Q.For given vectors, $\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k}$ and $\vec{b} = -\hat{i} + \hat{j} - \hat{k}$, find the unit vector in the direction of the vector $\vec{a} + \vec{b}$.

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 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If the line $\vec{r} = \vec{a} + t\vec{b}$ lies on the plane $\vec{r} \cdot \vec{n} = p$, then $p =$ (A) $|\vec{a} \times \vec{b}|$ (B) $\vec{a} \cdot \vec{n}$ (C) $|\vec{b} \times \vec{n}|$ (D) $|\vec{a}| + |\vec{b}|$

   ›Reveal solutionSolution

   If a line lies entirely in a plane, every point on the line must satisfy the plane equation. Substituting the line’s parametric form into the plane’s equation gives a condition that must hold for all $t$, leading to $p = \vec{a} \cdot \vec{n}$.

   The key idea is that a line is contained in a plane if and only if every point on the line satisfies the plane’s equation. The line is given in parametric vector form:

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

   where $\vec{a}$ is a point on the line and $\vec{b}$ is the direction vector. The plane is given by

   $\vec{r} \cdot \vec{n} = p$

   where $\vec{n}$ is the normal vector and $p$ is the constant (the distance from the origin times the magnitude of $\vec{n}$).

   For the line to lie in the plane, the equation must hold for all real $t$. That forces two conditions: one from the constant term and one from the coefficient of $t$.

   1. Substitute the line into the plane equation:

   $(\vec{a} + t\vec{b}) \cdot \vec{n} = p$

   This expands to:

   $\vec{a} \cdot \vec{n} + t (\vec{b} \cdot \vec{n}) = p$

   2. This must be true for every $t$.

      The left side is a linear function of $t$. For it to equal the constant $p$ for all $t$, the coefficient of $t$ must be zero, and the constant term must equal $p$.

      • Coefficient condition: $\vec{b} \cdot \vec{n} = 0$ (the line’s direction is perpendicular to the plane’s normal — i.e., the line is parallel to the plane).
      • Constant condition: $\vec{a} \cdot \vec{n} = p$.

   3. Interpretation: …

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3. 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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4. 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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5. 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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6. 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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7. 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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8. 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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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 $\vec c=2\hat i+\hat j-\hat k$: …

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10. 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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11. 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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12. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
    Q.A particle is travelling in clockwise direction on the ellipse $\dfrac{x^2}{100} + \dfrac{y^2}{25} = 1$. If the particle leaves the ellipse at the point $(-8, 3)$ on it and travels along the tangent to the ellipse at that point then the point where the particle crosses the Y-axis is (A) $\left(0, \dfrac{7}{3}\right)$ (B) $\left(0, \dfrac{25}{3}\right)$ (C) $(0, 9)$ (D) $\left(0, -\dfrac{25}{3}\right)$

    ›Reveal solutionSolution

    The particle leaves the ellipse along the tangent line at $(-8,3)$; the equation of that tangent is found using the standard ellipse tangent formula, and its y‑intercept gives the crossing point on the Y‑axis, which is $\left(0,\frac{25}{3}\right)$.

    We are given the ellipse $\frac{x^2}{100}+\frac{y^2}{25}=1$. A particle moving clockwise on it leaves at $(-8,3)$ and continues along the tangent line at that point. We need the point where this tangent line crosses the Y‑axis.

    Concept & Intuition

    For an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the tangent line at a point $(x_1,y_1)$ on it is $\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$. This is a direct, reliable formula derived from the fact that the ellipse is a quadratic curve and the tangent is the “linear approximation” at that point. Once we have the line’s equation, the Y‑axis crossing (x=0) is simply the y‑intercept.

    Step‑by‑step reasoning

    1. Identify parameters

       Here $a^2=100$ and $b^2=25$. The point is $(-8,3)$. Check it lies on the ellipse:

       $\frac{(-8)^2}{100}+\frac{3^2}{25}=\frac{64}{100}+\frac{9}{25}=0.64+0.36=1$. Good.

    2. Write the tangent line equation

       Using the formula $\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$:

    $\frac{x(-8)}{100}+\frac{y(3)}{25}=1$

    Simplify:

    $-\frac{8x}{100}+\frac{3y}{25}=1$

    Reduce $-\frac{8}{100}$ to $-\frac{2}{25}$:

    $-\frac{2x}{25}+\frac{3y}{25}=1$

    3. Multiply through by 25 to clear denominators: $$-2x+3y=25 …$$

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