Q.Find the direction cosines of the line passing through the two points $(-2, 4, -5)$ and $(1, 2, 3)$.

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1. 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: $\left(\dfrac{3}{\sqrt{77}}, \dfrac{-2}{\sqrt{77}}, \dfrac{8}{\sqrt{77}}\right)$.

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2. 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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3. 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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4. 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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5. 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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6. 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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7. 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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8. 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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9. 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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10. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
    Q.If $P(2,\beta,\alpha)$ lies on the plane $x + 2y - z - 2 = 0$ and $Q(\alpha,-1,\beta)$ lies on the plane $2x - y + 3z + 6 = 0$ then the direction cosines of the line PQ are (A) $\left(-\dfrac{4}{\sqrt{17}}, 0, \dfrac{1}{\sqrt{17}}\right)$ (B) $\left(+\dfrac{4}{\sqrt{17}}, 0, \dfrac{1}{\sqrt{17}}\right)$ (C) $\left(\dfrac{1}{\sqrt{17}}, 0, \dfrac{4}{\sqrt{17}}\right)$ (D) $\left(-\dfrac{1}{\sqrt{17}}, 0, \dfrac{4}{\sqrt{17}}\right)$

    ›Reveal solutionSolution

    Use the two plane conditions to pin down $\alpha,\beta$, locate $P$ and $Q$, then the direction cosines are the components of $\vec{PQ}$ divided by $|\vec{PQ}|$.

    Concept and Intuition

    A point lies on a plane $ax+by+cz+d=0$ exactly when its coordinates satisfy the plane equation. Once the two unknowns $\alpha,\beta$ are pinned down by the two plane conditions, the line $PQ$ is completely determined, and its direction cosines are just the direction ratios of $\vec{PQ}$ scaled to unit length.

    Step-by-Step Solution

    1. $P(2,\beta,\alpha)$ lies on $x+2y-z-2=0$:

    $2+2\beta-\alpha-2=0 \Rightarrow 2\beta=\alpha \Rightarrow \alpha=2\beta.$

    2. $Q(\alpha,-1,\beta)$ lies on $2x-y+3z+6=0$:

    $2\alpha-(-1)+3\beta+6=0 \Rightarrow 2\alpha+3\beta+7=0.$

    3. Substitute $\alpha=2\beta$ into step 2:

    $2(2\beta)+3\beta+7=0 \Rightarrow 7\beta=-7 \Rightarrow \beta=-1.$

    Then $\alpha=2\beta=-2$.

    4. So $P=(2,-1,-2)$ and $Q=(-2,-1,-1)$.

    5. Direction ratios of $PQ$: $\vec{PQ}=Q-P=(-2-2,\,-1-(-1),\,-1-(-2))=(-4,0,1)$.

    6. Magnitude: $|\vec{PQ}|=\sqrt{(-4)^2+0^2+1^2}=\sqrt{17}$. …

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11. 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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12. 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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