Q.Write the direction ratios of the vector $\vec{a}=\hat{i}+\hat{j}-2\hat{k}$ and hence calculate its direction cosines.

Showing the 12 most recent of 34 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If a vector $3\bar{i} - 6\bar{j} + 2\bar{k}$ makes angles $\alpha, \beta, \gamma$ with the positive X, Y, Z-axes respectively, then $\cos\alpha + \cos^2\beta + 7\cos^3\gamma =$ (A) $1$ (B) $\dfrac{65}{49}$ (C) $2$ (D) $\dfrac{-7}{49}$

   ›Reveal solutionSolution

   This is a direct application of direction cosines: $\cos\alpha,\cos\beta,\cos\gamma$ are the components of the unit vector along the given vector. The answer is (B).

   Concept and Intuition

   For any vector $a\bar i+b\bar j+c\bar k$, the direction cosines with the coordinate axes are simply its components divided by its magnitude: $\cos\alpha=\dfrac a{|\vec v|}$, $\cos\beta=\dfrac b{|\vec v|}$, $\cos\gamma=\dfrac c{|\vec v|}$. Once these are known, the requested expression is pure arithmetic.

   Step-by-Step Solution

   1. Magnitude: $|\vec v| = \sqrt{3^2+(-6)^2+2^2} = \sqrt{9+36+4}=\sqrt{49}=7$.
   2. Direction cosines: $\cos\alpha=\dfrac37$, $\cos\beta=\dfrac{-6}7$, $\cos\gamma=\dfrac27$.
   3. $\cos^2\beta = \dfrac{36}{49}$.
   4. $\cos^3\gamma = \dfrac{8}{343}$, so $7\cos^3\gamma = \dfrac{7\times8}{343} = \dfrac{56}{343}=\dfrac8{49}$ (since $343=7^3$).
   5. $\cos\alpha+\cos^2\beta+7\cos^3\gamma = \dfrac37+\dfrac{36}{49}+\dfrac8{49}$.
   6. Convert $\dfrac37$ to forty-ninths: $\dfrac37=\dfrac{21}{49}$.
   7. Sum: $\dfrac{21+36+8}{49} = \dfrac{65}{49}$.

   Common Mistakes …

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2. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.Let $A(1,-1,2)$, $B(6,11,2)$, $C(1,2,6)$ be three points. If $l_1,m_1,n_1$ are the direction cosines of AB and $l_2,m_2,n_2$ are the direction cosines of AC, then $|l_1l_2+m_1m_2+n_1n_2|=$ (A) $63/65$ (B) $36/65$ (C) $16/65$ (D) $13/64$

   ›Reveal solutionSolution

   Direction cosines of AB and AC are found from their displacement vectors divided by their magnitudes; their dot product is $36/65$, which is $\cos(\angle BAC)$.

   Concept and Intuition

   The direction cosines of a segment PQ are the components of the unit vector along PQ. The sum $l_1l_2+m_1m_2+n_1n_2$ is just the dot product of the two unit vectors, i.e. $\cos$ of the angle between AB and AC.

   Step-by-Step Solution

   1. $\vec{AB}=B-A=(6-1,11-(-1),2-2)=(5,12,0)$, $|AB|=\sqrt{25+144}=13$. So $(l_1,m_1,n_1)=(5/13,12/13,0)$.
   2. $\vec{AC}=C-A=(1-1,2-(-1),6-2)=(0,3,4)$, $|AC|=\sqrt{0+9+16}=5$. So $(l_2,m_2,n_2)=(0,3/5,4/5)$.
   3. $l_1l_2+m_1m_2+n_1n_2=\dfrac{5}{13}\cdot0+\dfrac{12}{13}\cdot\dfrac35+0\cdot\dfrac45=\dfrac{36}{65}$. …

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3. 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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4. 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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5. 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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6. 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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7. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.If a, b, c are the direction ratios of a line L and $\ell, m, n$ are its direction cosines, then $\dfrac{a^2}{b^2+c^2} =$ (A) $\dfrac{1-\ell^2}{\ell^2}$ (B) $\dfrac{\ell^2}{1+\ell^2}$ (C) $\dfrac{\ell^2}{\ell^2+m^2}$ (D) $\dfrac{\ell^2}{1-\ell^2}$

   ›Reveal solutionSolution

   Direction cosines are just a normalised version of direction ratios; the common scaling constant cancels in the ratio $a^2/(b^2+c^2)$, leaving $\ell^2/(1-\ell^2)$.

   Concept and Intuition

   Direction ratios $(a,b,c)$ and direction cosines $(\ell,m,n)$ of the same line are proportional: $\ell=ka,\ m=kb,\ n=kc$ for some constant $k$, chosen so that $\ell^2+m^2+n^2=1$. Since the ratio $a^2/(b^2+c^2)$ is scale-invariant, the constant $k$ drops out entirely, and we can substitute the direction-cosine identity directly.

   Step-by-Step Solution

   1. $\ell=ka,\ m=kb,\ n=kc$ for some $k\ne0$.
   2. $\dfrac{a^2}{b^2+c^2}=\dfrac{(\ell/k)^2}{(m/k)^2+(n/k)^2}=\dfrac{\ell^2}{m^2+n^2}$ (the $k^2$ cancels).
   3. Since $\ell^2+m^2+n^2=1$, we have $m^2+n^2=1-\ell^2$. …

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8. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.If $(\alpha,\beta,\gamma)$ are the Direction cosines of an angular bisector of two lines whose Direction ratios are $(2,2,1)$ and $(2,-1,-2)$, then $(\alpha+\beta+\gamma)^2=$ (A) 3 (B) 2 (C) 4 (D) 5

   ›Reveal solutionSolution

   The direction cosines of an angle bisector between two lines are proportional to the sum (or difference) of their unit direction vectors. Using the difference here (since both direction ratios have equal magnitude 3) gives $(\alpha+\beta+\gamma)^2=2$.

   Concept and Intuition

   Given two lines with direction vectors of equal magnitude, the two angle bisectors between them are along the sum and the difference of the corresponding unit vectors — one bisects the angle containing the two rays, the other the supplementary angle.

   Step-by-Step Solution

   1. Magnitude of $(2,2,1)$: $\sqrt{4+4+1}=3$; unit vector $\left(\dfrac23,\dfrac23,\dfrac13\right)$.
   2. Magnitude of $(2,-1,-2)$: $\sqrt{4+1+4}=3$; unit vector $\left(\dfrac23,-\dfrac13,-\dfrac23\right)$.
   3. Since the magnitudes are equal, an angular bisector direction is along the difference of these unit vectors: $\left(\dfrac23-\dfrac23,\ \dfrac23+\dfrac13,\ \dfrac13+\dfrac23\right)=(0,1,1)$. …

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9. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.If the angles between the sides of the triangle ABC formed by $A(2, 3, 5)$, $B(-1, 3, 2)$ and $C(3, 5, -2)$ are $\alpha$, $\beta$ and $\gamma$, then $\sin^2\alpha + \sin^2\beta + \sin^2\gamma =$ (A) $1$ (B) $2$ (C) $\dfrac{3}{2}$ (D) $\dfrac{1}{2}$

   ›Reveal solutionSolution

   The triangle turns out to be right-angled (at $B$), which makes $\sin^2\alpha+\sin^2\beta+\sin^2\gamma=2$ instantly via the Pythagorean identity.

   Concept and Intuition

   Rather than compute each angle separately via the dot-product/cosine formula, it pays to first check the side lengths for a Pythagorean relation — a right triangle immediately gives one $\sin^2=1$ term, and the other two angles are automatically complementary, so their sine-squares also sum to $1$ by $\sin^2\theta+\cos^2\theta=1$.

   Step-by-Step Solution

   1. $A(2,3,5)$, $B(-1,3,2)$, $C(3,5,-2)$.
   2. $\vec{AB}=(-3,0,-3)\Rightarrow AB^2=9+0+9=18$.
   3. $\vec{BC}=(4,2,-4)\Rightarrow BC^2=16+4+16=36$.
   4. $\vec{CA}=(-1,-2,7)\Rightarrow CA^2=1+4+49=54$.
   5. Check Pythagoras: $AB^2+BC^2=18+36=54=CA^2$. Since $CA$ is the side opposite vertex $B$, this means $\angle B=90^\circ$, i.e. $\beta=\pi/2$, so $\sin^2\beta=1$.
   6. In any triangle $\alpha+\beta+\gamma=180^\circ$; with $\beta=90^\circ$, $\alpha+\gamma=90^\circ\Rightarrow\gamma=90^\circ-\alpha$.
   7. $\sin\gamma=\sin(90^\circ-\alpha)=\cos\alpha\Rightarrow\sin^2\alpha+\sin^2\gamma=\sin^2\alpha+\cos^2\alpha=1$. …

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10. 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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11. 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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12. 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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