Q.Find the unit vector in the direction of the sum of the vectors, $\vec{a}=2\hat{i}+2\hat{j}-5\hat{k}$ and $\vec{b}=2\hat{i}+\hat{j}+3\hat{k}$.

Showing the 12 most recent of 16 on this concept.

1. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.Let $\vec a = 2\vec i+\vec j-\vec k$ and $\vec b=\vec i+3\vec j-5\vec k$ be two vectors, and $\vec r$ be a vector along the vector $3\vec a-2\vec b$ such that $|\vec r|=\sqrt{74}$. If the direction of $\vec r$ is opposite to that of $3\vec a-2\vec b$, then $\vec r=$ (A) $-7\vec i-4\vec j+3\vec k$ (B) $4\vec i+7\vec j-3\vec k$ (C) $-4\vec i+3\vec j-7\vec k$ (D) $4\vec i-3\vec j+7\vec k$

   ›Reveal solutionSolution

   $3\vec a-2\vec b=4\vec i-3\vec j+7\vec k$ already has magnitude $\sqrt{74}$, so $\vec r$ (same magnitude, opposite direction) is simply its negative.

   Concept and Intuition

   A vector "along" a given vector but "opposite in direction" with a specified magnitude is found by first computing the reference vector, checking whether its own magnitude already matches the target (a nice simplification here), and if so just negating it.

   Step-by-Step Solution

   1. $\vec a=2\vec i+\vec j-\vec k=(2,1,-1)$, $\vec b=\vec i+3\vec j-5\vec k=(1,3,-5)$.
   2. $3\vec a=(6,3,-3)$, $2\vec b=(2,6,-10)$. So $3\vec a-2\vec b=(6-2,3-6,-3-(-10))=(4,-3,7)$.
   3. $|3\vec a-2\vec b|=\sqrt{4^2+(-3)^2+7^2}=\sqrt{16+9+49}=\sqrt{74}$ — exactly the given $|\vec r|$. …

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2. 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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3. 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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4. 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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5. 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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6. 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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7. 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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8. 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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9. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
   Q.If (a, b, c) are the direction ratios of a line joining the points $(4, 3, -5)$ and $(-2, 1, -8)$ then the point P $(a, 3b, 2c)$ lies on the plane (A) $x + y + z = 0$ (B) $x + y - 2z = 0$ (C) $x + 2y + 3z = 0$ (D) $x - 2y + 3z = 0$

   ›Reveal solutionSolution

   Compute the direction ratios of the joining line, form $P(a,3b,2c)$, and test each candidate plane — only $x+y-2z=0$ is satisfied.

   Concept and Intuition

   Direction ratios of a line through two points are simply the differences of corresponding coordinates (up to any common scalar multiple). Once we have $(a,b,c)$, constructing the point $P(a,3b,2c)$ and checking it against each candidate plane equation is a direct substitution exercise.

   Step-by-Step Solution

   1. Points: $(4,3,-5)$ and $(-2,1,-8)$.
   2. Direction ratios: $(-2-4,\ 1-3,\ -8-(-5)) = (-6,-2,-3)$, i.e. proportional to $(6,2,3)$ (dividing by $-1$).
   3. Take $a=6, b=2, c=3$ (any nonzero scalar multiple works equally, since all four candidate planes pass through the origin).
   4. $P = (a, 3b, 2c) = (6,\ 3\times2,\ 2\times3) = (6,6,6)$.
   5. Test $x+y+z=0$: $6+6+6=18\neq0$. Fails.
   6. Test $x+y-2z=0$: $6+6-12=0$. Holds.
   7. Test $x+2y+3z=0$: $6+12+18=36\neq0$. Fails. …

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10. 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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11. 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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12. AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
    Q.If the line joining the points $(k, 2, 3)$ and $(1, 1, 2)$ is parallel to the line joining the points $(5, 4, -1)$ and $(3, 2, -3)$, then the value of $k =$______ (A) $1$ (B) $2$ (C) $-2$ (D) $3$

    ›Reveal solutionSolution

    Two lines are parallel exactly when their direction vectors are proportional; equate the ratios to solve for $k$.

    Concept and Intuition

    A line through two 3D points has direction vector equal to the difference of the points. Parallel lines have proportional (or equal, up to sign) direction vectors.

    Step-by-Step Solution

    1. Direction of line through $(k,2,3)$ and $(1,1,2)$: $(1-k,\,1-2,\,2-3)=(1-k,\,-1,\,-1)$.
    2. Direction of line through $(5,4,-1)$ and $(3,2,-3)$: $(3-5,\,2-4,\,-3-(-1))=(-2,-2,-2)$, i.e. proportional to $(1,1,1)$.
    3. For parallelism, $(1-k,-1,-1)$ must be proportional to $(1,1,1)$: since the $y$- and $z$-components already match the ratio $-1/1=-1$, the $x$-component must also give ratio $-1$: $\dfrac{1-k}{1}=-1$.
    4. $1-k=-1\Rightarrow k=2$. …

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