Q.Find the equation of a line parallel to x-axis and passing through the origin.

Showing the 12 most recent of 13 on this concept.

1. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.If a plane is at a distance of 6 units from the origin and the vector $2\vec{i} + 6\vec{j} - 3\vec{k}$ is its normal, then the equation of the plane in Cartesian form is (A) $2x + 3y - 6z - 35 = 0$ (B) $2x + 6y - 3z - 42 = 0$ (C) $2x + 6y - 3z - 35 = 0$ (D) $2x - 6y + 3z - 42 = 0$

   ›Reveal solutionSolution

   The plane’s normal vector gives the coefficients of $x$, $y$, $z$ in its Cartesian equation. Using the distance from the origin, we find the constant term. The correct equation is $2x + 6y - 3z - 42 = 0$, which is option (B).

   The key idea is that a plane with normal vector $\vec{n} = a\vec{i} + b\vec{j} + c\vec{k}$ has a Cartesian equation of the form $ax + by + cz = d$, where $d$ is determined by the plane’s distance from the origin. The distance from the origin to the plane $ax + by + cz = d$ is $\frac{|d|}{\sqrt{a^2 + b^2 + c^2}}$. Here we know the distance is 6, and the normal is given, so we can solve for $d$ and then write the full equation.

   1. Identify the normal vector and its magnitude. The normal vector is $2\vec{i} + 6\vec{j} - 3\vec{k}$, so $a = 2$, $b = 6$, $c = -3$. Its magnitude is

   $\sqrt{2^2 + 6^2 + (-3)^2} = \sqrt{4 + 36 + 9} = \sqrt{49} = 7.$

   2. Use the distance formula. For a plane $2x + 6y - 3z = d$, the perpendicular distance from the origin is

   $\frac{|d|}{\sqrt{2^2 + 6^2 + (-3)^2}} = \frac{|d|}{7}.$

   This distance is given as 6, so

   $\frac{|d|}{7} = 6 \quad \Rightarrow \quad |d| = 42.$

   Hence $d = 42$ or $d = -42$. Both give a plane at distance 6 from the origin, but on opposite sides.

   3. Write the equation and match the options. The plane’s equation is $2x + 6y - 3z = \pm 42$. Rearranging: $2x + 6y - 3z - 42 = 0$ or $2x + 6y - 3z + 42 = 0$. Looking at the choices, option (B) is $2x + 6y - 3z - 42 = 0$, which matches the first form. Option (C) has $-35$, which would give a distance of $35/7 = 5$, not 6. Option (A) has a different normal ($2,3,-6$), and option (D) has a different normal ($2,-6,3$). So only (B) fits.

   Watch out

   A common mistake is to forget the absolute value in the distance formula and assume $d$ is positive. Here $d = -42$ also gives distance 6, but the given options only include the $-42$ version, so it’s fine. Always check the sign against the options.

   ✓Final answer

   The correct option is (B): $2x + 6y - 3z - 42 = 0$.

2. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
   Q.If the cartesian equation of the plane passing through the point $\vec{i} + 2\vec{j} + \vec{k}$ and parallel to the vectors $2\vec{i} + 3\vec{j} + \vec{k}$ and $-\vec{i} + 2\vec{j} - 3\vec{k}$ is $ax + by + cz = 1$ then $18(a+b+c) =$ (A) $-3$ (B) $3$ (C) $4$ (D) $-4$

   ›Reveal solutionSolution

   The plane is found by taking a normal vector as the cross product of the two direction vectors, then using the given point to determine the constant. The final result gives $18(a+b+c) = -4$.

   The key idea: a plane parallel to two given vectors has a normal vector perpendicular to both. That normal is simply the cross product of the two direction vectors. Once we have the normal, we write the plane equation in point-normal form and then convert it to the required Cartesian form $ax+by+cz=1$. The coefficients $a,b,c$ are then read off, and we compute $18(a+b+c)$.

   Let’s work through it.

   1. Find a normal vector to the plane. The plane is parallel to $\vec{v}_1 = 2\vec{i} + 3\vec{j} + \vec{k}$ and $\vec{v}_2 = -\vec{i} + 2\vec{j} - 3\vec{k}$. A vector perpendicular to both is their cross product:

   $\vec{n} = \vec{v}_1 \times \vec{v}_2 = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 3 & 1 \\ -1 & 2 & -3 \end{vmatrix}$

   Compute:

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

   $= \vec{i}(-9 - 2) - \vec{j}(-6 + 1) + \vec{k}(4 + 3)$

   $= -11\vec{i} + 5\vec{j} + 7\vec{k}$

   So a normal vector is $\vec{n} = (-11, 5, 7)$.

   2. Write the plane equation using the given point. The plane passes through $P(1, 2, 1)$ (since $\vec{i} + 2\vec{j} + \vec{k}$). The point-normal form is:

   $-11(x - 1) + 5(y - 2) + 7(z - 1) = 0$

   Expand:

   $-11x + 11 + 5y - 10 + 7z - 7 = 0$

   $-11x + 5y + 7z - 6 = 0$

   So:

   $-11x + 5y + 7z = 6$

   3. Convert to the form $ax + by + cz = 1$. Divide the whole equation by 6:

   $-\frac{11}{6}x + \frac{5}{6}y + \frac{7}{6}z = 1$

   Hence $a = -\frac{11}{6}$, $b = \frac{5}{6}$, $c = \frac{7}{6}$.

   4. Compute $18(a+b+c)$. First find $a+b+c$:

   $a+b+c = -\frac{11}{6} + \frac{5}{6} + \frac{7}{6} = \frac{-11 + 5 + 7}{6} = \frac{1}{6}$

   Then:

   $18(a+b+c) = 18 \times \frac{1}{6} = 3$

   Watch out

   A common mistake is to forget to divide the constant term as well when converting to $ax+by+cz=1$. Here the constant was 6, so dividing by 6 gives 1 on the right — but the coefficients also get divided. If you mistakenly read $a=-11$, $b=5$, $c=7$ directly, you’d get $18(-11+5+7)=18$, which is not among the options.

   ✓Final answer

   The value is $\boxed{3}$, so the correct option is (B).

3. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
   Q.The Cartesian equation of a plane parallel to the plane $\vec{r} \cdot (2\hat{i} + 3\hat{j} - 4\hat{k}) = 1$ and at a distance of 2 units from it is (A) $2x + 3y - 4z = 3$ (B) $2x + 3y - 4z = 1 \pm 2\sqrt{29}$ (C) $2x + 3y - 4z = -1 \pm 2\sqrt{29}$ (D) $2x + 3y - 4z = -3$

   ›Reveal solutionSolution

   To find the equation of a plane parallel to a given plane and at a specific distance, we use the fact that parallel planes share the same normal vector. We then apply the formula for the distance between two parallel planes to find the constant term in the new plane's equation. The final equation is $\boxed{2x + 3y - 4z = 1 \pm 2\sqrt{29}}$.

   Concept and Intuition

   The equation of a plane can be expressed in vector form as $\vec{r} \cdot \vec{n} = d$ or in Cartesian form as $Ax + By + Cz = D$. In both forms, the vector $\vec{n} = A\hat{i} + B\hat{j} + C\hat{k}$ (or simply $(A, B, C)$) is the normal vector to the plane. This vector is perpendicular to every vector lying in the plane.

   When two planes are parallel, it means their normal vectors are parallel. If we consider the simplest case, their normal vectors can be taken as identical. Therefore, if a plane has the equation $Ax + By + Cz = D_1$, any plane parallel to it will have the equation $Ax + By + Cz = D_2$ for some different constant $D_2$. The coefficients $A, B, C$ remain the same.

   The distance between two parallel planes $Ax + By + Cz = D_1$ and $Ax + By + Cz = D_2$ is a standard result derived from projecting the vector connecting a point on one plane to a point on the other plane onto the normal vector.

   The distance $D$ between two parallel planes $Ax + By + Cz = D_1$ and $Ax + By + Cz = D_2$ is given by:

   $D = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$

   We will use this formula to find the possible values for $D_2$ (which we'll call $k$ in our solution) for the new plane. Since the new plane can be on either side of the original plane, there will be two possible equations.

   Step-by-step Derivation

   1. Identify the normal vector and Cartesian equation of the given plane.

      The given plane is in vector form: $\vec{r} \cdot (2\hat{i} + 3\hat{j} - 4\hat{k}) = 1$.

      Here, the normal vector is $\vec{n} = 2\hat{i} + 3\hat{j} - 4\hat{k}$.

      To convert this to Cartesian form, let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

      Then, $(x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} + 3\hat{j} - 4\hat{k}) = 1$.

      This simplifies to $2x + 3y - 4z = 1$.

      So, for the given plane, we have $A=2$, $B=3$, $C=-4$, and $D_1=1$.

   2. Determine the general form of the parallel plane.

      Since the required plane is parallel to the given plane, it must have the same normal vector. Therefore, its equation will be of the form:

      $2x + 3y - 4z = k$

      Here, $k$ is the constant $D_2$ that we need to find.

   3. Apply the distance formula between parallel planes.

      We are given that the distance between the two planes is 2 units. Using the formula for the distance between $2x + 3y - 4z = 1$ and $2x + 3y - 4z = k$:

      $D = \frac{|D_1 - D_2|}{\sqrt{A^2 + B^2 + C^2}}$

      Substitute the known values: $D=2$, $D_1=1$, $D_2=k$, $A=2$, $B=3$, $C=-4$.

      $2 = \frac{|1 - k|}{\sqrt{2^2 + 3^2 + (-4)^2}}$

   4. Calculate the magnitude of the normal vector and solve for $k$.

      First, calculate the denominator:

      $\sqrt{2^2 + 3^2 + (-4)^2} = \sqrt{4 + 9 + 16} = \sqrt{29}$

      Now, substitute this back into the distance equation:

      $2 = \frac{|1 - k|}{\sqrt{29}}$

      Multiply both sides by $\sqrt{29}$:

      $2\sqrt{29} = |1 - k|$

      This absolute value equation implies two possibilities:

      Case 1: $1 - k = 2\sqrt{29}$

      $k = 1 - 2\sqrt{29}$

      Case 2: $1 - k = -2\sqrt{29}$

      $k = 1 + 2\sqrt{29}$

      So, the possible values for $k$ are $1 \pm 2\sqrt{29}$.

   5. Write the equations of the planes.

      Substituting the values of $k$ back into the general equation $2x + 3y - 4z = k$, we get the two possible equations for the plane:

      $2x + 3y - 4z = 1 + 2\sqrt{29}$

      and

      $2x + 3y - 4z = 1 - 2\sqrt{29}$

      These can be combined into a single expression:

      $2x + 3y - 4z = 1 \pm 2\sqrt{29}$

   Comparing this result with the given options, it matches option (B).

   ✓Final answer

   The Cartesian equation of the plane is $\boxed{2x + 3y - 4z = 1 \pm 2\sqrt{29}}$.

4. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
   Q.Equation of the plane through the mid-point of the line segment joining the points $A(4,5,-10)$ and $B(-1,2,1)$ and perpendicular to AB is (A) $10x + 6y - 22z + 135 = 0$ (B) $10x + 6y - 22z - 135 = 0$ (C) $5x + 3y + 11z = 135$ (D) $10x + 6y - 22z + 185 = 0$

   ›Reveal solutionSolution

   The plane passes through the midpoint of AB and is perpendicular to AB, so AB’s direction vector is the plane’s normal. The midpoint is (3/2, 7/2, -9/2) and the normal is (5, 3, -11), giving the equation 10x + 6y - 22z - 135 = 0. The correct option is (B).

   Concept & Intuition

   A plane perpendicular to a line means the line’s direction vector is normal (perpendicular) to the plane. So if we find the vector from A to B, that vector is the plane’s normal. The plane also passes through the midpoint of AB, so we have a point and a normal — that’s all we need to write the plane’s equation.

   Step-by-step solution

   1. Find the direction vector of AB

      The line segment goes from $A(4,5,-10)$ to $B(-1,2,1)$.

      The vector $\overrightarrow{AB} = B - A = (-1-4,\; 2-5,\; 1-(-10)) = (-5, -3, 11)$.

      This vector is perpendicular to the required plane, so it can serve as the plane’s normal vector $\mathbf{n}$.

      (We could also use $(5,3,-11)$, the negative, since normals are only defined up to a scalar multiple.)

   2. Find the midpoint of AB

      Midpoint $M = \left( \frac{4+(-1)}{2},\; \frac{5+2}{2},\; \frac{-10+1}{2} \right) = \left( \frac{3}{2},\; \frac{7}{2},\; -\frac{9}{2} \right)$.

   3. Write the plane equation using point-normal form

      For a plane with normal $\mathbf{n} = (a,b,c)$ through point $(x_0,y_0,z_0)$, the equation is

   $a(x - x_0) + b(y - y_0) + c(z - z_0) = 0.$

   Using $\mathbf{n} = (-5, -3, 11)$ and $M\left(\frac{3}{2}, \frac{7}{2}, -\frac{9}{2}\right)$:

   $-5\left(x - \frac{3}{2}\right) - 3\left(y - \frac{7}{2}\right) + 11\left(z + \frac{9}{2}\right) = 0.$

   4. Simplify the equation Expand:

   $-5x + \frac{15}{2} - 3y + \frac{21}{2} + 11z + \frac{99}{2} = 0.$

   Combine constants: $\frac{15}{2} + \frac{21}{2} + \frac{99}{2} = \frac{135}{2}$.

   So:

   $-5x - 3y + 11z + \frac{135}{2} = 0.$

   Multiply through by 2 to clear fractions:

   $-10x - 6y + 22z + 135 = 0.$

   Multiply by $-1$ (optional, to make the leading coefficients positive):

   $10x + 6y - 22z - 135 = 0.$

   5. Match with the options This matches option (B) exactly.

   Watch out

   A common mistake is to forget the midpoint and use A or B instead — that gives a parallel plane but not the one through the midpoint. Also, be careful with signs when simplifying: the constant term must be $-135$, not $+135$.

   Tip

   If you use the normal $(5,3,-11)$ instead of $(-5,-3,11)$, you get the same equation after simplification — just check the constant sign carefully.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

5. TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
   Q.If $\alpha, \beta$ are scalars and $\vec{r} = (2+\alpha-3\beta)\hat{i} + (\beta-3)\hat{j} + (2\alpha-5\beta-1)\hat{k}$ is equation of a plane, then that equation in Cartesian form is (A) $2x + y - z + 2 = 0$ (B) $2x - y - z = 8$ (C) $2x - y - z + 8 = 0$ (D) $2x + y - z = 2$

   ›Reveal solutionSolution

   The parametric plane has base point $(2,-3,-1)$ and direction vectors $(1,0,2)$ and $(-3,1,-5)$; its Cartesian equation is $2x + y - z = 2$.

   The position vector $\vec{r} = (2+\alpha-3\beta)\hat{i} + (\beta-3)\hat{j} + (2\alpha-5\beta-1)\hat{k}$ can be split as:

   $\vec{r} = (2\hat{i} - 3\hat{j} - \hat{k}) + \alpha(\hat{i} + 2\hat{k}) + \beta(-3\hat{i} + \hat{j} - 5\hat{k}).$

   So a point on the plane is $A = (2, -3, -1)$, with direction vectors $\vec{u} = (1, 0, 2)$ and $\vec{v} = (-3, 1, -5)$.

   The normal is $\vec{n} = \vec{u} \times \vec{v}$:

   $\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\1 & 0 & 2 \\-3 & 1 & -5 \end{vmatrix} = \hat{i}(0\cdot(-5) - 2\cdot 1) - \hat{j}(1\cdot(-5) - 2\cdot(-3)) + \hat{k}(1\cdot 1 - 0\cdot(-3)).$

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

   Plane through $A(2,-3,-1)$ with normal $(-2,-1,1)$:

   $-2(x-2) - 1(y+3) + 1(z+1) = 0 \implies -2x - y + z + 2 = 0.$

   Multiplying by $-1$: $2x + y - z - 2 = 0$, i.e. $2x + y - z = 2$.

   ✓Final answer

   $2x + y - z = 2$ — option (D).

6. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.If $l, m, n$ are the d.c.'s of a normal to the plane passing through the points $(0, 1, 2)$, $(3, 0, 2)$, $(4, 5, 0)$ then $|l| + |m| + |n| =$ (A) $\dfrac{13}{\sqrt{91}}$ (B) $\dfrac{11}{\sqrt{57}}$ (C) $\dfrac{13}{\sqrt{77}}$ (D) $\dfrac{12}{\sqrt{74}}$

   ›Reveal solutionSolution

   To find the direction cosines of the normal to a plane, we first find two vectors lying in the plane and compute their cross product to get a normal vector. Then, we normalize this vector to obtain the direction cosines. The sum of the absolute values of the direction cosines is $\boxed{\dfrac{12}{\sqrt{74}}}$.

   Concept and Intuition

   A plane in 3D space is uniquely defined by three non-collinear points. To find the direction of the normal to this plane, we need a vector that is perpendicular to the plane.

   The key idea is that if we form two vectors using these three points, these two vectors will lie within the plane. For example, if the points are $A, B, C$, then vectors $\vec{AB}$ and $\vec{AC}$ both lie in the plane.

   The cross product of two vectors results in a new vector that is perpendicular to both original vectors. Therefore, the cross product of $\vec{AB}$ and $\vec{AC}$ (i.e., $\vec{AB} \times \vec{AC}$) will give us a vector that is perpendicular to the plane containing $A, B, C$. This vector is a normal vector to the plane.

   Let the normal vector be $\vec{N} = a\hat{i} + b\hat{j} + c\hat{k}$. The components $(a, b, c)$ are called the direction ratios of the normal.

   The direction cosines ($l, m, n$) are the cosines of the angles the vector makes with the positive $x, y, z$ axes, respectively. They are obtained by dividing each component of the vector by its magnitude.

   If a vector is $\vec{V} = a\hat{i} + b\hat{j} + c\hat{k}$, its magnitude is $|\vec{V}| = \sqrt{a^2 + b^2 + c^2}$.

   The direction cosines are $l = \frac{a}{|\vec{V}|}$, $m = \frac{b}{|\vec{V}|}$, $n = \frac{c}{|\vec{V}|}$.

   An important property is $l^2 + m^2 + n^2 = 1$.

   Step-by-step Derivation

   Let the given points be $A = (0, 1, 2)$, $B = (3, 0, 2)$, and $C = (4, 5, 0)$.

   1. Form two vectors lying in the plane.

      We can form vectors $\vec{AB}$ and $\vec{AC}$:

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

      $\vec{AC} = C - A = (4 - 0)\hat{i} + (5 - 1)\hat{j} + (0 - 2)\hat{k} = 4\hat{i} + 4\hat{j} - 2\hat{k}$

   2. Find the normal vector to the plane.

      The normal vector $\vec{N}$ is given by the cross product of $\vec{AB}$ and $\vec{AC}$:

   $\vec{N} = \vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 0 \\ 4 & 4 & -2 \end{vmatrix}$

   Expanding the determinant:
   

   $\vec{N} = \hat{i}((-1)(-2) - (0)(4)) - \hat{j}((3)(-2) - (0)(4)) + \hat{k}((3)(4) - (-1)(4))$

   $\vec{N} = \hat{i}(2 - 0) - \hat{j}(-6 - 0) + \hat{k}(12 - (-4))$

   $\vec{N} = 2\hat{i} + 6\hat{j} + 16\hat{k}$

   The direction ratios of the normal are $(2, 6, 16)$.
   

   3. Calculate the magnitude of the normal vector.

   The magnitude of $\vec{N}$ is:

   $|\vec{N}| = \sqrt{2^2 + 6^2 + 16^2}$

   $|\vec{N}| = \sqrt{4 + 36 + 256}$

   $|\vec{N}| = \sqrt{296}$

   We can simplify $\sqrt{296}$ by factoring out perfect squares: $296 = 4 \times 74$.
   

   $|\vec{N}| = \sqrt{4 \times 74} = 2\sqrt{74}$

   4. Determine the direction cosines ($l, m, n$). Using the formula for direction cosines:

   $l = \frac{2}{2\sqrt{74}} = \frac{1}{\sqrt{74}}$

   $m = \frac{6}{2\sqrt{74}} = \frac{3}{\sqrt{74}}$

   $n = \frac{16}{2\sqrt{74}} = \frac{8}{\sqrt{74}}$

   5. Calculate $|l| + |m| + |n|$. Since $l, m, n$ are all positive in this case, their absolute values are themselves:

   $|l| + |m| + |n| = \left|\frac{1}{\sqrt{74}}\right| + \left|\frac{3}{\sqrt{74}}\right| + \left|\frac{8}{\sqrt{74}}\right|$

   $|l| + |m| + |n| = \frac{1}{\sqrt{74}} + \frac{3}{\sqrt{74}} + \frac{8}{\sqrt{74}}$

   $|l| + |m| + |n| = \frac{1 + 3 + 8}{\sqrt{74}}$

   $|l| + |m| + |n| = \frac{12}{\sqrt{74}}$

   ✓Final answer

   The value of $|l| + |m| + |n|$ is $\boxed{\dfrac{12}{\sqrt{74}}}$.

7. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.The shortest distance between the lines $\vec{r} = (3\vec{i} - 5\vec{j} + 2\vec{k}) + t(4\vec{i} + 3\vec{j} - \vec{k})$ and $\vec{r} = (\vec{i} + 2\vec{j} - 4\vec{k}) + s(6\vec{i} + 3\vec{j} - 2\vec{k})$ is (A) 7 (B) 8 (C) 9 (D) 12

   ›Reveal solutionSolution

   The shortest distance between the two skew lines is $8$, so the correct option is (B).

   Setting up

   For skew lines $\vec r=\vec a_1+t\,\vec d_1$ and $\vec r=\vec a_2+s\,\vec d_2$, the shortest distance is

   $d=\frac{\big|(\vec a_2-\vec a_1)\cdot(\vec d_1\times\vec d_2)\big|}{|\vec d_1\times\vec d_2|}.$

   Here $\vec a_1=(3,-5,2),\ \vec d_1=(4,3,-1)$ and $\vec a_2=(1,2,-4),\ \vec d_2=(6,3,-2)$.

   Solution

   Connecting vector: $\vec a_2-\vec a_1=(-2,\,7,\,-6)$.

   Cross product:

   $\vec d_1\times\vec d_2=\begin{vmatrix}\vec i&\vec j&\vec k\\4&3&-1\\6&3&-2\end{vmatrix}=(-3,\,2,\,-6),\qquad |\vec d_1\times\vec d_2|=\sqrt{9+4+36}=7.$

   Numerator: $(-2)(-3)+(7)(2)+(-6)(-6)=6+14+36=56.$

   $d=\frac{|56|}{7}=8.$

   ✓Final answer

   Shortest distance $=8$ — option (B).

8. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If the shortest distance between the two skew lines $\vec{r} = \vec{i} + \vec{j} + \vec{k} + t(3\vec{i} + 2\vec{j} + \vec{k})$ and $\vec{r} = \vec{i} - \vec{j} + x\vec{k} + s(\vec{i} + 2\vec{j} + 3\vec{k})$ is at most $2\sqrt{6}$ units, then all the values of $x$ lie in the interval (A) $[-9, 15]$ (B) $[-15, 9]$ (C) $(-\infty, -15]$ (D) $\mathbb{R} - [-11, 9]$

   ›Reveal solutionSolution

   The shortest distance between two skew lines is given by the formula involving the cross product of direction vectors and the vector connecting a point on each line. Setting that distance ≤ 2√6 yields a quadratic inequality in x whose solution is the interval [−9, 15], so the correct option is (A).

   We have two skew lines in vector form:

   $L_1: \vec{r} = \vec{i} + \vec{j} + \vec{k} + t(3\vec{i} + 2\vec{j} + \vec{k})$

   $L_2: \vec{r} = \vec{i} - \vec{j} + x\vec{k} + s(\vec{i} + 2\vec{j} + 3\vec{k})$

   Here, $t$ and $s$ are scalar parameters. The direction vectors are:

   $\vec{d}_1 = (3, 2, 1), \quad \vec{d}_2 = (1, 2, 3)$

   A point on $L_1$ is $A(1, 1, 1)$ and a point on $L_2$ is $B(1, -1, x)$.

   The shortest distance between two skew lines is given by:

   $d = \frac{|(\vec{AB}) \cdot (\vec{d}_1 \times \vec{d}_2)|}{|\vec{d}_1 \times \vec{d}_2|}$

   1. Compute the cross product $\vec{d}_1 \times \vec{d}_2$:

   $\vec{d}_1 \times \vec{d}_2 = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 3 & 2 & 1 \\ 1 & 2 & 3 \end{vmatrix} = \vec{i}(2\cdot3 - 1\cdot2) - \vec{j}(3\cdot3 - 1\cdot1) + \vec{k}(3\cdot2 - 2\cdot1)$

   $= \vec{i}(6 - 2) - \vec{j}(9 - 1) + \vec{k}(6 - 2) = 4\vec{i} - 8\vec{j} + 4\vec{k}$

   So $\vec{d}_1 \times \vec{d}_2 = (4, -8, 4)$.

   2. Magnitude of the cross product:

   $|\vec{d}_1 \times \vec{d}_2| = \sqrt{4^2 + (-8)^2 + 4^2} = \sqrt{16 + 64 + 16} = \sqrt{96} = 4\sqrt{6}$

   3. Vector $\vec{AB}$ from point on $L_1$ to point on $L_2$:

   $\vec{AB} = B - A = (1-1, -1-1, x-1) = (0, -2, x-1)$

   4. Dot product $(\vec{AB}) \cdot (\vec{d}_1 \times \vec{d}_2)$:

   $(0, -2, x-1) \cdot (4, -8, 4) = 0\cdot4 + (-2)(-8) + (x-1)(4) = 16 + 4(x-1) = 16 + 4x - 4 = 4x + 12$

   5. Shortest distance formula:

   $d = \frac{|4x + 12|}{4\sqrt{6}} = \frac{|x + 3|}{\sqrt{6}}$

   6. Condition: This distance is at most $2\sqrt{6}$:

   $\frac{|x + 3|}{\sqrt{6}} \leq 2\sqrt{6}$

   Multiply both sides by $\sqrt{6}$:

   $|x + 3| \leq 2 \cdot 6 = 12$

   7. Solve the inequality:

   $-12 \leq x + 3 \leq 12$

   Subtract 3:

   $-15 \leq x \leq 9$

   So all values of $x$ lie in the interval $[-15, 9]$.

   Watch out

   A common mistake is to forget the absolute value or miscompute the cross product magnitude. Double-check the arithmetic: $4\sqrt{6}$ is correct, and the dot product simplifies neatly to $4x+12$.

   Tip

   Notice that the factor 4 cancels nicely, leaving a clean expression $\frac{|x+3|}{\sqrt{6}}$. Always simplify the fraction before squaring or solving.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

9. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
   Q.The direction cosines of the supporting line of the vector $\vec{i} + \vec{j} - 2\vec{k}$ are (A) $\left( \dfrac{1}{\sqrt{6}}, \dfrac{1}{\sqrt{6}}, \dfrac{-2}{\sqrt{6}} \right)$ (B) $\left( \dfrac{1}{2}, \dfrac{1}{2}, -1 \right)$ (C) $\left( \dfrac{1}{\sqrt{6}}, \dfrac{1}{\sqrt{6}}, \dfrac{2}{\sqrt{6}} \right)$ (D) $\left( \dfrac{-1}{2}, \dfrac{-1}{2}, -1 \right)$

   ›Reveal solutionSolution

   Direction cosines are the cosines of the angles a vector makes with the coordinate axes — they are simply the components of the unit vector in that direction. For $\vec{i} + \vec{j} - 2\vec{k}$, the correct direction cosines are $\left( \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{-2}{\sqrt{6}} \right)$, which is option (A).

   The idea is straightforward: direction cosines are not just the raw components of the vector — they must be the components of a unit vector along that line. Any vector can be scaled up or down, but the direction it points in stays the same. So to get the direction cosines, you take the original vector and divide each component by its magnitude.

   Let’s walk through it.

   1. Write the vector in component form.

      The given vector is $\vec{i} + \vec{j} - 2\vec{k}$, so its components are $(1, 1, -2)$.

   2. Find the magnitude of the vector.

      The magnitude is

   $|\vec{v}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{1 + 1 + 4} = \sqrt{6}.$

   3. Form the unit vector. Divide each component by $\sqrt{6}$:

   $\hat{v} = \left( \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{-2}{\sqrt{6}} \right).$

   4. Read off the direction cosines. By definition, the direction cosines $l, m, n$ are exactly these three components of the unit vector. So

   $l = \frac{1}{\sqrt{6}}, \quad m = \frac{1}{\sqrt{6}}, \quad n = \frac{-2}{\sqrt{6}}.$

   Watch out

   A common mistake is to forget to divide by the magnitude and just pick the raw components $(1, 1, -2)$ — that’s option (B), which is wrong because direction cosines must satisfy $l^2 + m^2 + n^2 = 1$. Check: $1^2 + 1^2 + (-2)^2 = 6 \neq 1$, so (B) is out. Option (C) has the wrong sign on the third component, and (D) has the wrong magnitude and signs.

   ✓Final answer

   The correct direction cosines are $\left( \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{-2}{\sqrt{6}} \right)$, which is option (A).

10. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
    Q.If $P(1,2,5)$, $Q(3,0,7)$, $R(6,-3,10)$ are three points on a line and $(\alpha, \beta, \gamma)$ is a point at a distance of $3$ units from $P$ on the same line, then value of $\alpha + \beta + \gamma$ that lies between $6$ and $7$ is (A) $13 - \sqrt{3}$ (B) $8 - \sqrt{3}$ (C) $16 + \sqrt{3}$ (D) $7 - \sqrt{2}$

    ›Reveal solutionSolution

    The three points are collinear, so we find the direction vector, paramaterize the line, locate the two points 3 units from P, and then pick the one whose sum of coordinates lies between 6 and 7. That point gives $\alpha+\beta+\gamma = 8 - \sqrt{3}$, which is option (B).

    Concept & Intuition

    When three points lie on a line, the vector from one to another is a scalar multiple of the direction vector. We can paramaterize the line as $P + t \cdot \vec{d}$, where $\vec{d}$ is a direction vector. The distance from $P$ to any point on the line is $|t| \cdot \|\vec{d}\|$. Setting this equal to 3 gives two possible $t$ values (one on each side of $P$). We then compute $\alpha+\beta+\gamma$ for both and see which lies between 6 and 7.

    Step-by-step solution

    1. Check collinearity

       Compute vectors:

       $\overrightarrow{PQ} = (3-1,\,0-2,\,7-5) = (2,-2,2)$

       $\overrightarrow{PR} = (6-1,\,-3-2,\,10-5) = (5,-5,5)$

       Since $\overrightarrow{PR} = \frac{5}{2} \overrightarrow{PQ}$, the points are collinear.

    2. Direction vector and its length

       Use $\vec{d} = \overrightarrow{PQ} = (2,-2,2)$.

       Its magnitude: $\|\vec{d}\| = \sqrt{2^2+(-2)^2+2^2} = \sqrt{12} = 2\sqrt{3}$.

    3. Parametric form of the line

       Any point on the line:

       $(x,y,z) = P + t\,\vec{d} = (1+2t,\;2-2t,\;5+2t)$.

    4. Distance condition

       Distance from $P$ to this point is $|t| \cdot \|\vec{d}\| = |t| \cdot 2\sqrt{3}$.

       Set equal to 3: $|t| \cdot 2\sqrt{3} = 3 \implies |t| = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2}$.

       So $t = \pm \frac{\sqrt{3}}{2}$.

    5. Find the two candidate points

       • For $t = \frac{\sqrt{3}}{2}$:

         $\alpha = 1 + 2\cdot\frac{\sqrt{3}}{2} = 1+\sqrt{3}$

         $\beta = 2 - 2\cdot\frac{\sqrt{3}}{2} = 2-\sqrt{3}$

         $\gamma = 5 + 2\cdot\frac{\sqrt{3}}{2} = 5+\sqrt{3}$

         Sum: $\alpha+\beta+\gamma = (1+\sqrt{3}) + (2-\sqrt{3}) + (5+\sqrt{3}) = 8 + \sqrt{3}$.

       • For $t = -\frac{\sqrt{3}}{2}$:

         $\alpha = 1 - \sqrt{3}$

         $\beta = 2 + \sqrt{3}$

         $\gamma = 5 - \sqrt{3}$

         Sum: $\alpha+\beta+\gamma = (1-\sqrt{3}) + (2+\sqrt{3}) + (5-\sqrt{3}) = 8 - \sqrt{3}$.

    6. Which sum lies between 6 and 7?

       $\sqrt{3} \approx 1.732$, so:

       $8 + \sqrt{3} \approx 9.732$ (too large)

       $8 - \sqrt{3} \approx 6.268$ (between 6 and 7).

       Hence the required point is the one with $t = -\frac{\sqrt{3}}{2}$.

    Watch out

    A common mistake is to forget the $\pm$ sign and only take the positive $t$, which gives the sum $8+\sqrt{3}$ — that is not between 6 and 7.

    Tip

    The sum $\alpha+\beta+\gamma$ for any point on this line is $8 + 2t$ (since $1+2t + 2-2t + 5+2t = 8+2t$). Then $|t| = \frac{\sqrt{3}}{2}$ gives sums $8 \pm \sqrt{3}$ directly — no need to compute coordinates individually.

    ✓Final answer

    The correct option is (B).

    ANSWER: B

11. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
    Q.For the circle $x-2=5\cos\theta$, $y+1=5\sin\theta$ where $\theta$ is the parameter, the line $x=1+\dfrac{r}{2}$, $y=-2+\dfrac{\sqrt{3}}{2}r$ where $r$ is the parameter, is a (A) Chord of the circle other than diameter (B) Tangent of the circle (C) Diameter of the circle (D) Line that does not meet the circle

    ›Reveal solutionSolution

    The line passes through the center of the circle, so it is a diameter. The correct option is (C).

    We are given a circle in parametric form:

    $x - 2 = 5\cos\theta,\quad y + 1 = 5\sin\theta$

    This means the circle’s center is at $(2, -1)$ and its radius is $5$.

    The line is given in parametric form:

    $x = 1 + \frac{r}{2},\quad y = -2 + \frac{\sqrt{3}}{2}r$

    We need to determine its relationship to the circle: chord (but not diameter), tangent, diameter, or no intersection.

    ---

    1. Find the direction vector of the line

    From the parametric equations, as $r$ varies, the coefficients of $r$ give the direction:

    $\text{direction} = \left( \frac{1}{2},\ \frac{\sqrt{3}}{2} \right)$

    This is a unit vector because

    $\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{4} + \frac{3}{4} = 1$

    So the line moves exactly 1 unit in that direction per unit increase in $r$.

    2. Find a point on the line

    When $r = 0$, we get the point

    $(1, -2)$

    So the line passes through $(1, -2)$.

    3. Check if this point lies inside, on, or outside the circle

    Distance from center $(2, -1)$ to $(1, -2)$:

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

    Since $\sqrt{2} < 5$, the point is inside the circle. So the line definitely meets the circle — it is not a tangent and not a line that misses the circle. That eliminates options (B) and (D).

    4. Determine if the line passes through the center

    The center is $(2, -1)$. Does this lie on the line?

    We need to find $r$ such that:

    $1 + \frac{r}{2} = 2 \quad\Rightarrow\quad \frac{r}{2} = 1 \quad\Rightarrow\quad r = 2$

    And then check $y$:

    $y = -2 + \frac{\sqrt{3}}{2} \cdot 2 = -2 + \sqrt{3}$

    For the center, $y$ should be $-1$. But $-2 + \sqrt{3} \approx -2 + 1.732 = -0.268$, not $-1$. So the center is not on the line.

    Watch out

    A common mistake: assuming that because a line passes through a point inside the circle, it must be a chord. But a chord is any segment joining two points on the circle. A diameter is a special chord through the center. Here the line does not go through the center, so it is a chord but not a diameter.

    5. Confirm it is a chord (intersects the circle at two points)

    Since the line passes through an interior point and extends infinitely, it will intersect the circle at exactly two points. That makes it a chord.

    Thus the line is a chord of the circle, but not a diameter.

    Tip

    To quickly check if a line through an interior point is a diameter, just test whether the center lies on the line. If it does, it’s a diameter; if not, it’s a non-diameter chord.

    ✓Final answer

    The correct option is (A).

    ANSWER: A

12. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
    Q.$\vec{n}$ is a unit vector normal to the plane $\pi$ containing the vectors $\vec{i} + 3\vec{k}$ and $2\vec{i} + \vec{j} - \vec{k}$. If this plane $\pi$ passes through the point $(-3, 7, 1)$ and $p$ is the perpendicular distance from the origin to this plane $\pi$, then $\sqrt{p^2 + 5} =$ (A) $59$ (B) $8$ (C) $64$ (D) $51$

    ›Reveal solutionSolution

    The key idea is to find the plane’s equation using its normal vector (cross product of the two given vectors) and the given point, then compute the perpendicular distance from the origin, and finally evaluate $\sqrt{p^2+5}$. The result is $8$, so the correct option is (B).

    We are given two vectors lying in the plane $\pi$:

    $\vec{a} = \vec{i} + 3\vec{k} = (1,0,3)$ and $\vec{b} = 2\vec{i} + \vec{j} - \vec{k} = (2,1,-1)$.

    A normal vector $\vec{n}$ to the plane is perpendicular to both $\vec{a}$ and $\vec{b}$, so we take their cross product. Since $\vec{n}$ is also given to be a unit vector, we will later normalize it.

    Step 1: Find a normal vector to the plane

    $\vec{n}_0 = \vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & 3 \\ 2 & 1 & -1 \end{vmatrix} = \vec{i}(0\cdot(-1) - 3\cdot1) - \vec{j}(1\cdot(-1) - 3\cdot2) + \vec{k}(1\cdot1 - 0\cdot2)$

    $= \vec{i}(0 - 3) - \vec{j}(-1 - 6) + \vec{k}(1 - 0) = -3\vec{i} + 7\vec{j} + \vec{k}$

    So $\vec{n}_0 = (-3, 7, 1)$.

    Step 2: Unit normal vector

    The magnitude is

    $|\vec{n}_0| = \sqrt{(-3)^2 + 7^2 + 1^2} = \sqrt{9 + 49 + 1} = \sqrt{59}.$

    Thus the unit normal is

    $\hat{n} = \frac{1}{\sqrt{59}}(-3, 7, 1).$

    Step 3: Equation of the plane

    The plane passes through point $P_0 = (-3, 7, 1)$. For any point $(x,y,z)$ on the plane, the vector from $P_0$ to $(x,y,z)$ is perpendicular to $\hat{n}$. So the plane equation is

    $\hat{n} \cdot (x+3,\; y-7,\; z-1) = 0.$

    Using the unnormalized normal for simplicity (since scaling doesn’t change the plane):

    $(-3,7,1) \cdot (x+3,\; y-7,\; z-1) = 0.$

    Compute:

    $-3(x+3) + 7(y-7) + 1(z-1) = 0$

    $-3x - 9 + 7y - 49 + z - 1 = 0$

    $-3x + 7y + z - 59 = 0.$

    So the plane equation is

    $-3x + 7y + z = 59.$

    Step 4: Perpendicular distance from origin

    The distance from origin $(0,0,0)$ to plane $Ax+By+Cz = D$ is

    $p = \frac{|D|}{\sqrt{A^2+B^2+C^2}}.$

    Here $A=-3, B=7, C=1, D=59$, so

    $p = \frac{|59|}{\sqrt{(-3)^2+7^2+1^2}} = \frac{59}{\sqrt{59}} = \sqrt{59}.$

    Step 5: Compute $\sqrt{p^2 + 5}$

    $p^2 = 59, \quad p^2 + 5 = 64, \quad \sqrt{p^2+5} = \sqrt{64} = 8.$

    Tip

    Notice that the point $(-3,7,1)$ given in the problem is exactly the normal vector’s components — that’s why the constant term $D$ turned out to be $59$, the square of the normal’s length. This is a neat check.

    Watch out

    A common mistake is to forget that the distance formula uses the absolute value of $D$ and the full magnitude of the normal vector, not the unit normal. Using the unit normal directly would give the same $p$ but requires extra care.

    ✓Final answer

    The correct option is (B).

    ANSWER: B