Q.Find the cartesian equation of the line which passes through the point $(-2, 4, -5)$ and parallel to the line given by $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$.

Showing the 12 most recent of 20 on this concept.

1. 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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2. 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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3. 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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4. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.A line L is parallel to both the planes $2x+3y+z=1$ and $x+3y+2z=2$. If line L makes an angle $\alpha$ with the positive direction of 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

   A line parallel to two planes must be perpendicular to both normals, so its direction vector is along the cross product of the normals. The cosine of its angle with the X-axis is then $\frac{1}{\sqrt{3}}$, option (A).

   The key idea is simple: if a line is parallel to a plane, its direction vector is perpendicular to the plane’s normal vector. Since the line is parallel to both planes, its direction vector must be perpendicular to both normals at once. That means it lies along the cross product of the two normals.

   Once we have the direction vector, finding the cosine of the angle it makes with the X-axis is just a dot product with the unit vector along the X-axis, divided by the magnitude.

   Let’s work it through.

   1. Identify the normal vectors

      Plane 1: $2x + 3y + z = 1$ has normal $\vec{n}_1 = (2, 3, 1)$.

      Plane 2: $x + 3y + 2z = 2$ has normal $\vec{n}_2 = (1, 3, 2)$.

   2. Find a direction vector for line L

      Since L is parallel to both planes, its direction vector $\vec{d}$ is perpendicular to both $\vec{n}_1$ and $\vec{n}_2$. So $\vec{d}$ is parallel to $\vec{n}_1 \times \vec{n}_2$.

      Compute the cross product:

   $\vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 1 & 3 & 2 \end{vmatrix} = \hat{i}(3\cdot 2 - 1\cdot 3) - \hat{j}(2\cdot 2 - 1\cdot 1) + \hat{k}(2\cdot 3 - 3\cdot 1)$

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

   So $\vec{d} = (3, -3, 3)$, or any scalar multiple. We can simplify to $(1, -1, 1)$.

   3. Find $\cos \alpha$ …

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5. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.If L is the line of intersection of two planes $x+2y+2z=15$ and $x-y+z=4$ and the direction ratios of the line L are (a, b, c), then $\dfrac{a^2+b^2+c^2}{b^2}=$ (A) 14 (B) 10 (C) 22 (D) 26

   ›Reveal solutionSolution

   The direction ratios of the line of intersection of two planes are given by the cross product of their normals. For the planes $x+2y+2z=15$ and $x-y+z=4$, the cross product yields $(4,1,-3)$, so $a=4, b=1, c=-3$. Then $\frac{a^2+b^2+c^2}{b^2} = \frac{16+1+9}{1} = 26$. The correct option is (D).

   ---

   Concept & Intuition

   When two planes intersect, their line of intersection lies in both planes. That means the direction vector of the line must be perpendicular to both normal vectors of the planes. The simplest way to find a vector perpendicular to two given vectors is to take their cross product. So the direction ratios $(a,b,c)$ of the line are exactly the components of the cross product of the normals.

   ---

   Step-by-step solution

   1. Identify the normal vectors

      For the plane $x+2y+2z=15$, the normal vector is $\vec{n}_1 = (1,2,2)$.

      For the plane $x-y+z=4$, the normal vector is $\vec{n}_2 = (1,-1,1)$.

   2. Compute the cross product

      The direction vector of the line is

   $\vec{d} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 2 \\ 1 & -1 & 1 \end{vmatrix}$

   Expanding:

   $\vec{d} = \mathbf{i}(2\cdot1 - 2\cdot(-1)) - \mathbf{j}(1\cdot1 - 2\cdot1) + \mathbf{k}(1\cdot(-1) - 2\cdot1)$

   $= \mathbf{i}(2 + 2) - \mathbf{j}(1 - 2) + \mathbf{k}(-1 - 2)$

   $= (4, 1, -3)$

   So $a=4$, $b=1$, $c=-3$.

   3. Compute the required expression $$$$

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6. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If A$(0,3,4)$, B$(1,5,6)$, C$(-2,0,-2)$ are the vertices of a triangle ABC and the bisector of angle A meets the side BC at D, then AD = (A) $\dfrac{\sqrt{21}}{5}$ (B) $\dfrac{\sqrt{42}}{10}$ (C) $10$ (D) $4$

   ›Reveal solutionSolution

   The key idea is to use the Angle Bisector Theorem to find the coordinates of point D on BC, then compute the distance AD. The result is $\frac{\sqrt{42}}{10}$, so the correct option is (B).

   ---

   Concept and Intuition

   When a triangle’s internal angle at A is bisected, it meets the opposite side BC at a point D that divides BC in the ratio of the adjacent sides: $BD:DC = AB:AC$. This is the Angle Bisector Theorem. Once we know D’s coordinates (using section formula), we can directly compute the length AD using the distance formula. The trick is to avoid messy algebra by carefully computing the side lengths first.

   ---

   Step-by-step solution

   1. Find the side lengths AB and AC
      • $A(0,3,4)$, $B(1,5,6)$

   $AB = \sqrt{(1-0)^2 + (5-3)^2 + (6-4)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3$

   • $A(0,3,4)$, $C(-2,0,-2)$

   $AC = \sqrt{(-2-0)^2 + (0-3)^2 + (-2-4)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$

   2. Apply the Angle Bisector Theorem Since AD bisects $\angle A$, we have

   $\frac{BD}{DC} = \frac{AB}{AC} = \frac{3}{7}$

   So D divides BC internally in the ratio $3:7$ (from B to C).

   3. Find coordinates of D using section formula
      • $B(1,5,6)$, $C(-2,0,-2)$
      • For internal division in ratio $m:n = 3:7$,

   $D = \left( \frac{m x_C + n x_B}{m+n}, \frac{m y_C + n y_B}{m+n}, \frac{m z_C + n z_B}{m+n} \right)$

    Here $m=3$ (for C) and $n=7$ (for B), so  
   

   $x_D = \frac{3(-2) + 7(1)}{10} = \frac{-6 + 7}{10} = \frac{1}{10}$

   $y_D = \frac{3(0) + 7(5)}{10} = \frac{0 + 35}{10} = \frac{35}{10} = \frac{7}{2}$

   $z_D = \frac{3(-2) + 7(6)}{10} = \frac{-6 + 42}{10} = \frac{36}{10} = \frac{18}{5}$

   4. Compute the distance AD
      • $A(0,3,4)$, $D\left(\frac{1}{10}, \frac{7}{2}, \frac{18}{5}\right)$

   $AD^2 = \left(\frac{1}{10} - 0\right)^2 + \left(\frac{7}{2} - 3\right)^2 + \left(\frac{18}{5} - 4\right)^2$

    Simplify each term:  
    - $\left(\frac{1}{10}\right)^2 = \frac{1}{100}$ …
   

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7. 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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8. 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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9. 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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10. 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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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 2026Set eng-2026-05-11-AN1 markMCQ
    Q.$\alpha, \beta, \gamma \; (\alpha > \beta > \gamma)$ are roots of the equation $x^3 - x^2 - 4x + 4 = 0$. The volume of the parallelepiped whose coterminous edges are $\alpha \vec{i} + \beta \vec{j} + \gamma \vec{k}, \; \beta \vec{i} + \gamma \vec{j} + \alpha \vec{k}, \; \gamma \vec{i} + \alpha \vec{j} + \beta \vec{k}$ is (A) $13$ (B) $3$ (C) $\dfrac{15}{6}$ (D) $\dfrac{13}{6}$

    ›Reveal solutionSolution

    The volume is the absolute value of the determinant formed by the three vectors. Using the cubic’s roots and symmetric sums, the determinant simplifies to $(\alpha - \beta)(\beta - \gamma)(\gamma - \alpha)$, whose square is the discriminant of the cubic. Computing the discriminant gives $13$, so the volume is $\sqrt{13}$, but the problem asks for the volume (scalar triple product magnitude) — careful: the determinant itself equals $(\alpha - \beta)(\beta - \gamma)(\gamma - \alpha)$, and its absolute value is $\sqrt{13}$. However, the given options are all rational numbers; re-checking shows the determinant’s value is actually $(\alpha - \beta)(\beta - \gamma)(\gamma - \alpha) = \pm \sqrt{13}$, so the volume is $\sqrt{13}$, which is not among the options. Wait — the problem likely expects the scalar triple product (not its absolute value) as a rational number? Let’s re-evaluate: the determinant of the matrix of coefficients is $(\alpha + \beta + \gamma)(\alpha\beta + \beta\gamma + \gamma\alpha) - \alpha\beta\gamma - (\alpha^3 + \beta^3 + \gamma^3)$? No — better compute directly. The correct volume is $|(\alpha - \beta)(\beta - \gamma)(\gamma - \alpha)| = \sqrt{13}$, but none of the options match. So perhaps the volume is the absolute value of the determinant of the vectors as given, which simplifies to $(\alpha - \beta)(\beta - \gamma)(\gamma - \alpha)$ and its square is $13$, so the volume is $\sqrt{13}$. Since $\sqrt{13}$ is not listed, maybe the problem means the scalar triple product (signed volume) equals $(\alpha - \beta)(\beta - \gamma)(\gamma - \alpha) = \pm \sqrt{13}$? Still not rational. Let’s check the cubic: $x^3 - x^2 - 4x + 4 = (x-1)(x-2)(x+2)$? Indeed, $1^3 - 1 - 4 + 4 = 0$, $2^3 - 4 - 8 + 4 = 0$, $(-2)^3 - 4 + 8 + 4 = 0$. So roots are $2, 1, -2$ with $\alpha > \beta > \gamma$ gives $\alpha=2, \beta=1, \gamma=-2$. Then the vectors are $(2,1,-2)$, $(1,-2,2)$, $(-2,2,1)$. The scalar triple product is the determinant:

    $> \begin{vmatrix} > 2 & 1 & -2 \\ > 1 & -2 & 2 \\ > -2 & 2 & 1 > \end{vmatrix} > = 2(-2\cdot1 - 2\cdot2) - 1(1\cdot1 - 2\cdot(-2)) + (-2)(1\cdot2 - (-2)\cdot(-2)) >$

    Compute: $2(-2 - 4) = 2(-6) = -12$; $-1(1 + 4) = -5$; $+(-2)(2 - 4) = (-2)(-2) = 4$; sum = $-12 -5 +4 = -13$. Absolute value $13$. So volume is $13$. The correct option is (A).

    The volume of a parallelepiped is the absolute value of the scalar triple product of its edge vectors. For the given vectors, the determinant simplifies to $(\alpha - \beta)(\beta - \gamma)(\gamma - \alpha)$, and using the actual roots $2,1,-2$ of the cubic, this equals $13$. Hence the volume is $13$, option (A).

    ---

    Concept & Intuition

    The volume of a parallelepiped formed by three vectors $\vec{u}, \vec{v}, \vec{w}$ is $|\vec{u} \cdot (\vec{v} \times \vec{w})|$, which is the absolute value of the determinant whose rows (or columns) are the components of the vectors. So we need to compute:

    $V = \left| \det \begin{pmatrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{pmatrix} \right|.$

    The cubic $x^3 - x^2 - 4x + 4 = 0$ has roots $\alpha, \beta, \gamma$ (with $\alpha > \beta > \gamma$). Instead of solving the cubic immediately, we can use symmetric sums to simplify the determinant. But here the cubic factors nicely, so we can also find the exact roots. Let’s do both to see the elegance.

    ---

    Step-by-step solution

    1. Find the roots of the cubic. The equation is $x^3 - x^2 - 4x + 4 = 0$. Try $x=1$: $1 - 1 - 4 + 4 = 0$, so $x=1$ is a root. Factor out $(x-1)$:

    $x^3 - x^2 - 4x + 4 = (x-1)(x^2 - 4) = (x-1)(x-2)(x+2).$

    Hence the roots are $1, 2, -2$. Since $\alpha > \beta > \gamma$, we have $\alpha = 2$, $\beta = 1$, $\gamma = -2$.

    2. Write the three vectors explicitly.

    $\vec{u} = 2\vec{i} + 1\vec{j} + (-2)\vec{k} = (2,1,-2),$

    $\vec{v} = 1\vec{i} + (-2)\vec{j} + 2\vec{k} = (1,-2,2),$

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

    3. Compute the scalar triple product (determinant).

    $\det = \begin{vmatrix} 2 & 1 & -2 \\ 1 & -2 & 2 \\ -2 & 2 & 1 \end{vmatrix}.$

    Expand using the first row:

    $\det = 2 \cdot \begin{vmatrix} -2 & 2 \\ 2 & 1 \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & 2 \\ -2 & 1 \end{vmatrix} + (-2) \cdot \begin{vmatrix} 1 & -2 \\ -2 & 2 \end{vmatrix}.$

    Compute each minor:

    • First minor: $(-2)(1) - (2)(2) = -2 - 4 = -6$.
    • Second minor: $(1)(1) - (2)(-2) = 1 + 4 = 5$.
    • Third minor: $(1)(2) - (-2)(-2) = 2 - 4 = -2$.

    So:

    $\det = 2(-6) - 1(5) + (-2)(-2) = -12 - 5 + 4 = -13.$

    4. Volume is the absolute value.

    $V = |\det| = 13.$

    Tip

    If you prefer a symmetric approach: For any cubic with roots $\alpha,\beta,\gamma$, the determinant

    $$$$

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