Q.The vector having initial and terminal points as $(2, 5, 0)$ and $(-3, 7, 4)$, respectively is
(A) $-\hat{i}+12\hat{j}+4\hat{k}$
(B) $5\hat{i}+2\hat{j}-4\hat{k}$
(C) $-5\hat{i}+2\hat{j}+4\hat{k}$
(D) $\hat{i}+\hat{j}+\hat{k}$

Showing the 12 most recent of 20 on this concept.

1. TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
   Q.If $\vec{r} \cdot (2\vec{i} + 3\vec{j} + 4\vec{k}) = 5$, $\vec{r} \cdot (\vec{i} + \vec{j} - \vec{k}) = 7$ are two planes and $(16, -9, 0)$ is a point common to both the planes then the vector equation of the line of intersection of the planes is $\vec{r} =$ (A) $(16 + 7\lambda)\vec{i} + (6\lambda + 9)\vec{j} + \lambda\vec{k}$ (B) $(16 - 7\lambda)\vec{i} + (6\lambda - 9)\vec{j} - \lambda\vec{k}$ (C) $16\vec{i} - 9\vec{j} + \lambda(\vec{i} - 7\vec{j} + 6\vec{k})$ (D) $16\vec{i} - 9\vec{j} + \lambda(6\vec{i} - \vec{j} - 7\vec{k})$

   ›Reveal solutionSolution

   The line of intersection of two planes is found by taking a known common point and adding a scalar multiple of the direction vector perpendicular to both normals. The correct option is (B).

   The key idea: two non-parallel planes intersect in a straight line. To write its vector equation, you need one point on the line (given) and the direction vector of the line. The direction vector must be perpendicular to the normal vectors of both planes — so it is parallel to the cross product of the two normals.

   Let’s work through it.

   1. Identify the normal vectors.

      The first plane is $\vec{r} \cdot (2\vec{i} + 3\vec{j} + 4\vec{k}) = 5$, so its normal is $\vec{n}_1 = 2\vec{i} + 3\vec{j} + 4\vec{k}$.

      The second plane is $\vec{r} \cdot (\vec{i} + \vec{j} - \vec{k}) = 7$, so its normal is $\vec{n}_2 = \vec{i} + \vec{j} - \vec{k}$.

   2. Find the direction vector of the line of intersection.

      The line lies in both planes, so its direction $\vec{d}$ must be perpendicular to both normals. Hence $\vec{d} = \vec{n}_1 \times \vec{n}_2$.

      Compute the cross product:

   $\vec{d} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 3 & 4 \\ 1 & 1 & -1 \end{vmatrix} = \vec{i}(3 \cdot (-1) - 4 \cdot 1) - \vec{j}(2 \cdot (-1) - 4 \cdot 1) + \vec{k}(2 \cdot 1 - 3 \cdot 1)$

   $= \vec{i}(-3 - 4) - \vec{j}(-2 - 4) + \vec{k}(2 - 3) = -7\vec{i} + 6\vec{j} - \vec{k}.$

   So $\vec{d} = -7\vec{i} + 6\vec{j} - \vec{k}$.

   Tip

   You can also take any scalar multiple of $\vec{d}$ as the direction. Here, multiplying by $-1$ gives $7\vec{i} - 6\vec{j} + \vec{k}$, which is equally valid — just check which option matches.

   3. Write the vector equation using the given point. …

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2. TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
   Q.If $2\vec{i} + 4\vec{j} - 5\vec{k},\ \vec{i} + \vec{j} + \vec{k},\ \vec{j} + 2\vec{k}$ are the position vectors of the vertices A, B, C of a triangle respectively, then a unit vector along the median drawn through the vertex A is (A) $\dfrac{1}{\sqrt{174}} (5\vec{i} + 10\vec{j} - 7\vec{k})$ (B) $\dfrac{1}{\sqrt{214}} (3\vec{i} + 6\vec{j} - 13\vec{k})$ (C) $\dfrac{1}{\sqrt{66}} (\vec{i} + \vec{j} - 8\vec{k})$ (D) $\dfrac{1}{7} (3\vec{i} + 6\vec{j} - 2\vec{k})$

   ›Reveal solutionSolution

   The median from A goes to the midpoint of BC. Find that midpoint, subtract A’s position vector to get the median vector, then divide by its magnitude to get the unit vector. The result matches option (A).

   The key idea: a median in a triangle joins a vertex to the midpoint of the opposite side. So the median through A goes from A to the midpoint of BC. Once we have that vector, making it a unit vector is just a matter of dividing by its length.

   Let’s work it step by step.

   1. Write the given position vectors clearly.

      $A = 2\vec{i} + 4\vec{j} - 5\vec{k}$

      $B = \vec{i} + \vec{j} + \vec{k}$

      $C = \vec{j} + 2\vec{k}$

   2. Find the midpoint M of BC.

      The midpoint’s position vector is the average of B and C:

      $M = \frac{B + C}{2} = \frac{(\vec{i} + \vec{j} + \vec{k}) + (0\vec{i} + \vec{j} + 2\vec{k})}{2}$

      Notice C has no $\vec{i}$ component, so it’s $0\vec{i} + \vec{j} + 2\vec{k}$.

      Adding: $B + C = (1+0)\vec{i} + (1+1)\vec{j} + (1+2)\vec{k} = \vec{i} + 2\vec{j} + 3\vec{k}$

      Hence $M = \frac{1}{2}\vec{i} + \vec{j} + \frac{3}{2}\vec{k}$.

   3. Get the median vector from A to M.

      The vector along the median (from A to M) is $\overrightarrow{AM} = M - A$.

      $M - A = \left(\frac{1}{2} - 2\right)\vec{i} + (1 - 4)\vec{j} + \left(\frac{3}{2} + 5\right)\vec{k}$

      Compute each:

      $\frac{1}{2} - 2 = -\frac{3}{2}$

      $1 - 4 = -3$

      $\frac{3}{2} + 5 = \frac{3}{2} + \frac{10}{2} = \frac{13}{2}$

      So $\overrightarrow{AM} = -\frac{3}{2}\vec{i} - 3\vec{j} + \frac{13}{2}\vec{k}$.

      Tip

      To avoid fractions, multiply the whole vector by 2: $2\overrightarrow{AM} = -3\vec{i} - 6\vec{j} + 13\vec{k}$. We can work with this scaled version and adjust at the end — just remember to divide the magnitude by 2 as well.

   4. Find the magnitude of $\overrightarrow{AM}$.

      Using the scaled vector: $|2\overrightarrow{AM}| = \sqrt{(-3)^2 + (-6)^2 + (13)^2} = \sqrt{9 + 36 + 169} = \sqrt{214}$.

      Therefore $|\overrightarrow{AM}| = \frac{\sqrt{214}}{2}$.

   5. Write the unit vector along the median.

      Unit vector = $\frac{\overrightarrow{AM}}{|\overrightarrow{AM}|} = \frac{-\frac{3}{2}\vec{i} - 3\vec{j} + \frac{13}{2}\vec{k}}{\frac{\sqrt{214}}{2}}$. …

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3. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.If $3\vec{i} - 5\vec{j} + 2\vec{k}$, $7\vec{i} + 2\vec{j} - 4\vec{k}$, $\vec{i} - 3\vec{j} + 4\vec{k}$ and $-7\vec{i} - 17\vec{j} + 16\vec{k}$ are position vectors of the points A, B, C and D respectively, then the angle between $\overrightarrow{AB}$ and $\overrightarrow{CD}$ is (A) $0^\circ$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{\pi}{2}$ (D) $\pi$

   ›Reveal solutionSolution

   The angle between two vectors is found using their dot product. Here, $\overrightarrow{AB}$ and $\overrightarrow{CD}$ turn out to be parallel and opposite, so the angle is $\pi$ (180°). The correct option is (D).

   The key idea: given position vectors of four points, you first find the vectors representing the directed segments $\overrightarrow{AB}$ and $\overrightarrow{CD}$ by subtracting the appropriate position vectors. Then the angle $\theta$ between them satisfies $\cos\theta = \frac{\overrightarrow{AB} \cdot \overrightarrow{CD}}{|\overrightarrow{AB}|\,|\overrightarrow{CD}|}$. If the dot product equals the negative of the product of the magnitudes, the vectors are anti-parallel and $\theta = \pi$.

   1. Find $\overrightarrow{AB}$.

      $\overrightarrow{AB} = \vec{B} - \vec{A}$

      $\vec{A} = 3\vec{i} - 5\vec{j} + 2\vec{k}$, $\vec{B} = 7\vec{i} + 2\vec{j} - 4\vec{k}$

      So $\overrightarrow{AB} = (7-3)\vec{i} + (2 - (-5))\vec{j} + (-4 - 2)\vec{k} = 4\vec{i} + 7\vec{j} - 6\vec{k}$.

   2. Find $\overrightarrow{CD}$.

      $\overrightarrow{CD} = \vec{D} - \vec{C}$

      $\vec{C} = \vec{i} - 3\vec{j} + 4\vec{k}$, $\vec{D} = -7\vec{i} - 17\vec{j} + 16\vec{k}$

      So $\overrightarrow{CD} = (-7 - 1)\vec{i} + (-17 - (-3))\vec{j} + (16 - 4)\vec{k} = -8\vec{i} - 14\vec{j} + 12\vec{k}$.

   3. Observe the relationship.

      Notice that $\overrightarrow{CD} = -2(4\vec{i} + 7\vec{j} - 6\vec{k}) = -2\,\overrightarrow{AB}$.

      This means $\overrightarrow{CD}$ is a scalar multiple of $\overrightarrow{AB}$ with a negative factor. Two vectors that are scalar multiples are collinear (parallel). A negative scalar means they point in exactly opposite directions.

   4. Determine the angle.

      For two vectors pointing in opposite directions, the angle between them is $180^\circ$, i.e., $\pi$ radians. …

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4. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.$\vec{i} - 2\vec{j} + \vec{k},\ 2\vec{i} + \vec{j} - \vec{k},\ \vec{i} - \vec{j} - 2\vec{k}$ are the position vectors of the vertices A, B, C of a triangle ABC respectively. If D and E are the mid points of BC and CA respectively, then the unit vector along $\overrightarrow{DE}$ is (A) $\dfrac{1}{7}(3\vec{i} - 2\vec{j} + 6\vec{k})$ (B) $\dfrac{1}{\sqrt{14}}(-\vec{i} - 3\vec{j} + 2\vec{k})$ (C) $\dfrac{1}{\sqrt{3}}(\vec{i} - \vec{j} - \vec{k})$ (D) $\dfrac{1}{13}(12\vec{i} + 3\vec{j} + 4\vec{k})$

   ›Reveal solutionSolution

   The key idea is that $\overrightarrow{DE}$ is half of $\overrightarrow{AB}$ (by the midsegment theorem in vector form). Computing $\overrightarrow{AB}$ from the given position vectors and halving it gives $\frac{1}{2}(\vec{i} - 3\vec{j} + 2\vec{k})$, whose unit vector is $\frac{1}{\sqrt{14}}(-\vec{i} - 3\vec{j} + 2\vec{k})$, matching option (B).

   Concept and intuition:

   In any triangle, the segment joining the midpoints of two sides is parallel to the third side and half its length. Here, D is the midpoint of BC and E is the midpoint of CA, so DE is parallel to BA (or AB) and exactly half its length. Therefore, instead of finding D and E separately and subtracting, we can directly compute $\overrightarrow{DE} = \frac{1}{2} \overrightarrow{BA}$ (or $-\frac{1}{2}\overrightarrow{AB}$). This saves work and avoids sign errors. Then we just need the unit vector along that result.

   Step-by-step solution:

   1. Write the position vectors clearly Let

   $\vec{A} = \vec{i} - 2\vec{j} + \vec{k}, \quad \vec{B} = 2\vec{i} + \vec{j} - \vec{k}, \quad \vec{C} = \vec{i} - \vec{j} - 2\vec{k}.$

   2. Find $\overrightarrow{AB}$

   $\overrightarrow{AB} = \vec{B} - \vec{A} = (2-1)\vec{i} + (1 - (-2))\vec{j} + (-1 - 1)\vec{k} = \vec{i} + 3\vec{j} - 2\vec{k}.$

   3. Apply the midpoint theorem D is midpoint of BC, E is midpoint of CA. In vector geometry,

   $\overrightarrow{DE} = \frac{1}{2}\overrightarrow{BA} = -\frac{1}{2}\overrightarrow{AB}.$

   So

   $\overrightarrow{DE} = -\frac{1}{2}(\vec{i} + 3\vec{j} - 2\vec{k}) = -\frac{1}{2}\vec{i} - \frac{3}{2}\vec{j} + \vec{k}.$

   4. Find the magnitude of $\overrightarrow{DE}$

   $|\overrightarrow{DE}| = \sqrt{\left(-\frac{1}{2}\right)^2 + \left(-\frac{3}{2}\right)^2 + (1)^2} = \sqrt{\frac{1}{4} + \frac{9}{4} + 1} = \sqrt{\frac{1+9+4}{4}} = \sqrt{\frac{14}{4}} = \frac{\sqrt{14}}{2}.$

   5. Compute the unit vector …

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5. TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
   Q.If $\vec{P} = \vec{i} - 2\vec{j} + 3\vec{k},\ 2\vec{i} + 3\vec{j} - 4\vec{k},\ 4\vec{i} + 13\vec{j} - 18\vec{k}$ are the position vectors of three collinear points A, B, C respectively, then the vector in the direction of $\overrightarrow{AB}$ of length $|P|$ units is (A) $\dfrac{2}{5\sqrt{3}}\left(\vec{i} + 5\vec{j} - 7\vec{k}\right)$ (B) $\dfrac{1}{\sqrt{83}}\left(3\vec{i} + 5\vec{j} - 7\vec{k}\right)$ (C) $\dfrac{1}{\sqrt{78}}\left(2\vec{i} + 5\vec{j} - 7\vec{k}\right)$ (D) $\dfrac{1}{5\sqrt{3}}\left(\vec{i} + 5\vec{j} - 7\vec{k}\right)$

   ›Reveal solutionSolution

   The key idea is to find the unit vector along $\overrightarrow{AB}$ and scale it by $|\vec{P}|$; the correct option is (D).

   The problem gives three collinear points A, B, C with position vectors $\vec{P}$, $2\vec{i}+3\vec{j}-4\vec{k}$, and $4\vec{i}+13\vec{j}-18\vec{k}$ respectively. Wait — careful: $\vec{P}$ itself is the position vector of A, given as $\vec{i} - 2\vec{j} + 3\vec{k}$. So A, B, C are collinear, meaning vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are parallel. We need the vector in the direction of $\overrightarrow{AB}$ whose length equals $|\vec{P}|$.

   1. Find $\overrightarrow{AB}$ and $\overrightarrow{AC}$.

      $\overrightarrow{AB} = \vec{B} - \vec{A} = (2\vec{i}+3\vec{j}-4\vec{k}) - (\vec{i}-2\vec{j}+3\vec{k}) = \vec{i} + 5\vec{j} - 7\vec{k}$.

      $\overrightarrow{AC} = \vec{C} - \vec{A} = (4\vec{i}+13\vec{j}-18\vec{k}) - (\vec{i}-2\vec{j}+3\vec{k}) = 3\vec{i} + 15\vec{j} - 21\vec{k}$.

      Notice $\overrightarrow{AC} = 3(\vec{i} + 5\vec{j} - 7\vec{k}) = 3\overrightarrow{AB}$, confirming collinearity. So the direction of $\overrightarrow{AB}$ is given by the vector $\vec{i} + 5\vec{j} - 7\vec{k}$.

   2. Find the unit vector along $\overrightarrow{AB}$.

      Magnitude of $\overrightarrow{AB}$: $|\overrightarrow{AB}| = \sqrt{1^2 + 5^2 + (-7)^2} = \sqrt{1 + 25 + 49} = \sqrt{75} = 5\sqrt{3}$.

      So the unit vector is $\hat{u} = \frac{\vec{i} + 5\vec{j} - 7\vec{k}}{5\sqrt{3}}$.

   3. Find $|\vec{P}|$.

      $\vec{P} = \vec{i} - 2\vec{j} + 3\vec{k}$, so $|\vec{P}| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$.

      Watch out

      A common mistake is to confuse $\vec{P}$ (the position vector of A) with the vector we need to scale. The required vector has length $|\vec{P}|$, not $\vec{P}$ itself.

   4. Scale the unit vector by $|\vec{P}|$.

      The required vector = $|\vec{P}| \cdot \hat{u} = \sqrt{14} \cdot \frac{\vec{i} + 5\vec{j} - 7\vec{k}}{5\sqrt{3}} = \frac{\sqrt{14}}{5\sqrt{3}} (\vec{i} + 5\vec{j} - 7\vec{k})$. …

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6. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.The unit vector perpendicular to the vector $\vec{i} - 2\vec{j} + 3\vec{k}$ and coplanar with the vectors $\vec{i} + \vec{j} + \vec{k}$ and $2\vec{i} - \vec{j} - \vec{k}$ is (A) $\pm \dfrac{1}{\sqrt{5}} (2\vec{i} + \vec{j})$ (B) $\pm \dfrac{1}{4\sqrt{5}} (3\vec{i} - 6\vec{j} - 5\vec{k})$ (C) $\pm \dfrac{1}{\sqrt{6}} (\vec{i} + 2\vec{j} + \vec{k})$ (D) $\pm \dfrac{1}{\sqrt{3}} (\vec{i} - \vec{j} - \vec{k})$

   ›Reveal solutionSolution

   Write the required vector as $\vec{u}+\lambda\vec{v}$; perpendicularity gives $\lambda=-2$, yielding $(-3,3,3)\parallel(1,-1,-1)$, so the unit vector is $\pm\tfrac{1}{\sqrt3}(\hat{i}-\hat{j}-\hat{k})$ — option (D).

   A vector coplanar with $\vec{u}=\hat{i}+\hat{j}+\hat{k}$ and $\vec{v}=2\hat{i}-\hat{j}-\hat{k}$ can be written

   $\vec{r}=\vec{u}+\lambda\vec{v}=(1+2\lambda)\hat{i}+(1-\lambda)\hat{j}+(1-\lambda)\hat{k}.$

   It must be perpendicular to $\vec{w}=\hat{i}-2\hat{j}+3\hat{k}$, so $\vec{r}\cdot\vec{w}=0$:

   $(1+2\lambda)(1)+(1-\lambda)(-2)+(1-\lambda)(3)=0.$

   $1+2\lambda-2+2\lambda+3-3\lambda=2+\lambda=0\;\Rightarrow\;\lambda=-2.$

   Then

   $\vec{r}=(1-4)\hat{i}+(1+2)\hat{j}+(1+2)\hat{k}=-3\hat{i}+3\hat{j}+3\hat{k},$ …

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7. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.$\vec{a} = 2\hat{i} - \hat{j},\ \vec{b} = 2\hat{j} - \hat{k},\ \vec{c} = 2\hat{k} - \hat{i}$ are three vectors and $\vec{d}$ is a unit vector perpendicular to $\vec{c}$. If $\vec{a}, \vec{b}, \vec{d}$ are coplanar vectors, then $|\vec{d} \cdot \vec{b}| =$ (A) $0$ (B) $\dfrac{1}{\sqrt{14}}$ (C) $\sqrt{\dfrac{2}{7}}$ (D) $\sqrt{\dfrac{7}{2}}$

   ›Reveal solutionSolution

   Coplanarity together with $\vec d\perp\vec c$ forces $\vec d\parallel(\vec a-\vec b)$; normalising and dotting with $\vec b$ gives $|\vec d\cdot\vec b|=\dfrac{7}{\sqrt{14}}=\sqrt{\dfrac72}$ — option (D).

   Coplanarity. $\vec a,\vec b,\vec d$ coplanar means $\vec d=\alpha\vec a+\beta\vec b$.

   Perpendicular to $\vec c=2\hat k-\hat i$. Here $\vec a\cdot\vec c=(2)(-1)+(-1)(0)+(0)(2)=-2$ and $\vec b\cdot\vec c=(0)(-1)+(2)(0)+(-1)(2)=-2$, so

   $\vec d\cdot\vec c=\alpha(-2)+\beta(-2)=0\ \Rightarrow\ \alpha+\beta=0,\qquad \vec d=\alpha(\vec a-\vec b).$

   Unit length. $\vec a-\vec b=2\hat i-3\hat j+\hat k$, so $|\vec a-\vec b|=\sqrt{4+9+1}=\sqrt{14}$ and $|\alpha|=\dfrac{1}{\sqrt{14}}$. …

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8. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
   Q.Let $\vec{a} = 2\vec{i} - \vec{j} + \vec{k}$ be the position vector of a point A. Let $\vec{b} = \vec{i} + 2\vec{j} - \vec{k}$ and $\vec{c} = \vec{i} + \vec{j} - 2\vec{k}$ be two vectors and $\vec{r}$ be a vector passing through the point $A(a)$ and parallel to the vector $\vec{b}$. If the projection of $\vec{r}$ on $\vec{c}$ is $\dfrac{9}{\sqrt{6}}$ then $|\vec{r}| =$ (A) $\sqrt{26}$ (B) $5$ (C) $\sqrt{5}$ (D) $\sqrt{34}$

   ›Reveal solutionSolution

   The vector $\vec{r}$ is a scalar multiple of $\vec{b}$ (since it’s parallel to $\vec{b}$) and passes through $A$. Using the given projection onto $\vec{c}$, we solve for the scalar and then compute $|\vec{r}|$, which turns out to be $\sqrt{26}$.

   We are told $\vec{r}$ passes through point $A$ (with position vector $\vec{a}$) and is parallel to $\vec{b}$. That means $\vec{r}$ is of the form

   $\vec{r} = \vec{a} + \lambda \vec{b}$

   for some scalar $\lambda$. The projection of $\vec{r}$ onto $\vec{c}$ is given as $\frac{9}{\sqrt{6}}$. The projection formula is

   $\text{proj}_{\vec{c}} \vec{r} = \frac{\vec{r} \cdot \vec{c}}{|\vec{c}|}.$

   We can set up an equation to find $\lambda$, then compute $|\vec{r}|$.

   1. Write $\vec{r}$ explicitly

   $\vec{a} = 2\vec{i} - \vec{j} + \vec{k}, \quad \vec{b} = \vec{i} + 2\vec{j} - \vec{k}$

   So

   $\vec{r} = (2 + \lambda)\vec{i} + (-1 + 2\lambda)\vec{j} + (1 - \lambda)\vec{k}.$

   2. Compute the dot product $\vec{r} \cdot \vec{c}$ $\vec{c} = \vec{i} + \vec{j} - 2\vec{k}$, so

   $\vec{r} \cdot \vec{c} = (2+\lambda)(1) + (-1+2\lambda)(1) + (1-\lambda)(-2)$

   Simplify:

   $= 2+\lambda -1 + 2\lambda -2 + 2\lambda$

   $= (2-1-2) + (\lambda + 2\lambda + 2\lambda) = -1 + 5\lambda.$

   3. Find $|\vec{c}|$

   $|\vec{c}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1+1+4} = \sqrt{6}.$

   4. Use the projection condition

   $\frac{\vec{r} \cdot \vec{c}}{|\vec{c}|} = \frac{-1 + 5\lambda}{\sqrt{6}} = \frac{9}{\sqrt{6}}.$

   Multiply both sides by $\sqrt{6}$:

   $-1 + 5\lambda = 9 \quad \Rightarrow \quad 5\lambda = 10 \quad \Rightarrow \quad \lambda = 2.$ …

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9. TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
   Q.Let $A(2, 3, -1), B(4, 1, 0), C(-1, -1, 11)$ be the vertices of a triangle $ABC$. Let $D$ be the point where the bisector of $\angle BAC$ meet the side $BC$. Then the direction ratios of $AD$ are (A) $(35, -19, 49)$ (B) $(17, -14, 49)$ (C) $(17, -38, 49)$ (D) $(17, -38, 23)$

   ›Reveal solutionSolution

   The angle bisector theorem in 3D gives the ratio in which $D$ divides $BC$ using the lengths of the adjacent sides $AB$ and $AC$. Computing these distances and applying the section formula yields the coordinates of $D$, from which the direction ratios of $AD$ are found to be $(17, -38, 49)$.

   The key idea is that the internal angle bisector of $\angle BAC$ meets the opposite side $BC$ at a point $D$ that divides $BC$ in the ratio of the lengths of the adjacent sides: $BD : DC = AB : AC$. This is the angle bisector theorem, and it works in 3D just as it does in 2D because it’s a purely metric property — it depends only on distances, not on the dimension of the space.

   Once we know the ratio, we can find $D$ using the section formula for a point dividing a line segment internally. Then the direction ratios of $AD$ are simply the differences of the coordinates of $D$ and $A$.

   Let’s go step by step.

   1. Find the lengths $AB$ and $AC$. $A(2,3,-1)$, $B(4,1,0)$, $C(-1,-1,11)$.

   $AB = \sqrt{(4-2)^2 + (1-3)^2 + (0+1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3.$

   $AC = \sqrt{(-1-2)^2 + (-1-3)^2 + (11+1)^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13.$

   2. Apply the angle bisector theorem. Since $AD$ bisects $\angle BAC$, we have

   $\frac{BD}{DC} = \frac{AB}{AC} = \frac{3}{13}.$

   So $D$ divides $BC$ internally in the ratio $3 : 13$, with $B$ as the first point and $C$ as the second.

   3. Find the coordinates of $D$ using the section formula. For internal division in the ratio $m:n$, the coordinates are

   $\left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n}, \frac{m z_2 + n z_1}{m+n} \right),$

   where the point dividing $B(x_1,y_1,z_1)$ and $C(x_2,y_2,z_2)$ in the ratio $m:n$ from $B$ to $C$.

   Here $m=3$, $n=13$, $B(4,1,0)$, $C(-1,-1,11)$.

   $x_D = \frac{3(-1) + 13(4)}{3+13} = \frac{-3 + 52}{16} = \frac{49}{16},$

   $y_D = \frac{3(-1) + 13(1)}{16} = \frac{-3 + 13}{16} = \frac{10}{16} = \frac{5}{8},$ …

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10. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
    Q.If $(1, -2, 2)$ and $(2, 6, -3)$ are the direction ratios of two straight lines then the direction cosines of the line bisecting an angle between these two lines are (A) $\left(\dfrac{1}{\sqrt{41}}, \dfrac{4}{\sqrt{41}}, \dfrac{5}{\sqrt{41}}\right)$ (B) $\left(\dfrac{13}{\sqrt{1218}}, \dfrac{32}{\sqrt{1218}}, \dfrac{5}{\sqrt{1218}}\right)$ (C) $\left(\dfrac{13}{\sqrt{210}}, \dfrac{4}{\sqrt{210}}, \dfrac{5}{\sqrt{210}}\right)$ (D) $\left(\dfrac{13}{\sqrt{714}}, \dfrac{4}{\sqrt{714}}, \dfrac{23}{\sqrt{714}}\right)$

    ›Reveal solutionSolution

    The direction cosines of the angle bisector are found by normalising the sum of the unit vectors along the two given lines. The correct answer is option (B).

    The key idea is simple: if you have two lines through the origin, the line that bisects the angle between them points in the direction of the sum of the unit vectors along the two lines. This works because adding two equal-length vectors gives a resultant that lies exactly halfway between them — like the diagonal of a rhombus.

    Let’s apply this cleanly.

    1. Find the unit vectors along each line. The given direction ratios are $(1, -2, 2)$ and $(2, 6, -3)$. Their magnitudes are:

    $|\vec{a}| = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$

    $|\vec{b}| = \sqrt{2^2 + 6^2 + (-3)^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7$

    So the unit vectors are:

    $\hat{a} = \left(\frac{1}{3}, -\frac{2}{3}, \frac{2}{3}\right), \quad \hat{b} = \left(\frac{2}{7}, \frac{6}{7}, -\frac{3}{7}\right)$

    2. Add the unit vectors to get the bisector direction. The bisector’s direction ratios are proportional to $\hat{a} + \hat{b}$:

    $\hat{a} + \hat{b} = \left(\frac{1}{3} + \frac{2}{7},\; -\frac{2}{3} + \frac{6}{7},\; \frac{2}{3} - \frac{3}{7}\right)$

    Compute each component:

    • First: $\frac{1}{3} + \frac{2}{7} = \frac{7}{21} + \frac{6}{21} = \frac{13}{21}$
    • Second: $-\frac{2}{3} + \frac{6}{7} = -\frac{14}{21} + \frac{18}{21} = \frac{4}{21}$
    • Third: $\frac{2}{3} - \frac{3}{7} = \frac{14}{21} - \frac{9}{21} = \frac{5}{21}$ So the bisector direction ratios are $\left(\frac{13}{21}, \frac{4}{21}, \frac{5}{21}\right)$, which is proportional to $(13, 4, 5)$.

    3. Normalise to get direction cosines. The magnitude of $(13, 4, 5)$ is: $$$$

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11. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
    Q.The direction cosines of the line making angles $\frac{\pi}{4}, \frac{\pi}{3}$ and $\theta \left(0 < \theta < \frac{\pi}{2}\right)$ respectively with $x, y$ and $z$ axes, are (A) $\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}$ (B) $\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{\sqrt{3}}{2}$ (C) $\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{\sqrt{2}}$ (D) $\frac{1}{\sqrt{2}}, \frac{\sqrt{3}}{2}, \frac{1}{\sqrt{2}}$

    ›Reveal solutionSolution

    The direction cosines are the cosines of the angles a line makes with the axes. Using the identity $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$, we find $\theta = \frac{\pi}{3}$, so the direction cosines are $\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}$ — option (A).

    The key idea is simple: direction cosines are literally the cosines of the angles the line makes with the $x$, $y$, and $z$ axes. If those angles are $\alpha$, $\beta$, and $\gamma$, then the direction cosines are $l = \cos\alpha$, $m = \cos\beta$, $n = \cos\gamma$.

    There’s a fundamental constraint: for any line in 3D space, the sum of the squares of its direction cosines is always exactly 1. That’s because they represent the components of a unit vector along the line. So if we know two of the angles, the third is forced — we don’t need to guess it.

    Here we’re given $\alpha = \frac{\pi}{4}$, $\beta = \frac{\pi}{3}$, and $\gamma = \theta$ (with $0 < \theta < \frac{\pi}{2}$). Let’s find $\theta$ and then the direction cosines.

    1. Write the known cosines.

       $\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$

       $\cos\frac{\pi}{3} = \frac{1}{2}$

       So $l = \frac{1}{\sqrt{2}}$, $m = \frac{1}{2}$.

    2. Apply the identity.

       $l^2 + m^2 + n^2 = 1$

       $\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{2}\right)^2 + \cos^2\theta = 1$

       $\frac{1}{2} + \frac{1}{4} + \cos^2\theta = 1$

       $\frac{3}{4} + \cos^2\theta = 1$

       $\cos^2\theta = \frac{1}{4}$

    3. Find $\theta$. …

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12. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
    Q.If $A = (1, -1, 2)$, $B = (3, 4, -2)$, $C = (0, 3, 2)$ \text{ and } $D = (3, 5, 6)$ then the angle between the lines $\overrightarrow{AB}$ and $\overrightarrow{CD}$ is (A) $30^\circ$ (B) $45^\circ$ (C) $60^\circ$ (D) $90^\circ$

    ›Reveal solutionSolution

    The angle between two lines in space is found from the dot product of their direction vectors. For $\overrightarrow{AB}$ and $\overrightarrow{CD}$, the cosine of the angle is zero, so the lines are perpendicular — the answer is $90^\circ$.

    The question gives four points and asks for the angle between the lines $AB$ and $CD$. In 3D geometry, the angle between two lines is defined as the angle between their direction vectors. So the first step is always to find those vectors, then use the dot product relation:

    $\cos\theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}|\,|\vec{v}|}$

    where $\vec{u}$ and $\vec{v}$ are the direction vectors of the two lines. The angle $\theta$ is taken between $0^\circ$ and $180^\circ$, and for lines we usually report the acute angle.

    Let’s work through it.

    1. Find $\overrightarrow{AB}$.

       $\overrightarrow{AB} = B - A = (3-1,\; 4-(-1),\; -2-2) = (2,\; 5,\; -4)$.

    2. Find $\overrightarrow{CD}$.

       $\overrightarrow{CD} = D - C = (3-0,\; 5-3,\; 6-2) = (3,\; 2,\; 4)$.

    3. Compute the dot product.

       $\overrightarrow{AB} \cdot \overrightarrow{CD} = (2)(3) + (5)(2) + (-4)(4) = 6 + 10 - 16 = 0$.

       A dot product of zero means the vectors are perpendicular.

    4. Check magnitudes (optional, but confirms).

       $|\overrightarrow{AB}| = \sqrt{2^2 + 5^2 + (-4)^2} = \sqrt{4 + 25 + 16} = \sqrt{45}$

       $|\overrightarrow{CD}| = \sqrt{3^2 + 2^2 + 4^2} = \sqrt{9 + 4 + 16} = \sqrt{29}$

       Neither is zero, so the zero dot product genuinely means $\cos\theta = 0$, i.e. $\theta = 90^\circ$. …

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