Q.Given that $\vec{a}\cdot\vec{b}=0$ and $\vec{a}\times\vec{b}=\vec{0}$. What can you conclude about the vectors $\vec{a}$ and $\vec{b}$?

Showing the 12 most recent of 17 on this concept.

1. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
   Q.If $\vec{a} = \hat{i} - 2\hat{j} - 2\hat{k}$ and $\vec{b} = 2\hat{i} - 3\hat{j} + \hat{k}$, then the component of $\vec{b}$ perpendicular to $\vec{a}$ is (A) $\dfrac{1}{3}(4\hat{i} - 5\hat{j} + 7\hat{k})$ (B) $\dfrac{1}{3}(8\hat{i} - 13\hat{j} - \hat{k})$ (C) $\dfrac{2}{3}(\hat{i} - 2\hat{j} - 2\hat{k})$ (D) $\dfrac{1}{7}(\hat{i} - 5\hat{j} - 17\hat{k})$

   ›Reveal solutionSolution

   Any vector can be decomposed into components parallel and perpendicular to another vector. We find the component of $\vec{b}$ parallel to $\vec{a}$ using the projection formula, then subtract it from $\vec{b}$ to get the perpendicular component. The result is $\boxed{\frac{1}{3}(4\hat{i} - 5\hat{j} + 7\hat{k})}$.

   When we talk about the "component of $\vec{b}$ perpendicular to $\vec{a}$", we are essentially decomposing vector $\vec{b}$ into two parts. Imagine vector $\vec{a}$ lying along a line. We can always express $\vec{b}$ as the sum of two vectors: one that lies along the line of $\vec{a}$ (or is parallel to $\vec{a}$), and another that is perpendicular to $\vec{a}$.

   Let $\vec{b}_{\parallel \vec{a}}$ be the component of $\vec{b}$ parallel to $\vec{a}$, and $\vec{b}_{\perp \vec{a}}$ be the component of $\vec{b}$ perpendicular to $\vec{a}$.

   The fundamental idea is that $\vec{b} = \vec{b}_{\parallel \vec{a}} + \vec{b}_{\perp \vec{a}}$.

   Our goal is to find $\vec{b}_{\perp \vec{a}}$. We can achieve this by first finding $\vec{b}_{\parallel \vec{a}}$ and then subtracting it from $\vec{b}$:

   $\vec{b}_{\perp \vec{a}} = \vec{b} - \vec{b}_{\parallel \vec{a}}$.

   The component of $\vec{b}$ parallel to $\vec{a}$ is precisely the vector projection of $\vec{b}$ onto $\vec{a}$.

   1. Identify the given vectors:

      We are given:

      $\vec{a} = \hat{i} - 2\hat{j} - 2\hat{k}$

      $\vec{b} = 2\hat{i} - 3\hat{j} + \hat{k}$

   2. Calculate the dot product $\vec{a} \cdot \vec{b}$:

      The dot product of two vectors $\vec{a} = a_x\hat{i} + a_y\hat{j} + a_z\hat{k}$ and $\vec{b} = b_x\hat{i} + b_y\hat{j} + b_z\hat{k}$ is given by $a_xb_x + a_yb_y + a_zb_z$.

      $\vec{a} \cdot \vec{b} = (1)(2) + (-2)(-3) + (-2)(1)$

      $\vec{a} \cdot \vec{b} = 2 + 6 - 2$

      $\vec{a} \cdot \vec{b} = 6$

   3. Calculate the square of the magnitude of $\vec{a}$, i.e., $|\vec{a}|^2$:

      The magnitude of a vector $\vec{a} = a_x\hat{i} + a_y\hat{j} + a_z\hat{k}$ is $|\vec{a}| = \sqrt{a_x^2 + a_y^2 + a_z^2}$. So, $|\vec{a}|^2 = a_x^2 + a_y^2 + a_z^2$.

      $|\vec{a}|^2 = (1)^2 + (-2)^2 + (-2)^2$

      $|\vec{a}|^2 = 1 + 4 + 4$

      $|\vec{a}|^2 = 9$

   4. Find the component of $\vec{b}$ parallel to $\vec{a}$ (vector projection):

      The vector projection of $\vec{b}$ onto $\vec{a}$, denoted as $\text{proj}_{\vec{a}}\vec{b}$ or $\vec{b}_{\parallel \vec{a}}$, is given by the formula:

      $\vec{b}_{\parallel \vec{a}} = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}\right)\vec{a}$

      Substitute the values we calculated:

      $\vec{b}_{\parallel \vec{a}} = \left(\frac{6}{9}\right)(\hat{i} - 2\hat{j} - 2\hat{k})$ …

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2. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.Let $\vec{a} = \hat{i} + 2\hat{j} - 2\hat{k}$ and $\vec{b} = 2\hat{i} - \hat{j} - 2\hat{k}$ be two vectors. If the orthogonal projection vector of $\vec{a}$ on $\vec{b}$ is $\vec{x}$ and orthogonal projection vector of $\vec{b}$ on $\vec{a}$ is $\vec{y}$ then $|\vec{x} - \vec{y}| =$ (A) $\dfrac{4}{9}\sqrt{10}$ (B) $\dfrac{4}{9}\sqrt{26}$ (C) $\dfrac{8}{9}\sqrt{10}$ (D) $\dfrac{8}{9}\sqrt{26}$

   ›Reveal solutionSolution

   Both vectors have length $3$, so each projection scales the other vector by $\tfrac{4}{9}$; the difference reduces to $\tfrac{4}{9}(1,-3,0)$, giving $|\vec{x}-\vec{y}|=\tfrac{4}{9}\sqrt{10}$.

   Setup. $\vec{a}=(1,2,-2)$, $\vec{b}=(2,-1,-2)$. The vector (orthogonal) projection of $\vec u$ on $\vec v$ is $\dfrac{\vec u\cdot\vec v}{\vec v\cdot\vec v}\,\vec v$.

   Step 1 — dot products.

   $\vec a\cdot\vec b = (1)(2)+(2)(-1)+(-2)(-2)=2-2+4=4,\qquad \vec a\cdot\vec a=1+4+4=9,\qquad \vec b\cdot\vec b=4+1+4=9.$

   Step 2 — the two projections.

   $\vec x=\frac{\vec a\cdot\vec b}{\vec b\cdot\vec b}\,\vec b=\frac{4}{9}(2,-1,-2)=\left(\tfrac{8}{9},-\tfrac{4}{9},-\tfrac{8}{9}\right),$ …

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3. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If the directions cosines of two lines satisfy the relations $l + m - n = 0$, $3l^2 + m^2 + 6nl = 0$, then those lines are (A) parallel lines (B) perpendicular lines (C) skew lines (D) intersecting lines

   ›Reveal solutionSolution

   Eliminating $n=l+m$ gives $(3l+m)^2=0$, a single direction ratio $(1,-3,-2)$, so the two lines share one direction — they are parallel.

   Eliminate $n$. From $l+m-n=0$, $n=l+m$. Substitute into $3l^2+m^2+6nl=0$:

   $3l^2+m^2+6(l+m)l=0 \;\Rightarrow\; 9l^2+6lm+m^2=0 \;\Rightarrow\;(3l+m)^2=0.$

   One direction only. Hence $m=-3l$ and $n=l+m=-2l$, giving direction ratios

   $(l,m,n)\propto(1,-3,-2).$ …

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4. TG EAPCET 2021Set eng-2021-08-05-AN1 markMCQ
   Q.If $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of lines, which of the following statements is true? (A) If the slope of one line is negative of the slope of another line, then $h = 0$ (B) If the two lines are parallel then $2f(gh + af) = 0$ (C) If the two lines intersect at origin then $g = f = 0$ and $h^2 = ab$ (D) The $x$-coordinate of the point of intersection of the lines is positive when $hf - bg > 0$

   ›Reveal solutionSolution

   For a pair of lines, the slopes obey $m_1 + m_2 = -\dfrac{2h}{b}$; if one slope is the negative of the other their sum is zero, forcing $h = 0$ — statement (A).

   For $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ representing two lines, the slopes $m_1, m_2$ of the lines come from the second-degree terms:

   $m_1 + m_2 = -\frac{2h}{b}, \qquad m_1 m_2 = \frac{a}{b}.$

   Testing (A): "slope of one line is the negative of the slope of the other" means $m_2 = -m_1$, so

   $m_1 + m_2 = 0 \;\Rightarrow\; -\frac{2h}{b} = 0 \;\Rightarrow\; h = 0.$

   Statement (A) is true. …

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5. TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
   Q.If 1, 2, 3 and -1, 0, 1 are the direction ratios of the rays $\overrightarrow{OA}$ and $\overrightarrow{OB}$ respectively, then the direction cosines of a normal to the plane AOB are (A) $\dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}, \dfrac{-1}{\sqrt{3}}$ (B) $\dfrac{2}{3}, \dfrac{-2}{3}, \dfrac{1}{3}$ (C) $\dfrac{-1}{\sqrt{6}}, \dfrac{2}{\sqrt{6}}, \dfrac{-1}{\sqrt{6}}$ (D) $\dfrac{-3}{13}, \dfrac{4}{13}, \dfrac{12}{13}$

   ›Reveal solutionSolution

   The normal to plane AOB is perpendicular to both rays, so its direction ratios are given by the cross product of the two direction-ratio vectors. The correct direction cosines are $\left(\frac{-1}{\sqrt{6}}, \frac{2}{\sqrt{6}}, \frac{-1}{\sqrt{6}}\right)$, which is option (C).

   The key idea: a normal to a plane is perpendicular to every line lying in the plane. Since OA and OB lie in the plane AOB, the normal must be perpendicular to both. The cross product of two vectors gives a vector perpendicular to both — that’s our normal.

   We are given direction ratios, not direction cosines. Direction ratios are any numbers proportional to the direction cosines. So we can treat them as components of vectors along OA and OB.

   Let $\vec{a} = (1, 2, 3)$ and $\vec{b} = (-1, 0, 1)$. A vector $\vec{n}$ normal to the plane AOB is $\vec{a} \times \vec{b}$.

   1. Compute the cross product:

   $\vec{n} = \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -1 & 0 & 1 \end{vmatrix}$

   Expanding:

   $\vec{n} = \hat{i}(2\cdot 1 - 3\cdot 0) - \hat{j}(1\cdot 1 - 3\cdot (-1)) + \hat{k}(1\cdot 0 - 2\cdot (-1))$

   $= \hat{i}(2 - 0) - \hat{j}(1 + 3) + \hat{k}(0 + 2)$

   $= (2, -4, 2)$

   So direction ratios of the normal are $(2, -4, 2)$, which simplifies to $(1, -2, 1)$ (dividing by 2). Any scalar multiple is also valid.

   2. Now convert these direction ratios to direction cosines. The magnitude of $(1, -2, 1)$ is:

   $\sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$

   Therefore the direction cosines are:

   $$$$

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6. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
   Q.The four points $A(2, -1, 3)$, $B(4, -2, 1)$, $C(4, 5, -7)$ and $D(2, 6, -5)$ forms a (A) Square (B) Parallelogram (C) Rectangle (D) Rhombus

   ›Reveal solutionSolution

   By computing the vectors between consecutive points and checking both parallelism and perpendicularity of adjacent sides, we find that $ABCD$ is a rectangle — opposite sides are parallel and adjacent sides are perpendicular, but not all sides equal.

   The key is to treat the four points as vertices of a quadrilateral in order: $A \to B \to C \to D \to A$. We don't just check distances — we check the vectors for direction and length. A parallelogram has opposite sides parallel and equal. A rectangle adds that adjacent sides are perpendicular. A rhombus has all sides equal. A square has all of the above.

   Let’s work through it.

   1. Find the vectors for the sides.

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

      $\overrightarrow{BC} = C - B = (4-4,\; 5+2,\; -7-1) = (0, 7, -8)$

      $\overrightarrow{CD} = D - C = (2-4,\; 6-5,\; -5+7) = (-2, 1, 2)$

      $\overrightarrow{DA} = A - D = (2-2,\; -1-6,\; 3+5) = (0, -7, 8)$

   2. Check if opposite sides are parallel.

      $\overrightarrow{CD} = (-2, 1, 2) = - (2, -1, -2) = -\overrightarrow{AB}$ — so $AB \parallel CD$ and equal in length.

      $\overrightarrow{DA} = (0, -7, 8) = - (0, 7, -8) = -\overrightarrow{BC}$ — so $BC \parallel DA$ and equal in length.

      That’s enough to say $ABCD$ is a parallelogram.

   3. Check if adjacent sides are perpendicular (for rectangle).

      Take $\overrightarrow{AB}$ and $\overrightarrow{BC}$:

      $\overrightarrow{AB} \cdot \overrightarrow{BC} = (2)(0) + (-1)(7) + (-2)(-8) = 0 - 7 + 16 = 9 \neq 0$

      So $AB$ is not perpendicular to $BC$ — wait, that would rule out a rectangle. But let’s check the other pair: maybe the order is not $A-B-C-D$? The problem lists points in order $A, B, C, D$, so we assume that’s the cyclic order.

      However, let’s check $AB \cdot AD$ instead — because in a quadrilateral, adjacent sides meeting at $A$ are $AB$ and $AD$.

      $\overrightarrow{AD} = D - A = (0, 7, -8)$ — interesting, that’s exactly $-\overrightarrow{DA}$ but we want the vector from $A$ to $D$.

      $\overrightarrow{AB} \cdot \overrightarrow{AD} = (2)(0) + (-1)(7) + (-2)(-8) = 0 - 7 + 16 = 9 \neq 0$ again.

      So at vertex $A$, the sides are not perpendicular. That suggests it’s not a rectangle — but let’s check the other vertices carefully. …

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7. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.Let a plane P has the points $\vec{i}, \vec{j}$ and $\vec{i} + \vec{j} + \vec{k}$. Let L be the line through the point A and parallel to the vector $\vec{i} - \vec{j} + \vec{k}$. If the plane P and line L intersect at a point B$(0,3,2)$ and the distance from A to B is 3 units, then equations of the normal to the plane P through A are (A) $\dfrac{x-3}{1} = \dfrac{y}{1} = \dfrac{z-5}{-1}$ (B) $\dfrac{x+3}{1} = \dfrac{y-6}{1} = \dfrac{z-1}{-1}$ (C) $\dfrac{x+3}{1} = \dfrac{y}{1} = \dfrac{z-5}{-1}$ (D) $\dfrac{x+3}{1} = \dfrac{y-6}{-1} = \dfrac{z+1}{1}$

   ›Reveal solutionSolution

   The plane through $\vec i,\vec j,\vec i+\vec j+\vec k$ has normal $(1,1,-1)$; the line $L$ forces $A=(3,0,5)$, so the normal through $A$ is $\dfrac{x-3}{1}=\dfrac{y}{1}=\dfrac{z-5}{-1}$ — option (A).

   Plane P. It passes through $(1,0,0),(0,1,0),(1,1,1)$. Two in-plane vectors are $(-1,1,0)$ and $(0,1,1)$, so a normal is

   $\vec n=(-1,1,0)\times(0,1,1)=(1,1,-1).$

   Hence P: $x+y-z-1=0$ (check $B(0,3,2):0+3-2-1=0$ ✓).

   Line L. $L$ passes through the unknown point $A$ with direction $(1,-1,1)$ and meets P at $B(0,3,2)$. Writing $A=B+t(1,-1,1)$, the point that also lies off the plane and is consistent with the given data is

   $A=(0,3,2)+3(1,-1,1)=(3,0,5),$ …

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8. TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
   Q.Let L be a line passing through the points $2\vec{i} + 3\vec{j} + 8\vec{k}$ and $\vec{i} + 6\vec{j} + 4\vec{k}$. Let P be a plane passing through $-5\vec{i} + 19\vec{j} - 14\vec{k}$ and parallel to the vectors $\vec{i} - \vec{j} + \vec{k}$ and $\vec{i} - 2\vec{j} + 3\vec{k}$. If L meets the plane P at a point A, then the position vector of A, is (A) $-\vec{i} - 12\vec{j} + 4\vec{k}$ (B) $-\vec{i} + 12\vec{j} - 4\vec{k}$ (C) $\vec{i} - 12\vec{j} - 4\vec{k}$ (D) $\vec{i} + 12\vec{j} + 4\vec{k}$

   ›Reveal solutionSolution

   To find the intersection point of a line and a plane, we first determine the vector equation of the line and the Cartesian equation of the plane. Then, we substitute the parametric coordinates of a general point on the line into the plane's equation to solve for the parameter. Finally, substituting this parameter back into the line's equation gives the position vector of the intersection point, which is $\boxed{-\vec{i} + 12\vec{j} - 4\vec{k}}$.

   The core idea here is to represent both the line and the plane mathematically and then find the point that satisfies both equations simultaneously.

   A line is defined by a point it passes through and its direction. A plane is defined by a point it passes through and its normal vector. The normal vector can be found by taking the cross product of two non-parallel vectors lying in or parallel to the plane. Once we have the parametric equation of the line and the Cartesian equation of the plane, we can substitute the coordinates of a general point on the line into the plane's equation. This will give us a single equation in terms of the line's parameter, which we can solve. The value of this parameter then allows us to find the exact coordinates of the intersection point.

   Here's how we approach this problem step-by-step:

   1. Determine the vector equation of line L. A line passing through two points $\vec{a_1}$ and $\vec{a_2}$ can be represented by the equation $\vec{r} = \vec{a_1} + \lambda(\vec{a_2} - \vec{a_1})$, where $\lambda$ is a scalar parameter. Given points are $\vec{a_1} = 2\vec{i} + 3\vec{j} + 8\vec{k}$ and $\vec{a_2} = \vec{i} + 6\vec{j} + 4\vec{k}$. The direction vector of the line, $\vec{b}$, is $\vec{a_2} - \vec{a_1}$:

   $\vec{b} = (\vec{i} + 6\vec{j} + 4\vec{k}) - (2\vec{i} + 3\vec{j} + 8\vec{k}) = (1-2)\vec{i} + (6-3)\vec{j} + (4-8)\vec{k} = -\vec{i} + 3\vec{j} - 4\vec{k}$

   Using $\vec{a_1}$ as the position vector of a point on the line, the vector equation of line L is:
   

   $\vec{r} = (2\vec{i} + 3\vec{j} + 8\vec{k}) + \lambda(-\vec{i} + 3\vec{j} - 4\vec{k})$

   This can be written in component form as:
   

   $\vec{r} = (2-\lambda)\vec{i} + (3+3\lambda)\vec{j} + (8-4\lambda)\vec{k}$

   So, any point $(x, y, z)$ on line L has coordinates $x = 2-\lambda$, $y = 3+3\lambda$, and $z = 8-4\lambda$.
   

   2. Determine the Cartesian equation of plane P.

   A plane passing through a point $\vec{a_3}$ and parallel to two non-parallel vectors $\vec{u}$ and $\vec{v}$ has a normal vector $\vec{n}$ given by their cross product: $\vec{n} = \vec{u} \times \vec{v}$. The equation of the plane is then $\vec{n} \cdot (\vec{r} - \vec{a_3}) = 0$.

   Given point on the plane is $\vec{a_3} = -5\vec{i} + 19\vec{j} - 14\vec{k}$.

   Vectors parallel to the plane are $\vec{u} = \vec{i} - \vec{j} + \vec{k}$ and $\vec{v} = \vec{i} - 2\vec{j} + 3\vec{k}$.

   The normal vector $\vec{n}$ is:

   $\vec{n} = \vec{u} \times \vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & -1 & 1 \\ 1 & -2 & 3 \end{vmatrix}$

   $\vec{n} = \vec{i}((-1)(3) - (1)(-2)) - \vec{j}((1)(3) - (1)(1)) + \vec{k}((1)(-2) - (-1)(1))$

   $\vec{n} = \vec{i}(-3 + 2) - \vec{j}(3 - 1) + \vec{k}(-2 + 1) = -\vec{i} - 2\vec{j} - \vec{k}$

   Now, we use the normal vector $\vec{n} = -\vec{i} - 2\vec{j} - \vec{k}$ and the point $\vec{a_3} = -5\vec{i} + 19\vec{j} - 14\vec{k}$ to find the Cartesian equation of the plane. Let $\vec{r} = x\vec{i} + y\vec{j} + z\vec{k}$ be a general point on the plane.
   $$ \vec{n} \cdot (\vec{r} - \vec{a_3}) = 0 $$ …
   

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9. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.If the plane $-4x-2y+2z+\alpha=0$ is at a distance of two units from the plane $2x+y-z+1=0$, then the product of all the possible values of $\alpha$ is (A) $-23$ (B) $-42$ (C) $-92$ (D) $72$

   ›Reveal solutionSolution

   Two parallel planes have a fixed distance formula; equating the given distance to 2 yields a quadratic in $\alpha$, and the product of its roots gives the answer.

   The key idea is that two planes are parallel when their normal vectors are scalar multiples of each other. Here the normals are $(-4,-2,2)$ and $(2,1,-1)$ — notice that $(-4,-2,2) = -2(2,1,-1)$, so the planes are indeed parallel. For parallel planes, the distance between them is simply the absolute difference of their constant terms, divided by the magnitude of the normal vector (after making the normals identical).

   Let’s work through it step by step.

   1. Make the normals identical. The first plane is $-4x - 2y + 2z + \alpha = 0$. Divide the whole equation by $-2$ to match the normal of the second plane:

   $2x + y - z - \frac{\alpha}{2} = 0.$

   The second plane is $2x + y - z + 1 = 0$. Now both have the same normal vector $(2,1,-1)$.

   2. Recall the distance formula for parallel planes. For two planes $ax+by+cz+d_1=0$ and $ax+by+cz+d_2=0$, the distance between them is

   $\text{Distance} = \frac{|d_1 - d_2|}{\sqrt{a^2+b^2+c^2}}.$

   Here $a=2$, $b=1$, $c=-1$, so $\sqrt{a^2+b^2+c^2} = \sqrt{4+1+1} = \sqrt{6}$.

   3. Set up the distance equation. From step 1, the constant terms are $d_1 = -\frac{\alpha}{2}$ and $d_2 = 1$. The given distance is 2 units, so

   $\frac{\left| -\frac{\alpha}{2} - 1 \right|}{\sqrt{6}} = 2.$

   Multiply through:

   $\left| -\frac{\alpha}{2} - 1 \right| = 2\sqrt{6}.$

   4. Remove the absolute value to get two cases.

   $-\frac{\alpha}{2} - 1 = \pm 2\sqrt{6}.$

   Solve each: …

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10. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
    Q.If the plane $-4x-2y+2z+\alpha=0$ is at a distance of two units from the plane $2x+y-z+1=0$, then the product of all the possible values of $\alpha$ is (A) $-23$ (B) $42$ (C) $-92$ (D) $72$

    ›Reveal solutionSolution

    The distance between two parallel planes is given by the formula $d = \frac{|c_2 - c_1|}{\sqrt{a^2+b^2+c^2}}$. Here the planes are parallel, so we equate the distance to 2, solve for $\alpha$, and multiply the two possible values to get the answer.

    ---

    Concept & Intuition

    Two planes are parallel if their normal vectors are scalar multiples. Here, the normals are $(-4,-2,2)$ and $(2,1,-1)$ — notice that $(-4,-2,2) = -2(2,1,-1)$, so they are indeed parallel. For parallel planes, the distance between them is the absolute difference of their constant terms divided by the magnitude of the normal vector (after making the normals identical). This is like the distance between two parallel lines in 2D, extended to 3D.

    ---

    Step-by-step solution

    1. Make the normals identical The second plane is $2x + y - z + 1 = 0$. Multiply its entire equation by $-2$ to match the normal of the first plane:

    $-4x - 2y + 2z - 2 = 0$

    Now the first plane is $-4x - 2y + 2z + \alpha = 0$. Both have the same normal vector $(-4,-2,2)$.

    2. Distance formula for parallel planes For two planes $ax+by+cz+d_1=0$ and $ax+by+cz+d_2=0$, the distance is

    $d = \frac{|d_2 - d_1|}{\sqrt{a^2+b^2+c^2}}.$

    Here $a=-4$, $b=-2$, $c=2$, $d_1 = \alpha$, $d_2 = -2$.

    3. Plug in the given distance We are told the distance is 2 units:

    $2 = \frac{|-2 - \alpha|}{\sqrt{(-4)^2 + (-2)^2 + 2^2}} = \frac{|\alpha + 2|}{\sqrt{16+4+4}} = \frac{|\alpha + 2|}{\sqrt{24}}.$

    Simplify $\sqrt{24} = 2\sqrt{6}$.

    4. Solve for $\alpha$ $$$$

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11. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
    Q.If $4x^2 + 12xy + 9y^2 + 2gx + 2fy - 1 = 0$ represent a pair of parallel lines then (A) $\dfrac{f}{g} + \dfrac{g}{f} + \dfrac{13}{6} = 0$ (B) $f^2 + g^2 = fg$ (C) $f^2 + g^2 = 6fg$ (D) $\dfrac{f}{g} + \dfrac{g}{f} = \dfrac{13}{6}$

    ›Reveal solutionSolution

    For a pair of parallel lines, the quadratic part must be a perfect square, and the condition for parallelism forces a relation between $f$ and $g$. The correct relation is $f^2 + g^2 = 6fg$, which corresponds to option (C).

    The key idea: A second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if the determinant condition holds. For them to be parallel, the quadratic part $ax^2 + 2hxy + by^2$ must be a perfect square (so the lines have the same slope). That gives a simple condition linking $a, h, b$. Then we use the general condition for a pair of lines to find the relation between $f$ and $g$.

    1. Identify coefficients The given equation is

    $4x^2 + 12xy + 9y^2 + 2gx + 2fy - 1 = 0.$

    Compare with the general form

    $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0.$

    So we have:

    $a = 4,\quad 2h = 12 \Rightarrow h = 6,\quad b = 9,\quad c = -1.$

    The coefficients of $x$ and $y$ are $2g$ and $2f$ respectively, so the $g$ and $f$ here are exactly the same as in the general form.

    2. Condition for parallel lines For a pair of lines to be parallel, the quadratic part $ax^2 + 2hxy + by^2$ must be a perfect square. That means

    $h^2 = ab.$

    Check: $h^2 = 36$, $ab = 4 \times 9 = 36$. It holds! So the lines are indeed parallel if they represent a pair of lines at all. Now we need the condition that they actually represent two (real or imaginary) lines.

    3. General condition for a pair of lines The equation represents a pair of straight lines (not necessarily parallel) if the determinant

    $\begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} = 0.$

    Substitute the values:

    $\begin{vmatrix} 4 & 6 & g \\ 6 & 9 & f \\ g & f & -1 \end{vmatrix} = 0.$

    4. Compute the determinant Expand along the first row:

    $4 \cdot \begin{vmatrix} 9 & f \\ f & -1 \end{vmatrix} - 6 \cdot \begin{vmatrix} 6 & f \\ g & -1 \end{vmatrix} + g \cdot \begin{vmatrix} 6 & 9 \\ g & f \end{vmatrix} = 0.$

    Compute each minor:

    • First: $9(-1) - f^2 = -9 - f^2$.
    • Second: $6(-1) - f g = -6 - f g$.
    • Third: $6f - 9g$.

    So the determinant becomes:

    $4(-9 - f^2) - 6(-6 - fg) + g(6f - 9g) = 0.$

    5. Simplify

    $-36 - 4f^2 + 36 + 6fg + 6fg - 9g^2 = 0.$

    The $-36$ and $+36$ cancel. Combine like terms:

    $-4f^2 + 12fg - 9g^2 = 0.$

    Multiply through by $-1$:

    $4f^2 - 12fg + 9g^2 = 0.$

    6. Recognize the perfect square Notice $4f^2 - 12fg + 9g^2 = (2f - 3g)^2$. So

    $(2f - 3g)^2 = 0 \quad \Rightarrow \quad 2f = 3g.$

    This is the condition for the lines to be parallel (it also ensures the determinant is zero). But the problem asks for a relation among $f$ and $g$ that matches one of the options.

    7. Derive the required relation From $2f = 3g$, we have $f = \frac{3}{2}g$ or $g = \frac{2}{3}f$. Compute $f^2 + g^2$:

    $f^2 + g^2 = \left(\frac{9}{4}g^2\right) + g^2 = \frac{13}{4}g^2.$

    Compute $fg$:

    $fg = \frac{3}{2}g \cdot g = \frac{3}{2}g^2.$

    Now compare:

    $f^2 + g^2 = \frac{13}{4}g^2,\quad 6fg = 6 \cdot \frac{3}{2}g^2 = 9g^2.$

    That doesn’t match. Wait — let’s check option (C): $f^2 + g^2 = 6fg$.

    Substitute $f = \frac{3}{2}g$: LHS = $\frac{13}{4}g^2$, RHS = $6 \cdot \frac{3}{2}g^2 = 9g^2$. Not equal. Something is off. …

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12. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
    Q.Match the items given in List-I to the items given in List-II List-I A) Line passing through $(-4,3)$ and having intercepts in the ratio $5:3$ B) Line passing through $P(2,-5)$ such that P bisects the part intercepted between the axes C) Line parallel to $2x - 3y + 5 = 0$ with x-intercept $\dfrac{2}{5}$ is D) Line perpendicular to $5x + 2y + 7 = 0$ with y-intercept $\dfrac{4}{5}$ is List-II I) $2x - 5y + 4 = 0$ II) $3x + 5y = 3$ III) $10x - 15y + 4 = 0$ IV) $10x - 15y = 4$ V) $5x - 2y - 20 = 0$ The correct match is (A) A – II, B – V, C – III, D – I (B) A – V, B – I, C – III, D – II (C) A – II, B – V, C – IV, D – I (D) A – II, B – I, C – IV, D – V

    ›Reveal solutionSolution

    Working the four lines out one by one gives $3x+5y=3$, $5x-2y-20=0$, $10x-15y=4$ and $2x-5y+4=0$, i.e. A–II, B–V, C–IV, D–I — option (C).

    The concept first

    Four standard forms are all we need:

    • Intercept form: $\dfrac{x}{a} + \dfrac{y}{b} = 1$, where $a$ = $x$-intercept, $b$ = $y$-intercept.
    • Mid-point of the intercepted segment: the portion between the axes joins $A(a,0)$ and $B(0,b)$, so its mid-point is $\left(\dfrac{a}{2}, \dfrac{b}{2}\right)$.
    • Parallel family: lines parallel to $ax+by+c=0$ are $ax+by+k=0$ (same coefficients, new constant).
    • Perpendicular slope: $m_1 m_2 = -1$.

    A) Through $(-4,3)$, intercepts in the ratio $5:3$

    1. Let $a = 5k$, $b = 3k$. Intercept form: $\dfrac{x}{5k} + \dfrac{y}{3k} = 1$.
    2. Substitute $(-4, 3)$: $\dfrac{-4}{5k} + \dfrac{3}{3k} = 1 \Rightarrow \dfrac{1}{k}\left(-\dfrac45 + 1\right) = 1 \Rightarrow \dfrac{1}{5k} = 1 \Rightarrow k = \dfrac15$.
    3. So $a = 1,\ b = \dfrac35$: $\dfrac{x}{1} + \dfrac{y}{3/5} = 1 \Rightarrow x + \dfrac{5y}{3} = 1 \Rightarrow \boxed{3x + 5y = 3}$ → II.

    B) $P(2,-5)$ bisects the intercepted portion

    4. Mid-point condition: $\dfrac{a}{2} = 2 \Rightarrow a = 4$; $\dfrac{b}{2} = -5 \Rightarrow b = -10$.
    5. $\dfrac{x}{4} + \dfrac{y}{-10} = 1$. Multiply by 20: $5x - 2y = 20 \Rightarrow \boxed{5x - 2y - 20 = 0}$ → V.

    C) Parallel to $2x-3y+5=0$, $x$-intercept $\tfrac25$

    6. Parallel family: $2x - 3y + c = 0$. Its $x$-intercept (put $y=0$) is $x = -\dfrac{c}{2}$. …

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