Q.Find the direction cosines of $x$, $y$ and $z$-axis.

Showing the 12 most recent of 34 on this concept.

1. 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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2. 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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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-2021-08-20-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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5. 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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6. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
   Q.If $-2, \dfrac{4}{3}, \dfrac{-4}{5}$ are the intercepts made by a plane on X, Y, Z - axes respectively then the direction cosines of a normal to this plane are (A) $\left(\dfrac{-1}{3}, \dfrac{2}{3}, \dfrac{-2}{3}\right)$ (B) $\left(\dfrac{2}{3\sqrt{5}}, \dfrac{-4}{3\sqrt{5}}, \dfrac{5}{3\sqrt{5}}\right)$ (C) $\left(\dfrac{-4}{\sqrt{57}}, \dfrac{4}{\sqrt{57}}, \dfrac{-5}{\sqrt{57}}\right)$ (D) $\left(\dfrac{2}{\sqrt{38}}, \dfrac{-3}{\sqrt{38}}, \dfrac{5}{\sqrt{38}}\right)$

   ›Reveal solutionSolution

   Convert the given intercepts into the plane's Cartesian equation, read off the normal's direction ratios as the coefficients of $x,y,z$, then normalize by dividing by the magnitude.

   Concept and Intuition

   A plane with $x,y,z$-intercepts $a,b,c$ (i.e. it meets the axes at $(a,0,0),(0,b,0),(0,0,c)$) has the intercept-form equation

   $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1.$

   If this is rewritten as $lx+my+nz=p$, then $(l,m,n)$ are direction ratios of the plane's normal (this falls straight out of comparing with the general plane equation $lx+my+nz=p$, whose normal is $(l,m,n)$). Direction cosines are just this direction-ratio vector scaled to unit length; note direction cosines are only defined up to an overall sign (the normal can point either way).

   Step-by-Step Solution

   1. Given intercepts: $a=-2,\ b=\dfrac{4}{3},\ c=-\dfrac{4}{5}$.
   2. Intercept form: $\dfrac{x}{-2}+\dfrac{y}{4/3}+\dfrac{z}{-4/5}=1$, i.e. $-\dfrac{x}{2}+\dfrac{3y}{4}-\dfrac{5z}{4}=1$.
   3. Multiply through by $4$: $-2x+3y-5z=4$.
   4. Direction ratios of the normal: $(-2,3,-5)$. Magnitude $=\sqrt{(-2)^2+3^2+(-5)^2}=\sqrt{4+9+25}=\sqrt{38}$. …

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7. 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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8. 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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9. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.Angle made by the position vector of the point $(5, -4, -3)$ with the positive direction of $X$-axis is ________ (A) $\dfrac{\pi}{2}$ (B) $\dfrac{\pi}{6}$ (C) $\dfrac{\pi}{4}$ (D) $\dfrac{\pi}{3}$

   ›Reveal solutionSolution

   The angle a position vector makes with the $X$-axis is $\cos^{-1}(x/|\vec r|)$; for $(5,-4,-3)$ this works out to $\pi/4$.

   Concept and Intuition

   The direction cosine along an axis is the cosine of the angle the vector makes with that axis, computed as the corresponding coordinate divided by the vector's magnitude. This follows directly from projecting the vector onto the axis.

   Step-by-Step Solution

   1. Magnitude: $|\vec r| = \sqrt{5^2+(-4)^2+(-3)^2} = \sqrt{25+16+9} = \sqrt{50} = 5\sqrt2$.
   2. Direction cosine along $X$: $l = \cos\theta = \dfrac{5}{5\sqrt2} = \dfrac{1}{\sqrt2}$. …

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10. 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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11. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
    Q.$A(-1,2,-3)$, $B(5,0,-6)$, $C(0,4,-1)$ are the vertices of a triangle $ABC$. The direction cosines of internal bisector of $\angle BAC$ are ______. (A) $\dfrac{25}{\sqrt{714}}, \dfrac{8}{\sqrt{714}}, \dfrac{-5}{\sqrt{714}}$ (B) $\dfrac{25}{\sqrt{714}}, \dfrac{8}{\sqrt{714}}, \dfrac{5}{\sqrt{714}}$ (C) $\dfrac{5}{\sqrt{74}}, \dfrac{6}{\sqrt{74}}, \dfrac{8}{\sqrt{74}}$ (D) $\dfrac{-5}{\sqrt{74}}, \dfrac{6}{\sqrt{74}}, \dfrac{-8}{\sqrt{74}}$

    ›Reveal solutionSolution

    The internal angle bisector direction at a vertex is the sum of unit vectors along the two adjacent sides. Answer: $(25,8,5)/\sqrt{714}$.

    Concept and Intuition

    For a triangle vertex $A$ with adjacent sides toward $B$ and $C$, the internal bisector of $\angle BAC$ points along $\hat{u}=\dfrac{\vec{AB}}{|\vec{AB}|}+\dfrac{\vec{AC}}{|\vec{AC}|}$, because this vector lies exactly midway (in direction) between the two unit vectors, and being a sum of unit vectors it always points into the angle (internal, not external).

    Step-by-Step Solution

    1. $A=(-1,2,-3)$, $B=(5,0,-6)$, $C=(0,4,-1)$.
    2. $\vec{AB}=B-A=(6,-2,-3)$, $|\vec{AB}|=\sqrt{36+4+9}=\sqrt{49}=7$.
    3. $\vec{AC}=C-A=(1,2,2)$, $|\vec{AC}|=\sqrt{1+4+4}=\sqrt9=3$.
    4. Unit vectors: $\hat{AB}=\left(\frac67,-\frac27,-\frac37\right)$, $\hat{AC}=\left(\frac13,\frac23,\frac23\right)$.
    5. Sum (common denominator 21): $\left(\frac{18+7}{21},\frac{-6+14}{21},\frac{-9+14}{21}\right)=\left(\frac{25}{21},\frac{8}{21},\frac{5}{21}\right)$, i.e. direction ratios $(25,8,5)$. …

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12. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
    Q.Let $\alpha, \beta, \gamma$ be the angles made by a vector $\bar{r}$ with the positive directions of $X, Y, Z$-axes respectively. If $\alpha = \tan^{-1}\left(\dfrac{3}{2}\right)$ and $\beta = \tan^{-1}\left(\dfrac{4}{3}\right)$, then $\cos\gamma =$ (A) $\dfrac{2}{3}$ (B) $\dfrac{3}{4}$ (C) $\dfrac{6\sqrt{3}}{5\sqrt{13}}$ (D) $\dfrac{3\sqrt{6}}{5\sqrt{13}}$

    ›Reveal solutionSolution

    Use the direction-cosine identity $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$ after converting each given tangent to a cosine via a right triangle. Answer: $\frac{6\sqrt3}{5\sqrt{13}}$.

    Concept and Intuition

    Any vector's angles with the three coordinate axes satisfy $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$ — this is just the statement that the direction cosines are the components of a unit vector along $\bar r$. So once two of the angles are pinned by their tangents, the third's cosine follows directly.

    Step-by-Step Solution

    1. $\tan\alpha=\frac32$ means a right triangle with opposite 3, adjacent 2, hypotenuse $\sqrt{13}$, so $\cos\alpha=\dfrac{2}{\sqrt{13}}$, and $\cos^2\alpha=\dfrac{4}{13}$.
    2. $\tan\beta=\frac43$ means opposite 4, adjacent 3, hypotenuse 5, so $\cos\beta=\dfrac35$, and $\cos^2\beta=\dfrac9{25}$.
    3. Identity: $\cos^2\gamma = 1-\cos^2\alpha-\cos^2\beta = 1-\dfrac4{13}-\dfrac9{25}$.
    4. Common denominator 325: $\dfrac{4}{13}=\dfrac{100}{325}$, $\dfrac{9}{25}=\dfrac{117}{325}$, so $\cos^2\gamma = 1-\dfrac{217}{325}=\dfrac{108}{325}$. …

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