Q.Write two different vectors having same direction.

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 $|\vec r|$.
   4. Since $\vec r$ points opposite to $3\vec a-2\vec b$ and has the same magnitude, $\vec r=-(3\vec a-2\vec b)=(-4,3,-7)=-4\vec i+3\vec j-7\vec k$.

   Common Mistakes

   • Forgetting to negate the vector after confirming the magnitude match (using $3\vec a-2\vec b$ itself instead of its negative).
   • Arithmetic slip in the vector subtraction (sign of the $z$-component especially).

   ✓Final answer

   The correct option is (C) — $-4\vec i+3\vec j-7\vec k$.

   ANSWER: C

2. 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: $C=(1,-4,6)$.
   5. Comparing to the options, $(1,8,-10)=\bar i+8\bar j-10\bar k$ matches option (A) exactly.

   Common Mistakes

   • Using $\overrightarrow{AB}$ itself (length 5) instead of the unit vector when scaling by the distance 10.
   • Forgetting there are two valid positions for $C$ (on either side of $B$) and not checking against the given options.

   ✓Final answer

   The correct option is (A) — $\bar i+8\bar j-10\bar k$.

   ANSWER: A

3. 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.)
   4. "Obtuse angle with the X-axis" means the angle $\alpha$ between $\overline{PQ}$ and the positive x-direction satisfies $\cos\alpha<0$. Since $\cos\alpha = \dfrac{x\text{-component}}{|\overline{PQ}|}$, this requires the x-component to be negative.
   5. Between $(27,-18,54)$ (x-component $+27$, acute) and $(-27,18,-54)$ (x-component $-27$, obtuse), the obtuse condition selects $(-27,18,-54)$.

   Common Mistakes

   • Picking the positive-x option out of habit, without checking the obtuse-angle condition.
   • Confusing "obtuse angle with the x-axis" with the magnitude of the direction ratios (only the sign of the x-component matters, once magnitude is already fixed by the length).

   ✓Final answer

   The correct option is (D) — $-27,18,-54$.

   ANSWER: D

4. 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 $\overline{AB}=\dfrac{-\bar i-4\bar j+3\bar k}{\sqrt{26}}$.

   Common Mistakes

   • Computing $A-B$ instead of $B-A$ (reversing the direction), which gives the wrong sign on every component.
   • Arithmetic slip subtracting the $j$-components ($3-(-1)=4$, not $3+1$ confusion errors).

   ✓Final answer

   The correct option is (B) — $\dfrac{1}{\sqrt{26}}(-\bar{i}-4\bar{j}+3\bar{k})$.

   ANSWER: B

5. 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 $(10,4,-1)$.
   5. $|(10,4,-1)|=\sqrt{100+16+1}=\sqrt{117}=3\sqrt{13}$ — this exactly matches the required $|\bar c|=3\sqrt{13}$!
   6. So $\bar c=(10,4,-1)=10\bar i+4\bar j-\bar k$ directly (scale factor 1, no further adjustment needed).

   Common Mistakes

   • Adding $\bar a+\bar b$ directly without first normalizing by their magnitudes — that does NOT generally bisect the angle unless $|\bar a|=|\bar b|$.
   • Sign errors combining fractions over the common denominator 117.

   ✓Final answer

   The correct option is (B) — $10\bar{i} + 4\bar{j} - \bar{k}$.

   ANSWER: B

6. 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: $\left(\dfrac13,-\dfrac23,-\dfrac23\right)$.
   3. $P = OP\times(l,m,n) = 3\left(\dfrac13,-\dfrac23,-\dfrac23\right) = (1,-2,-2)$.

   Common Mistakes

   • Forgetting to normalize the direction ratios before scaling by the distance (would wrongly give $(3,-6,-6)$, option B — that assumes the ratios were already unit vectors and then multiplies again by 3).

   ✓Final answer

   The correct option is (A) — $(1, -2, -2)$.

   ANSWER: A

7. 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. $1-k=-1\Rightarrow k=2$.

   Common Mistakes

   • Comparing components in the wrong order (mixing up which vector's components go in the numerator vs denominator of the ratio).
   • Forgetting to check that all three ratios are consistent (here the $y,z$ ratios confirm the overall proportionality constant is $-1$).

   ✓Final answer

   The correct option is (B) — $2$.

   ANSWER: B

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

   • Sign errors in expanding the cross-product determinant (especially the middle/$j$ component, which carries a negative sign).
   • Forgetting to normalize (dividing by the magnitude) before calling the result "direction cosines".

   ✓Final answer

   The correct option is (C) — $\left(\dfrac{3}{\sqrt{35}},\dfrac{1}{\sqrt{35}},\dfrac{-5}{\sqrt{35}}\right)$.

   ANSWER: C

9. 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 $|\bar a|^2 = x^2+y^2+z^2 = 50$.
   4. Substitute: $(2y)^2+y^2+(-5)^2=50 \Rightarrow 4y^2+y^2+25=50 \Rightarrow 5y^2=25 \Rightarrow y^2=5 \Rightarrow y=\sqrt5$ (taking the positive root to match option form).
   5. Then $x=2y=2\sqrt5$.
   6. So $\bar a = 2\sqrt5\,\bar i+\sqrt5\,\bar j-5\bar k$.

   Common Mistakes

   • Sign error on $\cos135^\circ$ (it's negative, giving $z=-5$, not $+5$) — this is what distinguishes option (C) from (D).
   • Forgetting to substitute $x=2y$ before solving for the magnitude equation.

   ✓Final answer

   The correct option is (C) — $2\sqrt5\bar i+\sqrt5\bar j-5\bar k$.

   ANSWER: C

10. 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 $\cos 90^\circ = 0$: $2\cos^2\alpha = 1 \Rightarrow \cos^2\alpha = \dfrac12$.
    4. $\cos\alpha = \pm\dfrac{1}{\sqrt2}$, so $\alpha = 45^\circ$ (positive cosine) or $\alpha=135^\circ$ (negative cosine, since $\cos135^\circ=-\tfrac{1}{\sqrt2}$).

    Common Mistakes

    • Forgetting that $\cos\alpha$ can be negative, missing the $135^\circ$ solution.
    • Confusing direction angles with direction ratios and skipping the normalization identity.

    ✓Final answer

    The correct option is (C) — $45^\circ$ or $135^\circ$.

    ANSWER: C

11. 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 $=\left(\frac{1}{21},\frac{5}{21},\frac{4}{21}\right)=\dfrac{1}{21}(\bar i+5\bar j+4\bar k)$.
    6. Magnitude of $(\bar i+5\bar j+4\bar k)$ is $\sqrt{1+25+16}=\sqrt{42}$, so the unit bisector direction is $\dfrac{\bar i+5\bar j+4\bar k}{\sqrt{42}}$.
    7. Since $OC=\sqrt{42}$: $\overline{OC}=\sqrt{42}\cdot\dfrac{\bar i+5\bar j+4\bar k}{\sqrt{42}}=\bar i+5\bar j+4\bar k$.

    Common Mistakes

    • Using the vectors $\overline{OA},\overline{OB}$ directly (not normalized) to find the bisector direction — this only works if the two vectors have equal magnitude.

    ✓Final answer

    The correct option is (B) — $\bar{i} + 5\bar{j} + 4\bar{k}$.

    ANSWER: B

12. 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

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

    Note

    The internal-bisector direction $\vec{BA}+\vec{BC}=(0,-5,3)$ is not among the printed choices; the only bisector direction offered is $(2,-3,-5)$, which matches the official key.

    ✓Final answer

    Direction ratios $(2,-3,-5)$ — option (D).