Q.If the direction cosines of a line are $k, k, k$, then
(A) $k > 0$
(B) $0 < k < 1$
(C) $k = 1$
(D) $k = \dfrac{1}{\sqrt{3}}$ or $-\dfrac{1}{\sqrt{3}}$

Showing the 12 most recent of 34 on this concept.

1. 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. $c^2=3 \Rightarrow c=\pm\sqrt3$. …

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2. 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 …

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3. 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 $n=\dfrac12$. …

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4. 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 $(l,m,n)=\left(\tfrac1{\sqrt3},\tfrac{-1}{\sqrt3},\tfrac{-1}{\sqrt3}\right)$:
      • $\cdot\hat u_A = \tfrac1{\sqrt3}\left(\tfrac13-\tfrac23-\tfrac23\right)=\tfrac1{\sqrt3}\left(-\tfrac33\right)=-\tfrac1{\sqrt3}$
      • $\cdot\hat u_B = \tfrac1{\sqrt3}\left(\tfrac27-\tfrac37-\tfrac67\right)=\tfrac1{\sqrt3}\left(-\tfrac77\right)=-\tfrac1{\sqrt3}$ …

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5. 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 …

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6. 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: $\left\langle 0,-\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right\rangle$. …

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7. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
   Q.Which of the following vector is equally inclined with the coordinate axes? (A) $\hat{i} + 2\hat{j} + 3\hat{k}$ (B) $2\hat{i} - 2\hat{j} + \hat{k}$ (C) $3\hat{i} + 3\hat{j} - 3\hat{k}$ (D) $4\hat{i} + 4\hat{j} + 4\hat{k}$

   ›Reveal solutionSolution

   Equal inclination to all three axes needs identical direction cosines, which only happens when all three components are equal in both size and sign — true only for $4\hat i+4\hat j+4\hat k$.

   Concept and Intuition

   The angle a vector makes with an axis depends on its direction cosine for that axis, $\cos\theta=\dfrac{\text{component}}{|\vec v|}$. For the angles to all be equal, the components themselves (not just their magnitudes) must be equal, since a negative component gives an obtuse angle, different from a positive component's acute angle even with the same magnitude.

   Step-by-Step Solution

   1. (A) $\hat i+2\hat j+3\hat k$: components $1,2,3$ — unequal, rejected.
   2. (B) $2\hat i-2\hat j+\hat k$: components $2,-2,1$ — unequal magnitudes, rejected. …

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8. 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: $\dfrac34+2b^2=1\Rightarrow 2b^2=\dfrac14\Rightarrow b^2=\dfrac18\Rightarrow b=\dfrac{1}{\sqrt8}$. …

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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 …

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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 $(1,0,-1)$. …

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11. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
    Q.If the direction cosines of two lines are $\left(\dfrac{2}{3}, \dfrac{2}{3}, \dfrac{1}{3}\right)$ and $\left(\dfrac{5}{13}, \dfrac{12}{13}, 0\right)$, then identify the direction ratios of a line which is bisecting one of the angle between them. (A) $\langle 40, 60, 13 \rangle$ (B) $\langle 41, 60, 10 \rangle$ (C) $\langle 41, 62, 13 \rangle$ (D) $\langle 1, 2, 3 \rangle$

    ›Reveal solutionSolution

    This tests the fact that the direction of the internal bisector between two lines through a common point is the (unit) sum of their unit direction vectors. Answer: $\langle 41,62,13\rangle$.

    Concept and Intuition

    If $\hat u$ and $\hat v$ are unit vectors along two lines through a point, then $\hat u + \hat v$ points along one internal angle bisector (and $\hat u - \hat v$ along the other), because the parallelogram built on two equal-length vectors has its diagonal bisecting the angle between them.

    Step-by-Step Solution

    1. Verify both given triples are unit vectors: $\left(\tfrac23\right)^2+\left(\tfrac23\right)^2+\left(\tfrac13\right)^2 = \tfrac49+\tfrac49+\tfrac19=1$; and $\left(\tfrac5{13}\right)^2+\left(\tfrac{12}{13}\right)^2+0^2 = \tfrac{25}{169}+\tfrac{144}{169}=1$. Good — both are direction cosines.
    2. Add them componentwise, using common denominator 39: $\tfrac23=\tfrac{26}{39}$, $\tfrac5{13}=\tfrac{15}{39}$, sum $=\tfrac{41}{39}$. …

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12. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
    Q.The direction cosines of two lines are connected by the relations $l+m-n=0$ and $lm-2mn+nl=0$. If $\theta$ is the acute angle between those lines then $\cos\theta=$ (A) $\dfrac{\pi}{6}$ (B) $\dfrac{1}{\sqrt{7}}$ (C) $\sqrt{\dfrac{5}{6}}$ (D) $\dfrac{\pi}{3}$

    ›Reveal solutionSolution

    Eliminate $n$ using the linear relation, reduce the quadratic relation to a simple ratio between $l$ and $m$, extract the two lines' direction ratios, and compute the angle between them.

    Concept and Intuition

    When two lines' direction cosines both satisfy a linear relation and a quadratic (pair-of-planes-like) relation, substituting the linear relation into the quadratic one typically collapses it into a simple relation between two of the three direction ratios, identifying the two specific lines.

    Step-by-Step Solution

    1. From $l+m-n=0$: $n=l+m$.
    2. Substitute into $lm-2mn+nl=0$: $lm-2m(l+m)+(l+m)l = lm-2lm-2m^2+l^2+lm = l^2-2m^2 + (1-2+1)lm = l^2-2m^2$ (the $lm$ terms cancel exactly).
    3. So $l^2=2m^2\Rightarrow l=\pm\sqrt2\,m$.
    4. Taking $m=1$: line 1 has $(l,m,n)=(\sqrt2,1,\sqrt2+1)$; line 2 has $(l,m,n)=(-\sqrt2,1,1-\sqrt2)$.
    5. Dot product: $\sqrt2(-\sqrt2)+1(1)+(\sqrt2+1)(1-\sqrt2) = -2+1+(-1) = -2$.
    6. $|v_1|^2=2+1+(\sqrt2+1)^2=3+3+2\sqrt2=6+2\sqrt2$; $|v_2|^2=2+1+(1-\sqrt2)^2=3+3-2\sqrt2=6-2\sqrt2$. …

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