Q.If a line makes angle $90^\circ$, $60^\circ$ and $30^\circ$ with the positive direction of $x$, $y$ and $z$-axis respectively, find its direction cosines.

Showing the 12 most recent of 34 on this concept.

1. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.If a line makes angles $90^\circ$, $135^\circ$ and $45^\circ$ with the positive x, y and z axes respectively, then its direction cosines are ______ (A) $\langle 0, \dfrac{1}{2}, \dfrac{1}{\sqrt{2}} \rangle$ (B) $\langle 0, \dfrac{-1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}} \rangle$ (C) $\langle 1, \dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}} \rangle$ (D) $\langle 1, \dfrac{-1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}} \rangle$

   ›Reveal solutionSolution

   Direction cosines are simply the cosines of the angles made with each axis; here that's $\langle 0,-1/\sqrt2,1/\sqrt2\rangle$.

   Concept and Intuition

   If a line makes angles $\alpha,\beta,\gamma$ with the positive x, y, z axes, its direction cosines are $(l,m,n)=(\cos\alpha,\cos\beta,\cos\gamma)$, and they always satisfy $l^2+m^2+n^2=1$.

   Step-by-Step Solution

   1. $\alpha=90^\circ \Rightarrow l=\cos90^\circ=0$.
   2. $\beta=135^\circ \Rightarrow m=\cos135^\circ=-\dfrac{1}{\sqrt2}$.
   3. $\gamma=45^\circ \Rightarrow n=\cos45^\circ=\dfrac{1}{\sqrt2}$.
   4. Direction cosines: $\left\langle 0,-\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right\rangle$.
   5. Verify: $0^2+\left(-\tfrac1{\sqrt2}\right)^2+\left(\tfrac1{\sqrt2}\right)^2=0+\tfrac12+\tfrac12=1$. Valid.

   Common Mistakes

   • Forgetting the sign of $\cos135^\circ$ (it's negative, not $+\tfrac1{\sqrt2}$).
   • Confusing direction cosines with direction ratios and skipping the normalization check.

   ✓Final answer

   The correct option is (B) — $\langle 0, \dfrac{-1}{\sqrt2}, \dfrac{1}{\sqrt2}\rangle$.

   ANSWER: B

2. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.The direction cosines of the line making angles $\dfrac{\pi}{4}$, $\dfrac{\pi}{3}$ and $\theta \left(0 < \theta < \dfrac{\pi}{2}\right)$ respectively with X, Y and Z axes are (A) $\dfrac{1}{\sqrt{2}}, \dfrac{1}{2}, \dfrac{1}{2}$ (B) $\dfrac{1}{\sqrt{2}}, \dfrac{1}{2}, \dfrac{\sqrt{3}}{2}$ (C) $\dfrac{1}{\sqrt{2}}, \dfrac{1}{2}, \dfrac{1}{\sqrt{2}}$ (D) $\dfrac{1}{\sqrt{2}}, \dfrac{\sqrt{3}}{2}, \dfrac{1}{\sqrt{2}}$

   ›Reveal solutionSolution

   Direction cosines of any line always satisfy $l^2+m^2+n^2=1$; use the two given angles to fix $l,m$ and solve for $n=\cos\theta$.

   Concept and Intuition

   If a line makes angles $\alpha,\beta,\gamma$ with the $X,Y,Z$ axes respectively, its direction cosines are $l=\cos\alpha$, $m=\cos\beta$, $n=\cos\gamma$, and these three numbers must always satisfy the fundamental identity $l^2+m^2+n^2=1$. This lets us solve for the third angle once two are known.

   Step-by-Step Solution

   1. $l=\cos\dfrac{\pi}{4}=\dfrac{1}{\sqrt2}$, so $l^2=\dfrac12$.
   2. $m=\cos\dfrac{\pi}{3}=\dfrac12$, so $m^2=\dfrac14$.
   3. Using $l^2+m^2+n^2=1$: $n^2=1-\dfrac12-\dfrac14=\dfrac14$.
   4. $n=\pm\dfrac12$; since $0<\theta<\pi/2$ means $\cos\theta>0$, take $n=\dfrac12$.
   5. So the direction cosines are $\dfrac{1}{\sqrt2},\ \dfrac12,\ \dfrac12$.

   Common Mistakes

   • Forgetting the constraint $0<\theta<\pi/2$ and taking the negative root for $n$.
   • Confusing direction cosines with direction ratios (the identity $l^2+m^2+n^2=1$ only applies to true direction cosines, not arbitrary ratios).

   ✓Final answer

   The correct option is (A) — $\dfrac{1}{\sqrt2},\ \dfrac12,\ \dfrac12$.

   ANSWER: A

3. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.The direction cosines of a line which makes equal angles with the co-ordinate axes are ____ (A) $\left\langle \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}} \right\rangle$ (B) $\left\langle \dfrac{-1}{\sqrt{3}}, \dfrac{-1}{\sqrt{3}}, \dfrac{-1}{\sqrt{3}} \right\rangle$ (C) $\left\langle \dfrac{\pm 1}{\sqrt{3}}, \dfrac{\pm 1}{\sqrt{3}}, \dfrac{\pm 1}{\sqrt{3}} \right\rangle$ (D) $\left\langle \dfrac{12}{13}, \dfrac{5}{13}, 0 \right\rangle$

   ›Reveal solutionSolution

   Equal angles with all three axes force $l=m=n$; combined with $l^2+m^2+n^2=1$ this gives $l=m=n=\pm\dfrac{1}{\sqrt3}$.

   Concept and Intuition

   Direction cosines are just the cosines of the angles a line makes with the positive $x$, $y$, $z$ axes. "Equal angles with the axes" literally means these three cosines are equal to each other, $l=m=n$. Since a line (as opposed to a directed ray) can be traversed in either of two opposite senses, both the all-positive and all-negative solutions represent the same line.

   Step-by-Step Solution

   1. Equal angles $\Rightarrow l=m=n$.
   2. Direction cosine identity: $l^2+m^2+n^2=1 \Rightarrow 3l^2=1 \Rightarrow l=\pm\dfrac{1}{\sqrt3}$.
   3. So $l=m=n=\dfrac{1}{\sqrt3}$ or $l=m=n=-\dfrac{1}{\sqrt3}$ — the two possible (opposite) orientations of the same line.

   Common Mistakes

   • Picking only the positive or only the negative triple and missing that the line has two possible orientations.
   • Confusing "equal angles" with "direction ratios summing to a fixed value" instead of the cosines being literally equal.

   ✓Final answer

   The correct option is (C) — $\left\langle \dfrac{\pm 1}{\sqrt{3}}, \dfrac{\pm 1}{\sqrt{3}}, \dfrac{\pm 1}{\sqrt{3}} \right\rangle$.

   ANSWER: C

4. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.A line $AB$ in three dimensions makes angles $45°$ and $120°$ with the positive $x$-axis and the positive $y$-axis respectively. If $AB$ makes an acute angle $\theta$ with the positive $z$-axis, then $\theta$ equals ________ (A) $30°$ (B) $45°$ (C) $60°$ (D) $75°$

   ›Reveal solutionSolution

   The direction angles a line makes with the three coordinate axes satisfy $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$; given two of the angles, the third (acute) angle follows directly.

   Concept and Intuition

   If a line makes angles $\alpha,\beta,\gamma$ with the positive $x$-, $y$-, $z$-axes respectively, its direction cosines are $\cos\alpha,\cos\beta,\cos\gamma$, and these always satisfy $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$.

   Step-by-Step Solution

   1. Given $\alpha=45°,\ \beta=120°$. So $\cos\alpha=\dfrac{\sqrt2}{2},\ \cos^2\alpha=\dfrac12$; $\cos\beta=-\dfrac12,\ \cos^2\beta=\dfrac14$.
   2. Apply the identity: $\dfrac12+\dfrac14+\cos^2\theta=1 \Rightarrow \cos^2\theta=1-\dfrac34=\dfrac14$.
   3. $\cos\theta=\pm\dfrac12 \Rightarrow \theta=60°$ or $\theta=120°$.
   4. Since $\theta$ is required to be acute, $\theta=60°$.

   Common Mistakes

   • Forgetting that $\cos120°=-\dfrac12$ is negative, but its square is still $\dfrac14$.
   • Picking $120°$ instead of the acute root $60°$.

   ✓Final answer

   The correct option is (C) — $60°$.

   ANSWER: C

5. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.If the direction cosines of a line $L$ are $(ab,b,b)$ and the angle between $L$ and X-axis is $\dfrac{\pi}{6}$, then a possible value of $(a,b)$ is (A) $\left(\sqrt{6},\sqrt{\dfrac{3}{8}}\right)$ (B) $\left(\sqrt{\dfrac{3}{8}},\sqrt{\dfrac{1}{8}}\right)$ (C) $\left(\sqrt{6},\dfrac{1}{\sqrt{8}}\right)$ (D) $\left(\dfrac{1}{\sqrt{8}},\sqrt{6}\right)$

   ›Reveal solutionSolution

   Solving the normalization condition together with the angle condition pins $(a,b)=(\sqrt6,\,1/\sqrt8)$.

   Concept and Intuition

   Direction cosines $(l,m,n)$ of any line must obey $l^2+m^2+n^2=1$. Also, if $\alpha$ is the angle the line makes with the X-axis, then $l=\cos\alpha$. Combining these two facts with the given form of the direction cosines determines $a$ and $b$.

   Step-by-Step Solution

   1. Normalization: $(ab)^2+b^2+b^2=1\Rightarrow a^2b^2+2b^2=1$.
   2. Angle with X-axis is $\pi/6$, and the X-direction cosine is the first component: $ab=\cos\dfrac{\pi}{6}=\dfrac{\sqrt3}{2}$.
   3. Substitute $a^2b^2=\left(\dfrac{\sqrt3}{2}\right)^2=\dfrac34$ into the normalization equation: $\dfrac34+2b^2=1\Rightarrow 2b^2=\dfrac14\Rightarrow b^2=\dfrac18\Rightarrow b=\dfrac{1}{\sqrt8}$.
   4. Then $a=\dfrac{\sqrt3/2}{b}=\dfrac{\sqrt3/2}{1/\sqrt8}=\dfrac{\sqrt3}{2}\cdot\sqrt8=\dfrac{\sqrt{24}}{2}=\dfrac{2\sqrt6}{2}=\sqrt6$.
   5. So $(a,b)=\left(\sqrt6,\dfrac{1}{\sqrt8}\right)$, matching option (C) exactly.

   Common Mistakes

   • Forgetting the factor of $2$ from the two equal components $b,b$ in the normalization sum.
   • Mixing up which product $(ab)$ equals $\cos(\pi/6)$ vs $\sin(\pi/6)$.

   ✓Final answer

   The correct option is (C) — $\left(\sqrt{6},\dfrac{1}{\sqrt{8}}\right)$.

   ANSWER: C

6. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
   Q.If a line L makes angles $\frac{\pi}{3}$ and $\frac{\pi}{4}$ with the positive X-axis and positive Y-axis respectively, then the angle made by L with the positive direction of Z-axis is (A) $\frac{\pi}{2}$ (B) $\frac{\pi}{3}$ (C) $\frac{\pi}{4}$ (D) $\frac{5\pi}{12}$

   ›Reveal solutionSolution

   A line's direction cosines with the three axes satisfy $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$; plugging in the given angles gives $\gamma=\pi/3$.

   Concept and Intuition

   For any line in 3D space, if $\alpha,\beta,\gamma$ are the angles it makes with the positive X, Y, Z axes respectively, the direction cosines $l=\cos\alpha, m=\cos\beta, n=\cos\gamma$ always satisfy $l^2+m^2+n^2=1$. This is simply the Pythagorean identity for a unit vector along the line.

   Step-by-Step Solution

   1. $\alpha=\pi/3\Rightarrow \cos\alpha=\frac12\Rightarrow \cos^2\alpha=\frac14$.
   2. $\beta=\pi/4\Rightarrow \cos\beta=\frac{1}{\sqrt2}\Rightarrow \cos^2\beta=\frac12$.
   3. Identity: $\cos^2\gamma = 1-\frac14-\frac12=\frac14$.
   4. So $\cos\gamma=\pm\frac12$, giving $\gamma=\pi/3$ or $2\pi/3$.
   5. Among the given options, $\pi/3$ matches (the acute solution corresponding to $\cos\gamma=+1/2$).

   Common Mistakes

   • Forgetting to square the cosines before adding (using $\cos\alpha+\cos\beta+\cos\gamma=1$, which is wrong).
   • Sign errors in inverting $\cos^2\gamma=1/4$.

   ✓Final answer

   The correct option is (B) — $\frac{\pi}{3}$.

   ANSWER: B

7. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.If the direction cosines of a line satisfy the relations $l - m + n = 0$ and $lm + mn - 4nl = 0$, then the direction cosines of the line are (A) $\left(\dfrac{-1}{\sqrt{6}}, \dfrac{2}{\sqrt{6}}, \dfrac{1}{\sqrt{6}}\right)$ (B) $\left(\dfrac{1}{\sqrt{6}}, \dfrac{-2}{\sqrt{6}}, \dfrac{1}{\sqrt{6}}\right)$ (C) $\left(\dfrac{1}{\sqrt{6}}, \dfrac{2}{\sqrt{6}}, \dfrac{-1}{\sqrt{6}}\right)$ (D) $\left(\dfrac{1}{\sqrt{6}}, \dfrac{2}{\sqrt{6}}, \dfrac{1}{\sqrt{6}}\right)$

   ›Reveal solutionSolution

   Eliminate one variable using the linear relation, reduce the quadratic relation to a perfect square, and normalize the resulting direction ratios.

   Concept and Intuition

   Direction cosines $(l,m,n)$ satisfy $l^2+m^2+n^2=1$; given two other relations among $l,m,n$, we solve for their ratio first and normalize at the end.

   Step-by-Step Solution

   1. From $l-m+n=0$: $m=l+n$.
   2. Substitute into $lm+mn-4nl=0$: $l(l+n)+(l+n)n-4nl=0\Rightarrow l^2+log+log+n^2-4nl=0\Rightarrow l^2-2ln+n^2=0$.
   3. This is $(l-n)^2=0\Rightarrow l=n$.
   4. Then $m=l+n=2l$. So $l:m:n=1:2:1$.
   5. Normalize: magnitude $=\sqrt{1^2+2^2+1^2}=\sqrt6$, giving direction cosines $\left(\dfrac1{\sqrt6},\dfrac2{\sqrt6},\dfrac1{\sqrt6}\right)$.

   Common Mistakes

   • Sign slip while substituting $m=l+n$ into the second (quadratic) relation.
   • Picking a direction-ratio set that doesn't actually satisfy both original relations (always a good idea to plug back in and check).

   ✓Final answer

   The correct option is (D) — $\left(\dfrac{1}{\sqrt{6}}, \dfrac{2}{\sqrt{6}}, \dfrac{1}{\sqrt{6}}\right)$.

   ANSWER: D

8. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
   Q.Given points $A(1, 2, 2)$, $B(2, 3, 6)$ and $C(3, 4, 12)$, find the direction cosines of a line which is equally inclined with $OA$, $OB$ and $OC$, where $O$ is the origin. (A) $\left\langle \dfrac{1}{\sqrt{2}}, \dfrac{-1}{\sqrt{2}}, 0 \right\rangle$ (B) $\left\langle \dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}}, 0 \right\rangle$ (C) $\left\langle \dfrac{1}{\sqrt{3}}, \dfrac{-1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}} \right\rangle$ (D) $\left\langle \dfrac{1}{\sqrt{3}}, \dfrac{-1}{\sqrt{3}}, \dfrac{-1}{\sqrt{3}} \right\rangle$

   ›Reveal solutionSolution

   A line equally inclined to three given lines makes the same cosine of angle (dot product with unit vectors) with all three — check each option's dot products with the unit vectors along $OA,OB,OC$. The answer is (D).

   Concept and Intuition

   "Equally inclined" to three directions means the direction cosines $(l,m,n)$ of the desired line give the same value of $\cos(\text{angle})$ when dotted with the unit vector along each of $OA,OB,OC$.

   Step-by-Step Solution

   1. $|OA|=\sqrt{1^2+2^2+2^2}=3$, unit vector $\hat{u}_A=\left(\tfrac13,\tfrac23,\tfrac23\right)$.
   2. $|OB|=\sqrt{2^2+3^2+6^2}=\sqrt{49}=7$, unit vector $\hat u_B=\left(\tfrac27,\tfrac37,\tfrac67\right)$.
   3. $|OC|=\sqrt{3^2+4^2+12^2}=\sqrt{169}=13$, unit vector $\hat u_C=\left(\tfrac3{13},\tfrac4{13},\tfrac{12}{13}\right)$.
   4. Test $(l,m,n)=\left(\tfrac1{\sqrt3},\tfrac{-1}{\sqrt3},\tfrac{-1}{\sqrt3}\right)$:
      • $\cdot\hat u_A = \tfrac1{\sqrt3}\left(\tfrac13-\tfrac23-\tfrac23\right)=\tfrac1{\sqrt3}\left(-\tfrac33\right)=-\tfrac1{\sqrt3}$
      • $\cdot\hat u_B = \tfrac1{\sqrt3}\left(\tfrac27-\tfrac37-\tfrac67\right)=\tfrac1{\sqrt3}\left(-\tfrac77\right)=-\tfrac1{\sqrt3}$
      • $\cdot\hat u_C = \tfrac1{\sqrt3}\left(\tfrac3{13}-\tfrac4{13}-\tfrac{12}{13}\right)=\tfrac1{\sqrt3}\left(-\tfrac{13}{13}\right)=-\tfrac1{\sqrt3}$
   5. All three dot products equal $-\tfrac1{\sqrt3}$, confirming equal inclination.

   Common Mistakes

   • Forgetting to normalize $OA,OB,OC$ to unit vectors before comparing dot products (unnormalized vectors of different lengths give misleadingly different-looking dot products).
   • Sign errors while summing the three terms in each dot product.

   ✓Final answer

   The correct option is (D) — $\left\langle \dfrac{1}{\sqrt{3}}, \dfrac{-1}{\sqrt{3}}, \dfrac{-1}{\sqrt{3}} \right\rangle$.

   ANSWER: D

9. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
   Q.The direction cosines of the line joining the points $(-2, 4, -5)$ and $(1, 2, 3)$ are ______ (A) $\left\langle \dfrac{3}{\sqrt{77}}, \dfrac{-2}{\sqrt{77}}, \dfrac{8}{\sqrt{77}} \right\rangle$ (B) $\left\langle \dfrac{3}{\sqrt{77}}, \dfrac{2}{\sqrt{77}}, \dfrac{8}{\sqrt{77}} \right\rangle$ (C) $\langle 1, 0, 0 \rangle$ (D) $\left\langle \dfrac{-3}{77}, \dfrac{-2}{77}, \dfrac{8}{77} \right\rangle$

   ›Reveal solutionSolution

   Direction cosines are the direction ratios divided by their magnitude; here that gives $\left(\frac{3}{\sqrt{77}},\frac{-2}{\sqrt{77}},\frac{8}{\sqrt{77}}\right)$.

   Concept and Intuition

   For a line joining two points, the direction ratios are simply the differences of corresponding coordinates. Dividing each ratio by the length of the direction vector (its magnitude) gives the direction cosines, which satisfy $l^2+m^2+n^2=1$.

   Step-by-Step Solution

   1. Direction ratios from $(-2,4,-5)$ to $(1,2,3)$: $(1-(-2),\ 2-4,\ 3-(-5))=(3,-2,8)$.
   2. Magnitude: $\sqrt{3^2+(-2)^2+8^2}=\sqrt{9+4+64}=\sqrt{77}$.
   3. Direction cosines: $\left(\dfrac{3}{\sqrt{77}}, \dfrac{-2}{\sqrt{77}}, \dfrac{8}{\sqrt{77}}\right)$.

   Common Mistakes

   • Reversing the order of subtraction (point 1 minus point 2 instead of point 2 minus point 1), which flips all signs.
   • Forgetting to take the square root when computing the magnitude.

   ✓Final answer

   The correct option is (A) — $\left\langle \dfrac{3}{\sqrt{77}}, \dfrac{-2}{\sqrt{77}}, \dfrac{8}{\sqrt{77}} \right\rangle$.

   ANSWER: A

10. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
    Q.The angle between the lines whose direction cosines satisfy the equations $l + m + n = 0$ and $l^2 + m^2 - n^2 = 0$ is _____ (A) $\dfrac{\pi}{2}$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{\pi}{6}$ (D) $\dfrac{\pi}{3}$

    ›Reveal solutionSolution

    The two constraint equations on the direction cosines pick out exactly two specific lines; compute the angle between those two explicit direction vectors. The answer is (D).

    Concept and Intuition

    The equations $l+m+n=0$ and $l^2+m^2-n^2=0$ together define (up to scaling) two specific directions, since substituting one into the other reduces to a simple factorable condition.

    Step-by-Step Solution

    1. From $l+m+n=0$: $n=-(l+m)$.
    2. Substitute into $l^2+m^2-n^2=0$: $l^2+m^2-(l+m)^2=0 \Rightarrow l^2+m^2-l^2-2lm-m^2=0 \Rightarrow -2lm=0 \Rightarrow lm=0$.
    3. Case $l=0$: then $n=-m$, giving direction ratios $(0,1,-1)$.
    4. Case $m=0$: then $n=-l$, giving direction ratios $(1,0,-1)$.
    5. Angle between $(0,1,-1)$ and $(1,0,-1)$: $\cos\theta=\dfrac{(0)(1)+(1)(0)+(-1)(-1)}{\sqrt{0^2+1^2+(-1)^2}\sqrt{1^2+0^2+(-1)^2}}=\dfrac{1}{\sqrt2\cdot\sqrt2}=\dfrac12$.
    6. $\theta=\dfrac\pi3$ (i.e., $60^\circ$).

    Common Mistakes

    • Trying to solve for $l,m,n$ directly as three unknowns instead of recognizing the factorization $lm=0$ that yields the two lines.
    • Sign error in the dot product, flipping $\cos\theta$ to $-\tfrac12$.

    ✓Final answer

    The correct option is (D) — $\dfrac{\pi}{3}$.

    ANSWER: D

11. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
    Q.If the direction cosines of a straight line are $\left(\dfrac{1}{c},\dfrac{1}{c},\dfrac{1}{c}\right)$, then $c=$________ (A) $\pm\sqrt{2}$ (B) $\pm\sqrt{3}$ (C) $\pm2$ (D) $\pm3$

    ›Reveal solutionSolution

    Direction cosines of any line always satisfy $l^2+m^2+n^2=1$; applying this to the given equal direction cosines gives $c$.

    Concept and Intuition

    If $(l,m,n)$ are the direction cosines of a line in 3D, they must satisfy the fundamental identity $l^2+m^2+n^2=1$.

    Step-by-Step Solution

    1. Given direction cosines: $l=m=n=\dfrac1c$.
    2. Apply the identity: $\left(\dfrac1c\right)^2+\left(\dfrac1c\right)^2+\left(\dfrac1c\right)^2=1 \Rightarrow \dfrac{3}{c^2}=1$.
    3. $c^2=3 \Rightarrow c=\pm\sqrt3$.

    Common Mistakes

    • Forgetting the $\pm$ sign (direction cosines' squares only fix $c^2$, not the sign of $c$).
    • Miscounting the number of terms (using 2 instead of 3).

    ✓Final answer

    The correct option is (B) — $\pm\sqrt3$.

    ANSWER: B

12. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
    Q.If the direction cosines of two lines satisfy the equations $l - 2m + n = 0$, $lm + 10mn - 2nl = 0$ and $\theta$ is the angle between the lines, then $\cos\theta =$ (A) $\dfrac{\pi}{6}$ (B) $\dfrac{8}{\sqrt{70}}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{20}{3\sqrt{70}}$

    ›Reveal solutionSolution

    Two families of lines are cut from a linear + a homogeneous quadratic relation in

    direction cosines; eliminate one variable to get a quadratic ratio equation, find

    both direction ratios, then apply the standard angle-between-lines formula.

    Concept and Intuition

    When direction cosines $(l,m,n)$ of a family of lines satisfy one linear and one

    homogeneous-quadratic relation, eliminating a variable between them gives a

    quadratic in the ratio of the remaining two — its two roots are exactly the two

    lines of the family. Once we have both direction ratios (not necessarily

    normalized), the angle between them is found from the dot-product formula

    $\cos\theta = \dfrac{\vec d_1\cdot\vec d_2}{|\vec d_1||\vec d_2|}$, which works with any

    scalar multiple of the direction cosines, not just the normalized ones.

    Step-by-Step Solution

    1. From $l-2m+n=0$, write $l=2m-n$.
    2. Substitute into $lm+10mn-2nl=0$: $(2m-n)m+10mn-2n(2m-n)=2m^2-mn+10mn-4mn+2n^2=2m^2+5mn+2n^2=0$.
    3. Divide by $n^2$ and set $t=m/n$: $2t^2+5t+2=0\Rightarrow t=\dfrac{-5\pm3}{4}=-\tfrac12,-2$.
    4. Case $t=-\tfrac12$: take $n=2,\ m=-1\Rightarrow l=2(-1)-2=-4$. Direction $(-4,-1,2)$.
    5. Case $t=-2$: take $n=1,\ m=-2\Rightarrow l=2(-2)-1=-5$. Direction $(-5,-2,1)$.
    6. $\cos\theta=\dfrac{(-4)(-5)+(-1)(-2)+(2)(1)}{\sqrt{16+1+4}\,\sqrt{25+4+1}}=\dfrac{24}{\sqrt{21}\sqrt{30}}=\dfrac{24}{\sqrt{630}}=\dfrac{24}{3\sqrt{70}}=\dfrac{8}{\sqrt{70}}$.

    Common Mistakes

    • Forgetting the direction cosines don't need to be normalized before applying the dot-product angle formula — the formula divides by the magnitudes anyway.
    • Picking only one root of the quadratic in $t$ and missing that the pair of lines is exactly the two roots.

    ✓Final answer

    The correct option is (B) — $\dfrac{8}{\sqrt{70}}$.

    ANSWER: B