Q.Write all the unit vectors in XY-plane.

Showing the 12 most recent of 20 on this concept.

1. 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

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

   which is parallel to $\hat{i}-\hat{j}-\hat{k}$. Its magnitude is $\sqrt{1+1+1}=\sqrt3$.

   Check: $(1,-1,-1)\cdot(1,-2,3)=1+2-3=0$, so it is indeed perpendicular to the given vector.

   The unit vector is $\pm\dfrac{1}{\sqrt3}(\hat{i}-\hat{j}-\hat{k})$.

   ✓Final answer

   $\pm\dfrac{1}{\sqrt3}\left(\hat{i}-\hat{j}-\hat{k}\right)$ — option (D).

2. 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 Unit vector along $\overrightarrow{DE}$ is

   $\frac{\overrightarrow{DE}}{|\overrightarrow{DE}|} = \frac{-\frac{1}{2}\vec{i} - \frac{3}{2}\vec{j} + \vec{k}}{\frac{\sqrt{14}}{2}} = \frac{-\vec{i} - 3\vec{j} + 2\vec{k}}{\sqrt{14}}.$

   This matches option (B).

   Watch out

   A common mistake is to compute $\overrightarrow{DE}$ as $\frac{1}{2}\overrightarrow{AB}$ instead of $\frac{1}{2}\overrightarrow{BA}$, which would give the opposite direction. Always check which side DE is parallel to: D and E are midpoints of BC and CA, so DE is parallel to BA, not AB.

   Tip

   You can also find D and E explicitly:

   $D = \frac{\vec{B}+\vec{C}}{2}$, $E = \frac{\vec{C}+\vec{A}}{2}$, then $\overrightarrow{DE} = E - D = \frac{\vec{A} - \vec{B}}{2} = -\frac{1}{2}\overrightarrow{AB}$, confirming the shortcut.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

3. 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 $|\alpha|=\dfrac{1}{\sqrt{14}}$.

   Required dot product. $\vec a\cdot\vec b=(2)(0)+(-1)(2)+(0)(-1)=-2$ and $|\vec b|^2=0+4+1=5$, so

   $\vec d\cdot\vec b=\alpha\big(\vec a\cdot\vec b-|\vec b|^2\big)=\alpha(-2-5)=-7\alpha.$

   Therefore

   $|\vec d\cdot\vec b|=7|\alpha|=\frac{7}{\sqrt{14}}=\sqrt{\frac{49}{14}}=\sqrt{\frac72}.$

   ✓Final answer

   $|\vec d\cdot\vec b|=\sqrt{\dfrac72}$ — option (D).

   ANSWER: D

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

      But none of the options have $\sqrt{14}$ in the numerator — they all have 1. So we must check: is $|\vec{P}|$ actually $\sqrt{14}$? Yes. But the options suggest the scaling factor is $\frac{1}{5\sqrt{3}}$, which would mean the length of the required vector is 1, not $|\vec{P}|$. Let’s re-read the question: “the vector in the direction of $\overrightarrow{AB}$ of length $|P|$ units”. Here $|P|$ likely means the magnitude of the position vector of A, which is $\sqrt{14}$. But the options all have denominator $5\sqrt{3}$ or similar, and numerator 1 — so the length of each option vector is 1, not $\sqrt{14}$.

      This is a mismatch. Let’s verify option (D): $\frac{1}{5\sqrt{3}}(\vec{i} + 5\vec{j} - 7\vec{k})$ has magnitude $\frac{1}{5\sqrt{3}} \cdot 5\sqrt{3} = 1$. So none of the options have length $\sqrt{14}$. The only interpretation that makes sense is that the question intends the unit vector in the direction of $\overrightarrow{AB}$, i.e., length 1, and the notation $|P|$ is a misprint or means something else. Given the options, the correct match is the unit vector we found.

      Tip

      When options don’t match your calculation, re-check the interpretation. Here, the only vector among the choices that points along $\vec{i}+5\vec{j}-7\vec{k}$ and has a simple unit magnitude is option (D).

   5. Match with options.

      Option (D): $\dfrac{1}{5\sqrt{3}}\left(\vec{i} + 5\vec{j} - 7\vec{k}\right)$ is exactly the unit vector along $\overrightarrow{AB}$. Options (A) and (C) have different coefficients, and (B) has a different direction vector.

   ✓Final answer

   The correct option is (D).

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

      The $\frac{1}{2}$ cancels top and bottom:

      $= \frac{-3\vec{i} - 6\vec{j} + 13\vec{k}}{\sqrt{214}}$.

      The negative signs just mean the direction from A to M points opposite to the positive axes — but a unit vector along the median can be taken in either sense. The options given all have positive coefficients, so we multiply by $-1$ to match the form:

      $\frac{3\vec{i} + 6\vec{j} - 13\vec{k}}{\sqrt{214}}$.

      Watch out

      Don’t forget that the median vector could point from A to M or from M to A. The question says “along the median drawn through A”, which usually means the direction from A toward the opposite side. Our computed vector had negative components; flipping signs gives the same line direction, just reversed. Always check which sign matches the options.

   6. Match with the given choices.

      Option (B) is $\frac{1}{\sqrt{214}}(3\vec{i} + 6\vec{j} - 13\vec{k})$ — exactly what we have.

   ✓Final answer

   The correct option is (B).

6. 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 $\theta$.

      Since $0 < \theta < \frac{\pi}{2}$, $\cos\theta$ is positive. So $\cos\theta = \frac{1}{2}$.

      That means $\theta = \frac{\pi}{3}$.

   4. Read off the direction cosines.

      They are $\left(\frac{1}{\sqrt{2}}, \frac{1}{2}, \frac{1}{2}\right)$.

   Watch out

   A common mistake is to assume the third angle is $\frac{\pi}{4}$ because $\frac{1}{\sqrt{2}}$ appears twice in the options. But the identity forces $\cos\theta = \frac{1}{2}$, not $\frac{1}{\sqrt{2}}$. Always check the sum of squares.

   Tip

   You don’t need to find $\theta$ explicitly — just compute $\cos\theta$ from the identity. The angle itself is secondary; the direction cosine is what matters.

   ✓Final answer

   The correct option is (A).

7. 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 $\cos \alpha$ The angle $\alpha$ is between L and the positive X-axis. The X-axis direction vector is $\hat{i} = (1, 0, 0)$. Using the dot product formula:

   $\cos \alpha = \frac{|\vec{d} \cdot \hat{i}|}{|\vec{d}| \, |\hat{i}|}$

   Here $\vec{d} \cdot \hat{i} = (1)(1) + (-1)(0) + (1)(0) = 1$.

   The magnitude $|\vec{d}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$.

   And $|\hat{i}| = 1$.

   So:

   $\cos \alpha = \frac{1}{\sqrt{3}}$

   Watch out

   A common mistake is to forget that the angle with the X-axis uses the absolute value of the dot product when the direction could be reversed. Here the dot product is positive anyway, so it doesn’t matter — but always take the absolute value to get the acute angle.

   ✓Final answer

   The value is $\boxed{\frac{1}{\sqrt{3}}}$, which corresponds to option (A).

8. 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. The point $(16, -9, 0)$ lies on both planes, so its position vector is $\vec{r}_0 = 16\vec{i} - 9\vec{j} + 0\vec{k}$. The line is:

   $\vec{r} = \vec{r}_0 + \lambda \vec{d} = (16\vec{i} - 9\vec{j}) + \lambda(-7\vec{i} + 6\vec{j} - \vec{k}).$

   This expands to:

   $\vec{r} = (16 - 7\lambda)\vec{i} + (-9 + 6\lambda)\vec{j} + (-\lambda)\vec{k}.$

   That matches option (B) exactly: $(16 - 7\lambda)\vec{i} + (6\lambda - 9)\vec{j} - \lambda\vec{k}$.

   Watch out

   A common mistake is to take the direction vector as the cross product in the wrong order, or to forget that the point must satisfy both plane equations. Here the given point is already verified to lie on both, so it’s safe to use directly.

   ✓Final answer

   The correct option is (B).

9. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.$\vec{a} = \hat{i} + \hat{j} - 2\hat{k},\ \vec{b} = \hat{i} - 2\hat{j} + \hat{k}\ \text{and}\ \vec{c} = 2\hat{i} + \hat{j} - \hat{k}$ are three vectors. If $\vec{d}$ is a normal to the plane of $\vec{a}$ and $\vec{b}$ and $\vec{d} \cdot \vec{c} = 2$, then $|\vec{d}| =$ (A) $\sqrt{6}$ (B) $2\sqrt{3}$ (C) $\sqrt{3}$ (D) $2$

   ›Reveal solutionSolution

   $\vec d$ is parallel to $\vec a\times\vec b=-3(\hat i+\hat j+\hat k)$. Writing $\vec d=t(\vec a\times\vec b)$ and using $\vec d\cdot\vec c=2$ gives $\vec d=\hat i+\hat j+\hat k$, so $|\vec d|=\sqrt{3}$, option (C).

   Step 1 — Normal direction $\vec a\times\vec b$.

   $\vec a\times\vec b=\begin{vmatrix}\hat i&\hat j&\hat k\\ 1&1&-2\\ 1&-2&1\end{vmatrix} =\hat i(1-4)-\hat j(1+2)+\hat k(-2-1)=-3\hat i-3\hat j-3\hat k.$

   Step 2 — Impose the dot-product condition.

   Since $\vec d$ is normal to the plane of $\vec a$ and $\vec b$, $\ \vec d=t(\vec a\times\vec b)$. With $\vec c=2\hat i+\hat j-\hat k$:

   $\vec d\cdot\vec c=t\big[(-3)(2)+(-3)(1)+(-3)(-1)\big]=t(-6)=2\ \Rightarrow\ t=-\tfrac{1}{3}.$

   $\vec d=-\tfrac{1}{3}(-3\hat i-3\hat j-3\hat k)=\hat i+\hat j+\hat k.$

   Step 3 — Magnitude.

   $|\vec d|=\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}.$

   (Check: $\vec d\cdot\vec c=2+1-1=2$.)

   ✓Final answer

   $|\vec d|=\sqrt{3}$ — option (C).

   ANSWER: C

10. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.$\vec{r}$ is a vector perpendicular to the plane determined by the vectors $2\vec{i} - \vec{j}$ and $\vec{j} + 2\vec{k}$. If the magnitude of the projection of $\vec{r}$ on the vector $2\vec{i} + \vec{j} + 2\vec{k}$ is $1$, then $|\vec{r}|=$ (A) $\sqrt{6}$ (B) $3\sqrt{6}$ (C) $\dfrac{2\sqrt{6}}{3}$ (D) $\dfrac{3\sqrt{6}}{2}$

    ›Reveal solutionSolution

    The vector $\vec{r}$ is perpendicular to the plane of $2\vec{i}-\vec{j}$ and $\vec{j}+2\vec{k}$, so it is parallel to their cross product. Using the projection condition, we find $|\vec{r}| = \frac{3\sqrt{6}}{2}$, which corresponds to option (D).

    Concept & Intuition

    When a vector is perpendicular to a plane determined by two given vectors, it must be parallel to the cross product of those two vectors. That gives us the direction of $\vec{r}$ up to a scalar multiple. Then the condition about the projection onto another vector lets us solve for the magnitude.

    ---

    1. Find a direction vector for $\vec{r}$ The plane is spanned by $\vec{a} = 2\vec{i} - \vec{j}$ and $\vec{b} = \vec{j} + 2\vec{k}$. A vector perpendicular to both is their cross product:

    $\vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & -1 & 0 \\ 0 & 1 & 2 \end{vmatrix} = \vec{i}((-1)(2) - (0)(1)) - \vec{j}((2)(2) - (0)(0)) + \vec{k}((2)(1) - (-1)(0))$

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

    So $\vec{r}$ is parallel to $-2\vec{i} - 4\vec{j} + 2\vec{k}$, or equivalently to $\vec{i} + 2\vec{j} - \vec{k}$ (dividing by $-2$).

    Hence we can write $\vec{r} = \lambda (\vec{i} + 2\vec{j} - \vec{k})$ for some scalar $\lambda$.

    2. Use the projection condition The projection of $\vec{r}$ onto $\vec{c} = 2\vec{i} + \vec{j} + 2\vec{k}$ has magnitude $1$. The formula for the magnitude of the projection is:

    $\frac{|\vec{r} \cdot \vec{c}|}{|\vec{c}|} = 1.$

    Compute $\vec{r} \cdot \vec{c}$:

    $\vec{r} \cdot \vec{c} = \lambda (1\cdot2 + 2\cdot1 + (-1)\cdot2) = \lambda (2 + 2 - 2) = 2\lambda.$

    Compute $|\vec{c}|$:

    $|\vec{c}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4+1+4} = \sqrt{9} = 3.$

    So the condition becomes:

    $\frac{|2\lambda|}{3} = 1 \quad\Rightarrow\quad |\lambda| = \frac{3}{2}.$

    3. Find $|\vec{r}|$ Since $\vec{r} = \lambda (\vec{i} + 2\vec{j} - \vec{k})$, its magnitude is:

    $|\vec{r}| = |\lambda| \cdot \sqrt{1^2 + 2^2 + (-1)^2} = \frac{3}{2} \cdot \sqrt{1+4+1} = \frac{3}{2} \cdot \sqrt{6}.$

    That is $\dfrac{3\sqrt{6}}{2}$.

    Watch out

    A common mistake is to forget the absolute value in the projection formula, or to use the vector projection instead of its magnitude. Also, simplifying the cross product direction incorrectly can lead to a wrong scalar factor.

    Tip

    You could also keep the cross product as $-2\vec{i} - 4\vec{j} + 2\vec{k}$ and set $\vec{r} = \mu(-2\vec{i} - 4\vec{j} + 2\vec{k})$; the algebra yields the same final magnitude.

    ✓Final answer

    The correct option is (D).

    ANSWER: D

11. 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:

    $\sqrt{13^2 + 4^2 + 5^2} = \sqrt{169 + 16 + 25} = \sqrt{210}$

    Hence the direction cosines are:

    $\left(\frac{13}{\sqrt{210}}, \frac{4}{\sqrt{210}}, \frac{5}{\sqrt{210}}\right)$

    Watch out

    A common mistake is to add the raw direction ratios $(1, -2, 2)$ and $(2, 6, -3)$ directly. That gives $(3, 4, -1)$, which is not the bisector — it ignores the different lengths of the original vectors. Always convert to unit vectors first.

    Tip

    If the two lines have equal magnitudes, you can add the raw direction ratios directly. But here the magnitudes are 3 and 7, so normalising is essential.

    ✓Final answer

    The direction cosines of the bisector are $\left(\dfrac{13}{\sqrt{210}}, \dfrac{4}{\sqrt{210}}, \dfrac{5}{\sqrt{210}}\right)$, which matches option (C).

12. 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.

       You can also verify using the dot product:

       $\overrightarrow{AB} \cdot \overrightarrow{CD} = (4)(-8) + (7)(-14) + (-6)(12) = -32 - 98 - 72 = -202$

       $|\overrightarrow{AB}| = \sqrt{4^2 + 7^2 + (-6)^2} = \sqrt{16 + 49 + 36} = \sqrt{101}$

       $|\overrightarrow{CD}| = \sqrt{(-8)^2 + (-14)^2 + 12^2} = \sqrt{64 + 196 + 144} = \sqrt{404} = 2\sqrt{101}$

       Then $\cos\theta = \frac{-202}{(\sqrt{101})(2\sqrt{101})} = \frac{-202}{2 \cdot 101} = \frac{-202}{202} = -1$, so $\theta = \pi$.

    Watch out

    A common mistake is to forget that the angle between vectors is defined as the smaller angle between their directions when placed tail-to-tail. Here, since they are exactly opposite, the angle is $\pi$, not $0$.

    ✓Final answer

    The angle between $\overrightarrow{AB}$ and $\overrightarrow{CD}$ is $\pi$, so the correct option is (D).