Q.State whether the following statement is True or False: The vector equation of the line $\dfrac{x-5}{3} = \dfrac{y+4}{7} = \dfrac{z-6}{2}$ is $\vec{r} = 5\hat{i} - 4\hat{j} + 6\hat{k} + \lambda(3\hat{i} + 7\hat{j} + 2\hat{k})$.

Showing the 12 most recent of 13 on this concept.

1. 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: …

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2. 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 $$$$

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3. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.If $A(2\vec{i} + \vec{j} - \vec{k})$, $B(\lambda \vec{i} + 5\vec{j} + 4\vec{k})$, $C(-4\vec{i} + 3\vec{j} + 2\vec{k})$ and $D(-\vec{i} - 2\vec{j} + 3\vec{k})$ are four points in space such that $\overrightarrow{AB} = x\overrightarrow{AC} + y\overrightarrow{AD}$ for some real numbers $x \neq 0, y \neq 0$ then $17(\lambda + 9) =$ (A) 5 (B) 3 (C) 7 (D) 9

   ›Reveal solutionSolution

   The given vector relationship implies that the four points are coplanar. By expressing the vectors in component form and equating coefficients, we solve for the unknown $\lambda$ and find that $17(\lambda + 9) = \boxed{7}$.

   The problem states a relationship between four vectors originating from a common point $A$: $\overrightarrow{AB} = x\overrightarrow{AC} + y\overrightarrow{AD}$. This equation means that the vector $\overrightarrow{AB}$ can be expressed as a linear combination of vectors $\overrightarrow{AC}$ and $\overrightarrow{AD}$.

   Geometrically, this implies that the vector $\overrightarrow{AB}$ lies in the plane formed by vectors $\overrightarrow{AC}$ and $\overrightarrow{AD}$. Since all three vectors share a common initial point $A$, this condition means that the four points $A, B, C, D$ are coplanar. Our strategy is to express each vector in terms of its components, substitute them into the given equation, and then equate the corresponding components to form a system of linear equations. Solving this system will allow us to find the value of $\lambda$.

   1. Determine the position vectors of the given points.

      The coordinates of the points are given in vector form. We can write their position vectors as:

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

      $\vec{b} = \lambda \vec{i} + 5\vec{j} + 4\vec{k}$

      $\vec{c} = -4\vec{i} + 3\vec{j} + 2\vec{k}$

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

   2. Calculate the vectors $\overrightarrow{AB}$, $\overrightarrow{AC}$, and $\overrightarrow{AD}$.

      A vector from point $P$ to point $Q$ is given by $\overrightarrow{PQ} = \vec{q} - \vec{p}$.

      $\overrightarrow{AB} = \vec{b} - \vec{a} = (\lambda \vec{i} + 5\vec{j} + 4\vec{k}) - (2\vec{i} + \vec{j} - \vec{k})$

      $\overrightarrow{AB} = (\lambda - 2)\vec{i} + (5 - 1)\vec{j} + (4 - (-1))\vec{k}$

      $\overrightarrow{AB} = (\lambda - 2)\vec{i} + 4\vec{j} + 5\vec{k}$

      $\overrightarrow{AC} = \vec{c} - \vec{a} = (-4\vec{i} + 3\vec{j} + 2\vec{k}) - (2\vec{i} + \vec{j} - \vec{k})$

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

      $\overrightarrow{AC} = -6\vec{i} + 2\vec{j} + 3\vec{k}$

      $\overrightarrow{AD} = \vec{d} - \vec{a} = (-\vec{i} - 2\vec{j} + 3\vec{k}) - (2\vec{i} + \vec{j} - \vec{k})$

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

      $\overrightarrow{AD} = -3\vec{i} - 3\vec{j} + 4\vec{k}$

   3. Substitute these vectors into the given relationship.

      The problem states $\overrightarrow{AB} = x\overrightarrow{AC} + y\overrightarrow{AD}$.

      $(\lambda - 2)\vec{i} + 4\vec{j} + 5\vec{k} = x(-6\vec{i} + 2\vec{j} + 3\vec{k}) + y(-3\vec{i} - 3\vec{j} + 4\vec{k})$

      Distribute $x$ and $y$ and group the components:

      $(\lambda - 2)\vec{i} + 4\vec{j} + 5\vec{k} = (-6x - 3y)\vec{i} + (2x - 3y)\vec{j} + (3x + 4y)\vec{k}$

   4. Equate the corresponding components of $\vec{i}$, $\vec{j}$, and $\vec{k}$.

      This gives us a system of three linear equations:

      (1) $\lambda - 2 = -6x - 3y$

      (2) $4 = 2x - 3y$

      (3) $5 = 3x + 4y$

   5. Solve the system of equations for $x$ and $y$.

      We use equations (2) and (3) to find $x$ and $y$:

      $2x - 3y = 4 \quad \text{(Eq. 2)}$

      $3x + 4y = 5 \quad \text{(Eq. 3)}$ …

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4. 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 $2x + 6y - 3z = \pm 42$. …

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5. 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 $t = -\frac{\sqrt{3}}{2}$: …

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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:
   

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

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

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

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7. 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. …

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8. 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 $\vec{n} = \vec{u} \times \vec{v}$: …

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9. 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: $$$$

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10. 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 …

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11. 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:

    $$$$

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12. 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: $-5x + \frac{15}{2} - 3y + \frac{21}{2} + 11z + \frac{99}{2} = 0.$ …

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