Q.Find the unit vector in the direction of the vector $\vec{a} = \hat{i} + \hat{j} + 2\hat{k}$.

Showing the 12 most recent of 20 on this concept.

1. 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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2. 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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3. 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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4. 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 $\vec c=2\hat i+\hat j-\hat k$: …

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5. 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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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 $\theta$. …

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7. 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:

   $$$$

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8. 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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9. 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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10. 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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11. 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).

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    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.

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    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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12. 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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