Q.Find the direction cosines of the vector $\hat{i} + 2\hat{j} + 3\hat{k}$.

Showing the 12 most recent of 34 on this concept.

1. 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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2. 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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3. 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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4. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If $l, m, n$ are the direction cosines of a normal drawn to the plane $2x - 3y + 6z - 7 = 0$ and $d$ is the length of the perpendicular drawn from origin to this plane then $7d\,|l + m + n| =$ (A) 3 (B) 2 (C) 5 (D) 4

   ›Reveal solutionSolution

   This tests converting a plane's equation to normal (direction-cosine) form to read off $l,m,n$ and $d$; the answer is $5$.

   Concept and Intuition

   For a plane $Ax+By+Cz=D$, dividing through by $\sqrt{A^2+B^2+C^2}$ turns the coefficients of $x,y,z$ into the direction cosines of the normal, and the resulting constant on the right is exactly the perpendicular distance from the origin to the plane (once the sign is arranged so this constant is non-negative).

   Step-by-Step Solution

   1. Plane: $2x-3y+6z=7$. Direction ratios of normal: $(2,-3,6)$.
   2. Magnitude $=\sqrt{2^2+(-3)^2+6^2}=\sqrt{4+9+36}=\sqrt{49}=7$.
   3. Direction cosines: $l=\dfrac27,\ m=-\dfrac37,\ n=\dfrac67$. …

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5. 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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6. 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 $(-5,-2,1)$. …

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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 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 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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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 2023Set eng-2023-05-16-AN1 markMCQ
    Q.If $(l_1, m_1, n_1)$, $(l_2, m_2, n_2)$ are the direction cosines of two lines, then $(l_1 m_2 - l_2 m_1)^2 + (m_1 n_2 - m_2 n_1)^2 + (n_1 l_2 - n_2 l_1)^2 + (l_1 l_2 + m_1 m_2 + n_1 n_2)^2 =$ (A) $0$ (B) $1$ (C) $2$ (D) $4$

    ›Reveal solutionSolution

    The expression is $|\vec a\times\vec b|^2+(\vec a\cdot\vec b)^2$ for unit vectors, which always equals $1$.

    Concept and Intuition

    Direction cosines $(l,m,n)$ define a unit vector. For two unit vectors $\vec a=(l_1,m_1,n_1)$, $\vec b=(l_2,m_2,n_2)$, the identity $|\vec a\times\vec b|^2 + (\vec a\cdot\vec b)^2 = |\vec a|^2|\vec b|^2 = 1$ always holds (this is just $\sin^2\theta+\cos^2\theta=1$ in vector form).

    Step-by-Step Solution

    1. $(l_1m_2-l_2m_1)^2+(m_1n_2-m_2n_1)^2+(n_1l_2-n_2l_1)^2 = |\vec a\times\vec b|^2$.
    2. $(l_1l_2+m_1m_2+n_1n_2)^2 = (\vec a\cdot\vec b)^2$. …

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12. 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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