Q.A vector $\vec{r}$ has magnitude 14 and direction ratios 2, 3, -6. Find the direction cosines and components of $\vec{r}$, given that $\vec{r}$ makes an acute angle with x-axis.

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 $\cos^2\gamma = 1-\dfrac{217}{325}=\dfrac{108}{325}$. …

   🔒 Start your 14-day free trial to see the full solution →
2. 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: $\dfrac34+2b^2=1\Rightarrow 2b^2=\dfrac14\Rightarrow b^2=\dfrac18\Rightarrow b=\dfrac{1}{\sqrt8}$. …

   🔒 Start your 14-day free trial to see the full solution →
3. 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. $|v_1|^2=2+1+(\sqrt2+1)^2=3+3+2\sqrt2=6+2\sqrt2$; $|v_2|^2=2+1+(1-\sqrt2)^2=3+3-2\sqrt2=6-2\sqrt2$. …

   🔒 Start your 14-day free trial to see the full solution →
4. 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 $(25,8,5)$. …

   🔒 Start your 14-day free trial to see the full solution →
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 …

   🔒 Start your 14-day free trial to see the full solution →
6. 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 …

   🔒 Start your 14-day free trial to see the full solution →
7. 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. $l_1l_2+m_1m_2+n_1n_2=\dfrac{5}{13}\cdot0+\dfrac{12}{13}\cdot\dfrac35+0\cdot\dfrac45=\dfrac{36}{65}$. …

   🔒 Start your 14-day free trial to see the full solution →
8. 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 $(l,m,n)=\left(\tfrac1{\sqrt3},\tfrac{-1}{\sqrt3},\tfrac{-1}{\sqrt3}\right)$:
      • $\cdot\hat u_A = \tfrac1{\sqrt3}\left(\tfrac13-\tfrac23-\tfrac23\right)=\tfrac1{\sqrt3}\left(-\tfrac33\right)=-\tfrac1{\sqrt3}$
      • $\cdot\hat u_B = \tfrac1{\sqrt3}\left(\tfrac27-\tfrac37-\tfrac67\right)=\tfrac1{\sqrt3}\left(-\tfrac77\right)=-\tfrac1{\sqrt3}$ …

   🔒 Start your 14-day free trial to see the full solution →
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 $=\tfrac{41}{39}$. …

   🔒 Start your 14-day free trial to see the full solution →
10. 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 $X$: $l = \cos\theta = \dfrac{5}{5\sqrt2} = \dfrac{1}{\sqrt2}$. …

    🔒 Start your 14-day free trial to see the full solution →
11. 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 …

    🔒 Start your 14-day free trial to see the full solution →
12. 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: $|\vec{PQ}|=\sqrt{(-4)^2+0^2+1^2}=\sqrt{17}$. …

    🔒 Start your 14-day free trial to see the full solution →