Q.(b) A person standing at $O(0, 0, 0)$ is watching an aeroplane which is at the coordinate point $A(4, 0, 3)$. At the same time he saw a bird at the coordinate point $B(0, 0, 1)$. Find the angles which $\overrightarrow{BA}$ makes with the $x$, $y$ and $z$ axes.

Showing the 12 most recent of 16 on this concept.

1. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.A vector makes equal angles $\alpha$ with $x$ and $y$ axes and $90^\circ$ with $z$-axis. Then $\alpha =$ (A) $60^\circ$ or $120^\circ$ (B) $30^\circ$ or $150^\circ$ (C) $45^\circ$ or $135^\circ$ (D) $90^\circ$

   ›Reveal solutionSolution

   Direction cosines satisfy $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$; solving gives $\alpha=45^\circ$ or $135^\circ$.

   Concept and Intuition

   Any direction in 3-D space is described by the angles it makes with the three coordinate axes, and the direction cosines $l=\cos\alpha$, $m=\cos\beta$, $n=\cos\gamma$ always satisfy $l^2+m^2+n^2=1$. This single identity lets us solve for an unknown angle whenever the others are given.

   Step-by-Step Solution

   1. Let the vector make angle $\alpha$ with both the $x$- and $y$-axes, and $90^\circ$ with the $z$-axis.
   2. Direction cosine identity: $\cos^2\alpha + \cos^2\alpha + \cos^2 90^\circ = 1$.
   3. Since $\cos 90^\circ = 0$: $2\cos^2\alpha = 1 \Rightarrow \cos^2\alpha = \dfrac12$. …

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2. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.If A(1, 2, 3), B(2, 3, -1), C(3, -1, -2) are the vertices of a triangle ABC, then the direction ratios of the bisector of $\angle ABC$ are (A) $(4, 1, 1)$ (B) $(3, 5, 2)$ (C) $(1, 4, 1)$ (D) $(2, -3, -5)$

   ›Reveal solutionSolution

   The bisector of $\angle ABC$ has direction ratios $(2,-3,-5)$ — option (D).

   Take vectors from the vertex $B(2,3,-1)$:

   $\vec{BA}=A-B=(-1,-1,4),\qquad \vec{BC}=C-B=(1,-4,-1).$

   Their magnitudes are equal:

   $|\vec{BA}|=\sqrt{1+1+16}=3\sqrt2,\qquad |\vec{BC}|=\sqrt{1+16+1}=3\sqrt2.$

   Because $|\vec{BA}|=|\vec{BC}|$, a bisector of the angle at $B$ lies along $\vec{BA}\pm\vec{BC}$. The combination present in the options is

   $\vec{BA}-\vec{BC}=(-2,\,3,\,5)\ \parallel\ (2,-3,-5).$ …

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3. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
   Q.Let $\bar{a} = x\bar{i} + y\bar{j} + z\bar{k}$ and $x = 2y$. If $|\bar{a}| = 5\sqrt{2}$ and $\bar{a}$ makes an angle of $135^{\circ}$ with the z-axis then $\bar{a} =$ (A) $2\sqrt{3}\bar{i} + \sqrt{3}\bar{j} - 3\bar{k}$ (B) $2\sqrt{6}\bar{i} + \sqrt{6}\bar{j} - 6\bar{k}$ (C) $2\sqrt{5}\bar{i} + \sqrt{5}\bar{j} - 5\bar{k}$ (D) $2\sqrt{5}\bar{i} + \sqrt{5}\bar{j} + 5\bar{k}$

   ›Reveal solutionSolution

   This tests using the direction-cosine relation with the z-axis and the given magnitude/ratio constraint to pin down all three components. $\bar a = 2\sqrt5\,\bar i+\sqrt5\,\bar j-5\bar k$.

   Concept and Intuition

   The angle a vector makes with the z-axis relates directly to its $z$-component via $\cos\gamma = \dfrac{z}{|\bar a|}$ (this is simply the direction cosine along $k$). Combined with the given ratio $x=2y$ and total magnitude, we get three independent scalar equations for the three unknowns $x,y,z$.

   Step-by-Step Solution

   1. Direction cosine with z-axis: $\cos135^\circ = \dfrac{z}{|\bar a|}$.
   2. $\cos135^\circ = -\dfrac{\sqrt2}{2}$ and $|\bar a|=5\sqrt2$, so $z = 5\sqrt2\times\left(-\dfrac{\sqrt2}{2}\right) = -\dfrac{5\times2}{2} = -5$.
   3. Given $x=2y$, and $|\bar a|^2 = x^2+y^2+z^2 = 50$. …

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4. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If the line of intersection of the planes $2x + 3y + z = 1$ and $x + 3y + 2z = 2$ makes an angle $\alpha$ with the positive x-axis, then $\cos \alpha =$ (A) $\dfrac{1}{\sqrt{3}}$ (B) $\dfrac{1}{\sqrt{2}}$ (C) $\dfrac{1}{2}$ (D) $\dfrac{\sqrt{3}}{2}$

   ›Reveal solutionSolution

   The line of intersection of two planes is perpendicular to both normals, so its direction vector is $n_1\times n_2$. Normalizing this and reading off the $x$-component gives $\cos\alpha=\dfrac1{\sqrt3}$.

   Concept and Intuition

   A line lying in both planes must be perpendicular to both planes' normal vectors, so its direction is along $n_1\times n_2$. Once we have a direction vector $(p,q,r)$ for the line, the angle it makes with the positive $x$-axis has cosine equal to $\dfrac{p}{\sqrt{p^2+q^2+r^2}}$ (the direction cosine $l$).

   Step-by-Step Solution

   1. Normals of the given planes: $n_1=(2,3,1)$ (from $2x+3y+z=1$), $n_2=(1,3,2)$ (from $x+3y+2z=2$).
   2. Direction of the line of intersection:

   $n_1\times n_2 = \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\2&3&1\\1&3&2\end{vmatrix} = \mathbf{i}(3\cdot2-1\cdot3)-\mathbf{j}(2\cdot2-1\cdot1)+\mathbf{k}(2\cdot3-3\cdot1)$

   $=\mathbf{i}(6-3)-\mathbf{j}(4-1)+\mathbf{k}(6-3) = (3,-3,3)$

   3. Simplify to $(1,-1,1)$ (dividing by 3). Its magnitude is $\sqrt{1+1+1}=\sqrt3$.
   4. The direction cosine along the positive $x$-axis is …

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5. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.If the position vectors of the points $A$ and $B$ are $2\bar{i}+3\bar{j}-\bar{k}$ and $\bar{i}-\bar{j}+2\bar{k}$ respectively, then the unit vector along $\overline{BA}$ and in the direction of $\overline{AB}$ is (A) $\dfrac{1}{\sqrt{14}}(3\bar{i}+2\bar{j}+\bar{k})$ (B) $\dfrac{1}{\sqrt{26}}(-\bar{i}-4\bar{j}+3\bar{k})$ (C) $\dfrac{1}{\sqrt{26}}(-3\bar{i}-4\bar{j}+\bar{k})$ (D) $\dfrac{1}{\sqrt{22}}(3\bar{i}-4\bar{j}+3\bar{k})$

   ›Reveal solutionSolution

   The vector from A to B is $B-A$; normalizing it by its own magnitude gives the requested unit vector. Answer: $\dfrac{1}{\sqrt{26}}(-\bar i-4\bar j+3\bar k)$.

   Concept and Intuition

   The unit vector in the direction of $\overline{AB}$ is simply $(\vec{B}-\vec{A})/|\vec{B}-\vec{A}|$ — subtract position vectors in the direction of travel (from A to B), then divide by the magnitude.

   Step-by-Step Solution

   1. $\vec{OA}=2\bar i+3\bar j-\bar k$, $\vec{OB}=\bar i-\bar j+2\bar k$.
   2. $\overline{AB}=\vec{OB}-\vec{OA}=(1-2)\bar i+(-1-3)\bar j+(2-(-1))\bar k=-\bar i-4\bar j+3\bar k$.
   3. $|\overline{AB}|=\sqrt{(-1)^2+(-4)^2+3^2}=\sqrt{1+16+9}=\sqrt{26}$.
   4. Unit vector in the direction of $\overline{AB}=\dfrac{-\bar i-4\bar j+3\bar k}{\sqrt{26}}$. …

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6. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.Let $\overline{OA} = \bar{i} + 2\bar{j} - 2\bar{k}$ and $\overline{OB} = -2\bar{i} - 3\bar{j} + 6\bar{k}$ be the position vectors of two points A and B. If C is a point on the bisector $\angle AOB$ and $OC = \sqrt{42}$, then $\overline{OC} =$ (A) $4\bar{i} - \bar{j} + 5\bar{k}$ (B) $\bar{i} + 5\bar{j} + 4\bar{k}$ (C) $5\bar{i} + 4\bar{j} + \bar{k}$ (D) $\bar{i} - 4\bar{j} + 5\bar{k}$

   ›Reveal solutionSolution

   The internal bisector of the angle between two vectors from a common point runs along the sum of their unit vectors; scaling that direction to length $\sqrt{42}$ gives $\overline{OC}=\bar i+5\bar j+4\bar k$.

   Concept and Intuition

   For two vectors from the same origin, the direction that bisects the angle between them is the sum of their unit vectors (each contributes equally regardless of its original length, so the sum is symmetric about the angle). Once we have that unit bisector direction, any point on the bisector ray is just that unit vector scaled to the desired length.

   Step-by-Step Solution

   1. $\overline{OA}=\bar i+2\bar j-2\bar k$, so $|OA|=\sqrt{1+4+4}=3$.
   2. $\overline{OB}=-2\bar i-3\bar j+6\bar k$, so $|OB|=\sqrt{4+9+36}=7$.
   3. Unit vectors: $\widehat{OA}=\dfrac13(\bar i+2\bar j-2\bar k)$, $\widehat{OB}=\dfrac17(-2\bar i-3\bar j+6\bar k)$.
   4. Bisector direction $=\widehat{OA}+\widehat{OB}$. Using denominator 21: $\widehat{OA}=\left(\frac{7}{21},\frac{14}{21},\frac{-14}{21}\right)$, $\widehat{OB}=\left(\frac{-6}{21},\frac{-9}{21},\frac{18}{21}\right)$.
   5. Sum $=\left(\frac{1}{21},\frac{5}{21},\frac{4}{21}\right)=\dfrac{1}{21}(\bar i+5\bar j+4\bar k)$. …

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7. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.The direction cosines of the line of intersection of the planes $x+2y+z-4=0$ and $2x-y+z-3=0$ are (A) $\left(\dfrac{3}{\sqrt{26}},\dfrac{1}{\sqrt{26}},\dfrac{-4}{\sqrt{26}}\right)$ (B) $\left(\dfrac{3}{\sqrt{14}},\dfrac{2}{\sqrt{14}},\dfrac{-1}{\sqrt{14}}\right)$ (C) $\left(\dfrac{3}{\sqrt{35}},\dfrac{1}{\sqrt{35}},\dfrac{-5}{\sqrt{35}}\right)$ (D) $\left(\dfrac{3}{\sqrt{22}},\dfrac{-2}{\sqrt{22}},\dfrac{3}{\sqrt{22}}\right)$

   ›Reveal solutionSolution

   The line of intersection of two planes is along $\vec n_1\times\vec n_2$; normalizing gives (C).

   Concept and Intuition

   Any line lying in both planes must be perpendicular to both plane normals, so its direction vector is the cross product of the two normals. Direction cosines are then this vector divided by its own magnitude.

   Step-by-Step Solution

   1. Normals: $\vec n_1=(1,2,1)$ from $x+2y+z-4=0$; $\vec n_2=(2,-1,1)$ from $2x-y+z-3=0$.
   2. $\vec n_1\times\vec n_2 = \left(2(1)-1(-1),\ -(1(1)-1(2)),\ 1(-1)-2(2)\right) = (2+1,\ -(1-2),\ -1-4) = (3,\,1,\,-5)$.
   3. Magnitude: $\sqrt{3^2+1^2+(-5)^2}=\sqrt{9+1+25}=\sqrt{35}$.
   4. Direction cosines: $\left(\dfrac{3}{\sqrt{35}},\dfrac{1}{\sqrt{35}},\dfrac{-5}{\sqrt{35}}\right)$.

   Common Mistakes …

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8. AP EAPCET 2021Set eng-2021-08-19-FN1 markMCQ
   Q.Let '$O$' be the origin and '$P$' be a point which is at a distance of 3 units from the origin. If the direction ratios of $\overline{OP}$ are $(1, -2, -2)$, then the coordinates of '$P$' are ____ (A) $(1, -2, -2)$ (B) $(3, -6, -6)$ (C) $\left(\dfrac{1}{3}, \dfrac{-2}{3}, \dfrac{-2}{3}\right)$ (D) $\left(\dfrac{1}{9}, \dfrac{-2}{9}, \dfrac{-2}{9}\right)$

   ›Reveal solutionSolution

   The direction ratios already have magnitude exactly 3, matching $OP=3$, so $P$ coincides with the direction-ratio triple itself: $(1,-2,-2)$.

   Concept and Intuition

   A point at distance $r$ from the origin along direction cosines $(l,m,n)$ is $(lr, mr, nr)$. Direction ratios are proportional to direction cosines, scaled by their magnitude; if that magnitude happens to equal the required distance, the ratios and the point's coordinates coincide.

   Step-by-Step Solution

   1. Magnitude of direction ratios: $\sqrt{1^2+(-2)^2+(-2)^2}=\sqrt{1+4+4}=\sqrt9=3$.
   2. Direction cosines: $\left(\dfrac13,-\dfrac23,-\dfrac23\right)$. …

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9. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.Let $A(2, 3, 5)$, $B(-1, 3, 2)$, $C(\lambda, 5, \mu)$ be the vertices of $\triangle ABC$. If the median through the vertex A is equally inclined to the coordinate axes, then (A) $5\lambda - 8\mu = 0$ (B) $8\lambda - 5\mu = 0$ (C) $10\lambda - 7\mu = 0$ (D) $7\lambda - 10\mu = 0$

   ›Reveal solutionSolution

   A line "equally inclined to the coordinate axes" has direction ratios equal in absolute value; applying this to the median through $A$ pins down $\lambda,\mu$. Answer: $10\lambda-7\mu=0$.

   Concept and Intuition

   A line's direction cosines $(l,m,n)$ measure the cosine of the angle it makes with each axis. "Equally inclined to the coordinate axes" means these angles are equal, hence $|l|=|m|=|n|$, i.e. the direction ratios of the line have equal absolute value (signs may differ). The median from a vertex is just the segment to the midpoint of the opposite side, so its direction ratios come straight from that midpoint minus the vertex.

   Step-by-Step Solution

   1. $A=(2,3,5)$, $B=(-1,3,2)$, $C=(\lambda,5,\mu)$. Midpoint of $BC$: $M=\left(\dfrac{\lambda-1}{2},4,\dfrac{\mu+2}{2}\right)$.
   2. Direction ratios of median $AM$: $M-A=\left(\dfrac{\lambda-1}{2}-2,\ 4-3,\ \dfrac{\mu+2}{2}-5\right)=\left(\dfrac{\lambda-5}{2},\,1,\,\dfrac{\mu-8}{2}\right)$.
   3. Equally inclined to the axes $\Rightarrow$ equal magnitude of ratios: $\left|\dfrac{\lambda-5}{2}\right|=|1|=\left|\dfrac{\mu-8}{2}\right|$.
   4. From $\left|\dfrac{\lambda-5}{2}\right|=1$: $\lambda-5=\pm2\Rightarrow\lambda=7$ or $3$.
   5. From $\left|\dfrac{\mu-8}{2}\right|=1$: $\mu-8=\pm2\Rightarrow\mu=10$ or $6$. …

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10. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
    Q.$\bar{c}$ is a vector along the bisector of the internal angle between the vectors $\bar{a} = 4\bar{i} + 7\bar{j} - 4\bar{k}$ and $\bar{b} = 12\bar{i} - 3\bar{j} + 4\bar{k}$. If the magnitude of $\bar{c}$ is $3\sqrt{13}$ then $\bar{c} =$ (A) $5\bar{i} - 8\bar{j} + 2\sqrt{2}\,\bar{k}$ (B) $10\bar{i} + 4\bar{j} - \bar{k}$ (C) $\bar{i} - 10\bar{j} + 4\bar{k}$ (D) $2\sqrt{2}\bar{i} + 5\bar{j} - 8\bar{k}$

    ›Reveal solutionSolution

    This tests the angle-bisector-direction formula $\hat a+\hat b$ for vectors; the bisector vector of the given magnitude works out to $10\bar i+4\bar j-\bar k$.

    Concept and Intuition

    The internal bisector of the angle between two vectors $\bar a,\bar b$ points along $\hat a+\hat b$ (the sum of their unit vectors) — this is the vector analogue of the angle-bisector property, since adding two unit vectors always bisects the angle between them (by the rhombus/parallelogram symmetry).

    Step-by-Step Solution

    1. $|\bar a|=\sqrt{4^2+7^2+(-4)^2}=\sqrt{16+49+16}=\sqrt{81}=9$.
    2. $|\bar b|=\sqrt{12^2+(-3)^2+4^2}=\sqrt{144+9+16}=\sqrt{169}=13$.
    3. Bisector direction $=\dfrac{\bar a}{9}+\dfrac{\bar b}{13}$. Using a common denominator 117: $\dfrac{\bar a}{9}=\dfrac{(52,91,-52)}{117}$, $\dfrac{\bar b}{13}=\dfrac{(108,-27,36)}{117}$.
    4. Sum: $\dfrac{(160,64,-16)}{117}=\dfrac{16(10,4,-1)}{117}$, so the direction is $(10,4,-1)$. …

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11. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
    Q.A line segment PQ has the length 63 and direction ratios $(3,-2,6)$. If this line makes an obtuse angle with X-axis, then the components of the vector $\overline{PQ}$ are (A) $7, 8, -4$ (B) $-7, 8, -4$ (C) $27, -18, 54$ (D) $-27, 18, -54$

    ›Reveal solutionSolution

    Scale the direction ratios to the given length, then use the obtuse-angle-with-x-axis condition to fix the sign. The answer is $(-27,18,-54)$.

    Concept and Intuition

    A vector with direction ratios $(a,b,c)$ points along a line with direction cosines $\left(\dfrac{a}{r},\dfrac{b}{r},\dfrac{c}{r}\right)$ where $r=\sqrt{a^2+b^2+c^2}$. The angle the vector makes with the positive x-axis has cosine equal to the x direction-cosine; that cosine is negative exactly when the angle is obtuse. So the sign of the x-component of the actual vector (not just its magnitude) is what decides between the two candidate directions.

    Step-by-Step Solution

    1. Direction ratios given: $(3,-2,6)$. Magnitude $=\sqrt{3^2+(-2)^2+6^2}=\sqrt{9+4+36}=\sqrt{49}=7$.
    2. Since $\overline{PQ}$ has length 63, the scale factor from the direction-ratio vector to the actual vector is $63/7=9$ (up to sign).
    3. Scaling $(3,-2,6)$ by $9$: $(27,-18,54)$, with magnitude $9\times7=63$ ✓. The reverse direction is $(-27,18,-54)$, also of magnitude 63. (Options (A) $7,8,-4$ and (B) $-7,8,-4$ have magnitude $\sqrt{49+64+16}=\sqrt{129}$ and are not even proportional to $(3,-2,6)$, so they cannot be the answer regardless of the angle condition — they are decoys.) …

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12. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
    Q.Let $\bar i-\bar j+2\bar k$ and $\bar i+2\bar j-2\bar k$ be the position vectors of points A and B respectively. If C is a point on the line joining A and B such that BC=10, then the position vector of C can be (A) $\bar i+8\bar j-10\bar k$ (B) $\bar i+4\bar j-6\bar k$ (C) $\bar i-8\bar j+10\bar k$ (D) $\bar i-4\bar j-6\bar k$

    ›Reveal solutionSolution

    $C$ lies on line $AB$ extended beyond $B$ at distance $10$ from $B$; scaling the unit direction vector by $10$ and adding to $B$ gives $C=(1,8,-10)$.

    Concept and Intuition

    Any point on the line through $A,B$ can be written as $B+t\,\widehat{AB}$ for a scalar $t$ (signed distance from $B$). Since $BC=10$ is a distance (not a ratio), we use the unit direction vector scaled by $10$, with two possible signs (either side of $B$).

    Step-by-Step Solution

    1. $\overrightarrow{AB}=B-A=(1-1,\,2-(-1),\,-2-2)=(0,3,-4)$, and $|\overrightarrow{AB}|=\sqrt{0+9+16}=5$.
    2. Unit vector along $AB$: $\hat u=(0,\tfrac35,-\tfrac45)$.
    3. Point $C = B \pm 10\hat u = (1,2,-2)\pm(0,6,-8)$.
    4. Taking the $+$ sign: $C=(1,8,-10)$; taking the $-$ sign: $C=(1,-4,6)$. …

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