $x = 3 + 2\sqrt{3}$ at $(3 + 2\sqrt{3}, 3\sqrt{3} + k)$, then $k =$ (A) $2$ (B) $-2$ (C) $1$ (D) $-1$

   ›Reveal solutionSolution

   This tests the angle-between-two-lines formula to pin down the slope of a line through the origin, then substituting a specific $x$-value. Answer: $k=-2$.

   Concept and Intuition

   A line through the origin is $y=mx$. "Making $60°$" with a given line of slope $m_1$ means the angle-between-lines formula $\tan\theta=\left|\dfrac{m-m_1}{1+mm_1}\right|$ equals $\tan60°=\sqrt3$. Once $m$ is known, plugging in the given $x$-coordinate immediately gives the corresponding $y$, which we match against the stated form $3\sqrt3+k$ to isolate $k$ — notice the awkward-looking $x=3+2\sqrt3$ is chosen precisely so it cancels the denominator of $m$.

   Step-by-Step Solution

   1. Slope of $2x+3y-4=0$: $y=-\tfrac23x+\tfrac43$, so $m_1=-\tfrac23$.
   2. Set $\tan60°=\sqrt3=\left|\dfrac{m+2/3}{1-2m/3}\right|=\left|\dfrac{3m+2}{3-2m}\right|$.
   3. Solve the "$+$" branch: $3m+2=\sqrt3(3-2m) \Rightarrow 3m+2\sqrt3 m = 3\sqrt3-2 \Rightarrow m=\dfrac{3\sqrt3-2}{3+2\sqrt3}$. …

   🔒 Start your 14-day free trial to see the full solution →
4. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.The equation of the lines through the point $(3,2)$ which makes an angle of $45^\circ$ with the line $x - 2y = 3$ are (A) $3x - y = 7$ & $x + 3y = 9$ (B) $x - 3y = 7$ & $3x + y = 9$ (C) $x - y = 3$ & $x + y = 2$ (D) $2x + y = 7$ & $x - 2y = 9$

   ›Reveal solutionSolution

   This tests finding lines through a point that make a given angle with a reference line, using the tangent-of-angle-between-lines formula (which typically yields two solutions). The answer is (A).

   Concept and Intuition

   Given a fixed reference slope $m_0$ and a target angle $45^\circ$, there are generally two lines through any point making that angle with the reference line — one on each "side". Solving $\tan45^\circ = \left|\dfrac{m-m_0}{1+mm_0}\right|=1$ gives two values of $m$ (from the $+1$ and $-1$ cases), and each is used to write a line through $(3,2)$.

   Step-by-Step Solution

   1. Reference line $x-2y=3 \Rightarrow y=\dfrac12x-\dfrac32$, slope $m_0=\dfrac12$.
   2. Set $\left|\dfrac{m-1/2}{1+m/2}\right|=1$.
   3. Case $+1$: $m-\dfrac12=1+\dfrac{m}{2} \Rightarrow \dfrac{m}{2}=\dfrac32 \Rightarrow m=3$.
   4. Case $-1$: $m-\dfrac12=-1-\dfrac{m}{2} \Rightarrow \dfrac{3m}{2}=-\dfrac12 \Rightarrow m=-\dfrac13$.
   5. Line with slope $3$ through $(3,2)$: $y-2=3(x-3) \Rightarrow y=3x-7 \Rightarrow 3x-y=7$. …

   🔒 Start your 14-day free trial to see the full solution →
5. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.A line L passing through the point (2, 0) makes an angle $60^0$ with the line $2x-y+3=0$. If L makes an acute angle with the positive X-axis in the anticlockwise direction, then the Y-intercept of the line L is (A) $\dfrac{10\sqrt{3}-16}{11}$ (B) $\dfrac{3\sqrt{2}}{\sqrt{7}}$ (C) $\dfrac{16-10\sqrt{3}}{11}$ (D) $2$

   ›Reveal solutionSolution

   Use the angle-between-lines formula to get two candidate slopes, keep the positive one (acute angle with the positive $x$-axis), then find the $y$-intercept of the line through $(2,0)$.

   Concept and Intuition

   The tangent of the angle between two lines of slopes $m_1,m_2$ is $\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|$. This always gives two solutions for the unknown slope (the two lines making the given angle on either side); the extra condition ("acute angle with positive $x$-axis", i.e. $m>0$) picks out the correct one.

   Step-by-Step Solution

   1. Line $2x-y+3=0$ has slope $m_1=2$.
   2. $\tan60^\circ=\sqrt3=\left|\dfrac{m-2}{1+2m}\right|$.
   3. Case $\dfrac{m-2}{1+2m}=\sqrt3$: $m-2=\sqrt3+2\sqrt3 m\Rightarrow m(1-2\sqrt3)=\sqrt3+2\Rightarrow m=\dfrac{\sqrt3+2}{1-2\sqrt3}\approx-1.52$ (negative — rejected, since $L$ must make an acute angle with positive $x$-axis).
   4. Case $\dfrac{m-2}{1+2m}=-\sqrt3$: $m-2=-\sqrt3-2\sqrt3m\Rightarrow m(1+2\sqrt3)=2-\sqrt3\Rightarrow m=\dfrac{2-\sqrt3}{1+2\sqrt3}$. …

   🔒 Start your 14-day free trial to see the full solution →
6. 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: $l_1l_2+m_1m_2+n_1n_2 = (-2\sqrt2-3)(2\sqrt2-3) + (\sqrt2)(-\sqrt2) + 1\cdot1$. …

   🔒 Start your 14-day free trial to see the full solution →
7. 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 $(1,0,-1)$. …

   🔒 Start your 14-day free trial to see the full solution →
8. 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: $(-1,2,-1)\propto(1,-2,1)$. …

   🔒 Start your 14-day free trial to see the full solution →
9. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.Let $\theta$ be the angle between the line $\frac{x+1}{1} = \frac{y-1}{2} = \frac{z-2}{2}$ and the plane $2x - y + \sqrt{\lambda}z + 4 = 0$. If $\sin\theta = \frac{1}{3}$, then $\lambda =$ (A) $\frac{4}{3}$ (B) $\frac{5}{3}$ (C) $\frac{2}{3}$ (D) $\frac{7}{3}$

   ›Reveal solutionSolution

   This tests the line-plane angle formula. Setting up $\sin\theta$ in terms of $\lambda$ and solving the resulting equation gives $\lambda=\frac53$.

   Concept and Intuition

   The angle $\theta$ between a line with direction $\vec d$ and a plane with normal $\vec n$ satisfies $\sin\theta=\frac{|\vec d\cdot\vec n|}{|\vec d||\vec n|}$ (since the angle between the line and the normal is $90^\circ-\theta$). Plugging in the components and the given $\sin\theta$ gives a single equation in $\lambda$.

   Step-by-Step Solution

   1. Direction of the line: $\vec d=(1,2,2)$, so $|\vec d|=\sqrt{1+4+4}=3$.
   2. Normal of the plane: $\vec n=(2,-1,\sqrt\lambda)$, so $|\vec n|=\sqrt{4+1+\lambda}=\sqrt{5+\lambda}$.
   3. $\vec d\cdot\vec n=1(2)+2(-1)+2(\sqrt\lambda)=2-2+2\sqrt\lambda=2\sqrt\lambda$.
   4. $\sin\theta=\frac{2\sqrt\lambda}{3\sqrt{5+\lambda}}=\frac13$.
   5. Cross-multiplying: $2\sqrt\lambda=\sqrt{5+\lambda}$. …

   🔒 Start your 14-day free trial to see the full solution →
10. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
    Q.If the angle $\theta$ between the line $\dfrac{x+1}{1} = \dfrac{y-1}{2} = \dfrac{z-2}{2}$ and the plane $2x - y + \sqrt{\lambda}z + 4 = 0$ is such that $\sin\theta = \dfrac{1}{3}$, then the value of $\lambda$ is (A) $\dfrac{5}{3}$ (B) $-\dfrac{3}{5}$ (C) $\dfrac{3}{4}$ (D) $-\dfrac{4}{3}$

    ›Reveal solutionSolution

    Using $\sin\theta=\dfrac{|d\cdot n|}{|d||n|}$ for the line-plane angle and solving for $\lambda$ gives $\lambda=\dfrac53$. Answer: (A).

    Concept and Intuition

    The angle $\theta$ between a line (direction vector $d$) and a plane (normal vector $n$) is the complement of the angle between $d$ and $n$, so $\sin\theta=\cos(\text{angle between }d,n)=\dfrac{|d\cdot n|}{|d||n|}$. This directly converts the given $\sin\theta$ value into an equation for the unknown in the normal vector.

    Step-by-Step Solution

    1. Line: $\dfrac{x+1}{1}=\dfrac{y-1}{2}=\dfrac{z-2}{2}$ has direction $d=(1,2,2)$, $|d|=\sqrt{1+4+4}=3$.
    2. Plane: $2x-y+\sqrt\lambda z+4=0$ has normal $n=(2,-1,\sqrt\lambda)$, $|n|=\sqrt{4+1+\lambda}=\sqrt{5+\lambda}$.
    3. $d\cdot n = 1(2)+2(-1)+2(\sqrt\lambda)=2-2+2\sqrt\lambda=2\sqrt\lambda$
    4. $\sin\theta=\frac{|2\sqrt\lambda|}{3\sqrt{5+\lambda}}=\frac13$ …

    🔒 Start your 14-day free trial to see the full solution →
11. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
    Q.Let the curve $x^2 + 2y^2 = 2$ intersect the line $x + y = 1$ at two points P and Q and O be the origin. If $\theta$ is the acute angle between the lines OP and OQ, then $\tan\theta =$ (A) $\dfrac{1}{4}$ (B) $4$ (C) $\sqrt{3}$ (D) $\dfrac{1}{\sqrt{3}}$

    ›Reveal solutionSolution

    Find the two intersection points, then use the cross/dot product formula for the angle between $OP$ and $OQ$; $\tan\theta=4$.

    Concept and Intuition

    The angle between two lines from the origin to points $P,Q$ can be found directly from the vectors $OP,OQ$ using $\tan\theta = \dfrac{|OP\times OQ|}{OP\cdot OQ}$.

    Step-by-Step Solution

    1. Substitute $x=1-y$ into $x^2+2y^2=2$: $(1-y)^2+2y^2=2 \Rightarrow 3y^2-2y-1=0$.
    2. Solve: $y=\dfrac{2\pm4}{6}$, so $y=1$ or $y=-1/3$.
    3. $y=1\Rightarrow x=0$, giving $P=(0,1)$. $y=-1/3\Rightarrow x=4/3$, giving $Q=(4/3,-1/3)$.
    4. Cross product magnitude: $|0\cdot(-1/3)-(4/3)\cdot1| = 4/3$. …

    🔒 Start your 14-day free trial to see the full solution →
12. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
    Q.The angle subtended by the chord $x+y-1=0$ of the circle $x^2+y^2-2x+4y+4=0$ at the origin is (A) $\cos^{-1}\left(\dfrac{6}{\sqrt{34}}\right)$ (B) $\dfrac{\pi}{2}$ (C) $\cos^{-1}\left(\dfrac{2}{\sqrt{13}}\right)$ (D) $\dfrac{\pi}{3}$

    ›Reveal solutionSolution

    Make the circle's equation homogeneous of degree 2 using the chord to get the pair of lines from the origin to the chord's endpoints, then apply the angle formula. The angle is $\cos^{-1}\!\left(\dfrac{6}{\sqrt{34}}\right)$ — option (A).

    Step 1 — Homogenise the circle with the chord.

    Chord: $x+y=1$. Circle: $x^2+y^2-2x+4y+4=0$. Replace each $1$ by $(x+y)$ to make every term degree 2:

    $x^2+y^2-2x(x+y)+4y(x+y)+4(x+y)^2=0$

    Expanding and collecting terms:

    $3x^2+10xy+9y^2=0$

    This is the pair of straight lines joining the origin to the two ends of the chord.

    Step 2 — Angle between the pair of lines.

    For $ax^2+2hxy+by^2=0$ the angle $\theta$ between the lines satisfies

    $\cos\theta=\frac{|a+b|}{\sqrt{(a-b)^2+4h^2}}$

    Here $a=3,\ b=9,\ h=5$:

    $\cos\theta=\frac{|3+9|}{\sqrt{(3-9)^2+4(5)^2}}=\frac{12}{\sqrt{36+100}}=\frac{12}{\sqrt{136}}=\frac{12}{2\sqrt{34}}=\frac{6}{\sqrt{34}}$ …

    🔒 Start your 14-day free trial to see the full solution →