Q.Find the values of $p$ so that the lines $\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2}$ and $\frac{7-7x}{3p} = \frac{y-5}{1} = \frac{6-z}{5}$ are at right angles.

Showing the 12 most recent of 19 on this concept.

1. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.If the points $A(1, 3, 5)$, $B(2, 4, 6)$, $C(4, 5, k)$ form a right angled triangle then the number of possible values of $k$ is (A) 2 (B) 3 (C) 0 (D) 1

   ›Reveal solutionSolution

   For three points to form a right triangle, the dot product of the vectors along two sides must be zero. Checking all three possible right-angle vertices gives a quadratic in $k$ with two real solutions, so the number of possible values of $k$ is 2.

   The key idea is that a right angled triangle has one angle equal to $90^\circ$. In coordinate geometry, the condition for a right angle at a vertex is that the dot product of the vectors representing the two sides meeting at that vertex is zero. Since we don't know which vertex holds the right angle, we must test all three possibilities.

   Let's work through it systematically.

   1. Write the position vectors of the points

      $A(1,3,5)$, $B(2,4,6)$, $C(4,5,k)$.

      We'll use vector notation: $\vec{A} = \hat{i} + 3\hat{j} + 5\hat{k}$, $\vec{B} = 2\hat{i} + 4\hat{j} + 6\hat{k}$, $\vec{C} = 4\hat{i} + 5\hat{j} + k\hat{k}$.

   2. Form the side vectors for each possible right angle

      Case 1: Right angle at $A$

      Vectors along sides meeting at $A$:

      $\vec{AB} = \vec{B} - \vec{A} = (2-1)\hat{i} + (4-3)\hat{j} + (6-5)\hat{k} = \hat{i} + \hat{j} + \hat{k}$

      $\vec{AC} = \vec{C} - \vec{A} = (4-1)\hat{i} + (5-3)\hat{j} + (k-5)\hat{k} = 3\hat{i} + 2\hat{j} + (k-5)\hat{k}$

      Dot product: $\vec{AB} \cdot \vec{AC} = (1)(3) + (1)(2) + (1)(k-5) = 3 + 2 + k - 5 = k$

      Setting to zero: $k = 0$.

      Case 2: Right angle at $B$

      Vectors:

      $\vec{BA} = \vec{A} - \vec{B} = -\hat{i} - \hat{j} - \hat{k}$

      $\vec{BC} = \vec{C} - \vec{B} = (4-2)\hat{i} + (5-4)\hat{j} + (k-6)\hat{k} = 2\hat{i} + \hat{j} + (k-6)\hat{k}$

      Dot product: $\vec{BA} \cdot \vec{BC} = (-1)(2) + (-1)(1) + (-1)(k-6) = -2 -1 -k + 6 = 3 - k$

      Setting to zero: $3 - k = 0 \implies k = 3$.

      Case 3: Right angle at $C$

      Vectors:

      $\vec{CA} = \vec{A} - \vec{C} = (1-4)\hat{i} + (3-5)\hat{j} + (5-k)\hat{k} = -3\hat{i} -2\hat{j} + (5-k)\hat{k}$

      $\vec{CB} = \vec{B} - \vec{C} = (2-4)\hat{i} + (4-5)\hat{j} + (6-k)\hat{k} = -2\hat{i} - \hat{j} + (6-k)\hat{k}$

      Dot product: $\vec{CA} \cdot \vec{CB} = (-3)(-2) + (-2)(-1) + (5-k)(6-k)$ …

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2. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
   Q.Suppose $L_1$ and $L_2$ are two lines having the direction ratios $1, -2, -2$ and $0, 2, 1$ respectively. If the direction cosines of a line perpendicular to both $L_1$ and $L_2$ are $l, m, n$ then $|l| + |m| + |n| =$ (A) $3$ (B) $\dfrac{5}{3}$ (C) $\sqrt{3}$ (D) $\dfrac{7}{3}$

   ›Reveal solutionSolution

   The line perpendicular to both given lines is parallel to the cross product of their direction vectors. Computing that cross product and normalising gives direction cosines whose absolute values sum to $\frac{7}{3}$.

   The key idea is geometric: a line perpendicular to two given lines is parallel to the vector that is perpendicular to both direction vectors — that is, their cross product. Once we have that vector, its direction cosines are just its components divided by its magnitude. The question asks for the sum of the absolute values of those cosines.

   Let’s work through it.

   1. Write the direction vectors.

      For $L_1$, direction ratios $1, -2, -2$ give the vector $\vec{a} = (1, -2, -2)$.

      For $L_2$, direction ratios $0, 2, 1$ give $\vec{b} = (0, 2, 1)$.

   2. Find a vector perpendicular to both.

      The cross product $\vec{a} \times \vec{b}$ is perpendicular to both. Compute:

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

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

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

   $= 2\hat{i} - 1\hat{j} + 2\hat{k}$

   So the vector is $(2, -1, 2)$.

   3. Find its magnitude. $|\vec{v}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$ …

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3. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If $\vec{a} = \hat{i} + p\hat{j} - 3\hat{k},\ \vec{b} = 2\hat{i} - 3\hat{j} + q\hat{k},\ \vec{c} = \hat{i} + 2\hat{j} + 2\hat{k}\ (p < 0, q > 0)$ are three vectors such that the magnitude of projection of $\vec{a}$ on $\vec{c}$ is $3$ and the magnitude of projection of $\vec{b}$ on $\vec{c}$ is $2$, then the magnitude of projection of $\vec{a}$ on $\vec{b}$ is (A) $\dfrac{11}{\sqrt{38}}$ (B) $\dfrac{34}{\sqrt{38}}$ (C) $\dfrac{7}{\sqrt{38}}$ (D) $\dfrac{16}{\sqrt{38}}$

   ›Reveal solutionSolution

   Use the projection formula $\text{proj}_{\vec{c}}(\vec{a}) = \frac{|\vec{a}\cdot\vec{c}|}{|\vec{c}|}$ to set up equations for $p$ and $q$, then compute $\frac{|\vec{a}\cdot\vec{b}|}{|\vec{b}|}$ to get the answer $\frac{7}{\sqrt{38}}$.

   The key idea is that the magnitude of projection of one vector onto another is simply the absolute value of their dot product divided by the length of the vector you’re projecting onto. That’s a direct, no-nonsense formula — no angles, no geometry beyond the dot product.

   We’re given three vectors:

   $\vec{a} = \hat{i} + p\hat{j} - 3\hat{k},\quad \vec{b} = 2\hat{i} - 3\hat{j} + q\hat{k},\quad \vec{c} = \hat{i} + 2\hat{j} + 2\hat{k}$

   with $p < 0$ and $q > 0$. The projections give us two equations to solve for $p$ and $q$. Once we have them, the third projection is straightforward.

   1. Magnitude of projection of $\vec{a}$ on $\vec{c}$ is $3$. The formula:

   $\frac{|\vec{a}\cdot\vec{c}|}{|\vec{c}|} = 3$

   Compute $\vec{a}\cdot\vec{c} = (1)(1) + (p)(2) + (-3)(2) = 1 + 2p - 6 = 2p - 5$.

   Compute $|\vec{c}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3$.

   So:

   $\frac{|2p - 5|}{3} = 3 \quad\Rightarrow\quad |2p - 5| = 9$

   This gives $2p - 5 = 9$ or $2p - 5 = -9$.

   • If $2p - 5 = 9$, then $2p = 14$, $p = 7$. But $p < 0$, so discard.
   • If $2p - 5 = -9$, then $2p = -4$, $p = -2$. This satisfies $p < 0$. Hence $p = -2$.

   2. Magnitude of projection of $\vec{b}$ on $\vec{c}$ is $2$.

   $\frac{|\vec{b}\cdot\vec{c}|}{|\vec{c}|} = 2$

   Compute $\vec{b}\cdot\vec{c} = (2)(1) + (-3)(2) + (q)(2) = 2 - 6 + 2q = 2q - 4$.

   $|\vec{c}| = 3$ as before. So:

   $\frac{|2q - 4|}{3} = 2 \quad\Rightarrow\quad |2q - 4| = 6$

   This gives $2q - 4 = 6$ or $2q - 4 = -6$.

   • If $2q - 4 = 6$, then $2q = 10$, $q = 5$. This satisfies $q > 0$.
   • If $2q - 4 = -6$, then $2q = -2$, $q = -1$. This violates $q > 0$, so discard. Hence $q = 5$. …

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4. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.Let A be a point having position vector $\vec{i} - 3\vec{j}$ and $\vec{r} = (\vec{i} - 3\vec{j}) + t(\vec{j} - 2\vec{k})$ be a line. If P is a point on this line and is at a minimum distance from the plane $\vec{r} \cdot (2\vec{i} + 3\vec{j} + 5\vec{k}) = 0$, then the equation of the plane through P and perpendicular to AP, is (A) $\vec{r} \cdot (-\vec{j} + 2\vec{k}) = 8$ (B) $\vec{r} \cdot (\vec{j} + \vec{k}) = 4$ (C) $\vec{r} \cdot (\vec{i} + \vec{j} + \vec{k}) = 8$ (D) $\vec{r} \cdot (\vec{i} - \vec{j}) = 12$

   ›Reveal solutionSolution

   The point P on the line that is closest to the given plane is found by projecting the line’s direction onto the plane’s normal; then the required plane through P perpendicular to AP has normal vector AP, and its equation matches option (B).

   ---

   Concept & Intuition

   We have a line and a plane. The point on the line that is closest to the plane is the one where the line’s direction is “parallel” to the plane — more precisely, where the vector from a point on the line to the plane is perpendicular to the line’s direction. That’s equivalent to saying the line’s direction vector is orthogonal to the plane’s normal at the point of minimum distance. Once we find P, we need the plane through P whose normal is AP (since it’s perpendicular to AP). Then we match its equation to the options.

   ---

   Step-by-step solution

   1. Identify given vectors

      Point A: $\vec{a} = \hat{i} - 3\hat{j}$

      Line: $\vec{r} = \vec{a} + t(\hat{j} - 2\hat{k})$, so direction vector $\vec{d} = \hat{j} - 2\hat{k}$.

      Plane: $\vec{r} \cdot (2\hat{i} + 3\hat{j} + 5\hat{k}) = 0$, so normal vector $\vec{n} = 2\hat{i} + 3\hat{j} + 5\hat{k}$.

   2. Condition for minimum distance from a point on the line to the plane

      The distance from a point $\vec{r}(t)$ on the line to the plane is

   $D(t) = \frac{|\vec{r}(t) \cdot \vec{n}|}{|\vec{n}|}$

   (since the plane passes through origin).

   Minimising $D(t)$ is equivalent to minimising $|\vec{r}(t) \cdot \vec{n}|$.

   The minimum occurs when the line is parallel to the plane at that point — i.e., when the direction vector $\vec{d}$ is perpendicular to $\vec{n}$. But here $\vec{d} \cdot \vec{n} = (0)(2) + (1)(3) + (-2)(5) = 3 - 10 = -7 \neq 0$, so the line is not parallel to the plane.

   The point of minimum distance is where the line’s position vector’s component along $\vec{n}$ is as small as possible in absolute value. That happens when the derivative of $\vec{r}(t) \cdot \vec{n}$ with respect to $t$ is zero? Actually, $\vec{r}(t) \cdot \vec{n} = \vec{a} \cdot \vec{n} + t (\vec{d} \cdot \vec{n})$ is linear in $t$. Its absolute value is minimised when the linear expression equals zero (if possible). So set:

   $\vec{a} \cdot \vec{n} + t (\vec{d} \cdot \vec{n}) = 0.$

   Compute:

   $\vec{a} \cdot \vec{n} = (1)(2) + (-3)(3) + (0)(5) = 2 - 9 = -7$.

   $\vec{d} \cdot \vec{n} = -7$ (as above).

   So equation: $-7 + t(-7) = 0 \Rightarrow -7(1 + t) = 0 \Rightarrow t = -1$.

   3. Find point P Substitute $t = -1$ into line equation:

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

   4. Determine the required plane The plane passes through P and is perpendicular to AP. So its normal vector is $\overrightarrow{AP} = \vec{p} - \vec{a}$. $$$$

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5. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.Let $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} - \hat{j} + \hat{k}$ be two vectors. If the vector $\vec{r} = x\hat{i} + y\hat{j} + 2\hat{k}$ is along the bisector of the angle between $\vec{a}$ and $\vec{b}$, then $|\vec{r}| =$ (A) $\sqrt{14}$ (B) $\sqrt{6}$ (C) $3$ (D) $7$

   ›Reveal solutionSolution

   Since $|\vec a| = |\vec b| = \sqrt6$, the bisector is along $\vec a + \vec b = (3,1,2)$; matching the given $z$-component $2$ gives $\vec r = (3,1,2)$ and $|\vec r| = \sqrt{14}$.

   Equal magnitudes.

   $|\vec a| = \sqrt{1^2+2^2+1^2} = \sqrt6, \qquad |\vec b| = \sqrt{2^2+(-1)^2+1^2} = \sqrt6.$

   Because the two vectors have equal length, the internal angle bisector is simply along their sum (no need to normalise separately):

   $\vec a + \vec b = (1+2,\;2-1,\;1+1) = (3,\,1,\,2).$ …

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6. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.If (2, -1, 3) is the foot of the perpendicular drawn from the origin to a plane, then the equation of that plane is (A) $2x + y - 3z + 6 = 0$ (B) $2x - y + 3z - 14 = 0$ (C) $2x - y + 3z - 13 = 0$ (D) $2x + y + 3z - 10 = 0$

   ›Reveal solutionSolution

   The foot of the perpendicular from the origin gives the normal vector and a point on the plane; using the point-normal form yields the plane equation $2x - y + 3z - 14 = 0$, which matches option (B).

   Concept & Intuition

   When the foot of the perpendicular from the origin to a plane is known, that foot is the point on the plane closest to the origin. The vector from the origin to that point is perpendicular to the plane — it is the plane’s normal vector. So we have both a normal vector and a point on the plane, which is exactly what we need to write the plane’s equation in point-normal form.

   Step-by-step solution

   1. Identify the normal vector

      The foot of the perpendicular from the origin to the plane is $(2, -1, 3)$. The vector from the origin to this point is $\vec{n} = (2, -1, 3)$. Since this line is perpendicular to the plane, $\vec{n}$ is a normal vector to the plane.

   2. Write the point-normal form

      For a plane with normal vector $(a, b, c)$ passing through point $(x_0, y_0, z_0)$, the equation is

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

   Here $(a, b, c) = (2, -1, 3)$ and the point is the foot itself: $(x_0, y_0, z_0) = (2, -1, 3)$.

   3. Substitute and simplify

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

   Expand:

   $2x - 4 - y - 1 + 3z - 9 = 0.$

   Combine constants: $-4 - 1 - 9 = -14$, so

   $2x - y + 3z - 14 = 0.$

   4. Match with options …

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7. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
   Q.The perpendicular distance from the origin to the plane containing the points having position vectors $\vec{i} + 2\vec{j} + 3\vec{k},\ 2\vec{i} + 3\vec{j} - 4\vec{k},\ 3\vec{i} - 4\vec{j} + 5\vec{k}$, is (A) $\dfrac{10}{\sqrt{60}}$ (B) $\dfrac{12}{\sqrt{30}}$ (C) $\dfrac{15}{\sqrt{127}}$ (D) $\dfrac{25}{\sqrt{57}}$

   ›Reveal solutionSolution

   Build the plane through the three points from a normal vector (cross product), then apply the origin-to-plane distance formula: the distance is $\dfrac{12}{\sqrt{30}}$, option (B).

   The three position vectors give the points $A(1,2,3)$, $B(2,3,-4)$, $C(3,-4,5)$. The perpendicular distance from the origin to their plane is $\dfrac{|d|}{|\vec n|}$, where $\vec n\cdot\vec r=d$ is the plane's equation.

   1. Two in-plane vectors.

   $\vec{AB}=(1,1,-7),\qquad \vec{AC}=(2,-6,2).$

   2. Normal vector $=\vec{AB}\times\vec{AC}$. $\vec n=\begin{vmatrix}\vec i&\vec j&\vec k\\1&1&-7\\2&-6&2\end{vmatrix} =\vec i\big(1\cdot2-(-7)(-6)\big)-\vec j\big(1\cdot2-(-7)\cdot2\big)+\vec k\big(1\cdot(-6)-1\cdot2\big).$ …

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8. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.If $\theta$ is the angle between the vectors $4\vec{i} - \vec{j} + 2\vec{k}$ and $\vec{i} + 3\vec{j} - 2\vec{k}$ then $\sin 2\theta =$ (A) $\dfrac{\sqrt{3}}{\sqrt{95}}$ (B) $-\dfrac{\sqrt{3}}{\sqrt{95}}$ (C) $-\dfrac{\sqrt{285}}{49}$ (D) $\dfrac{\sqrt{258}}{49}$

   ›Reveal solutionSolution

   The key idea is to compute $\sin 2\theta$ using the dot and cross products of the two vectors, then apply the double-angle identity. The final value is $-\dfrac{\sqrt{285}}{49}$, so the correct option is (C).

   ---

   We are given two vectors:

   $\vec{a} = 4\vec{i} - \vec{j} + 2\vec{k}$ and $\vec{b} = \vec{i} + 3\vec{j} - 2\vec{k}$.

   We need $\sin 2\theta$, where $\theta$ is the angle between them.

   Concept and intuition:

   To find $\sin 2\theta$, we can use $\sin 2\theta = 2 \sin\theta \cos\theta$.

   We can get $\cos\theta$ from the dot product and $\sin\theta$ from the magnitude of the cross product.

   This avoids needing to find $\theta$ itself — we just compute the necessary quantities directly.

   ---

   1. Compute the dot product

   $\vec{a} \cdot \vec{b} = (4)(1) + (-1)(3) + (2)(-2) = 4 - 3 - 4 = -3$

   2. Compute magnitudes

   $|\vec{a}| = \sqrt{4^2 + (-1)^2 + 2^2} = \sqrt{16 + 1 + 4} = \sqrt{21}$

   $|\vec{b}| = \sqrt{1^2 + 3^2 + (-2)^2} = \sqrt{1 + 9 + 4} = \sqrt{14}$

   3. Find $\cos\theta$

   $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|\,|\vec{b}|} = \frac{-3}{\sqrt{21} \cdot \sqrt{14}} = \frac{-3}{\sqrt{294}}$

   Simplify $\sqrt{294} = \sqrt{49 \cdot 6} = 7\sqrt{6}$, so

   $\cos\theta = \frac{-3}{7\sqrt{6}}$

   4. Find $\sin\theta$ using the cross product magnitude Compute $\vec{a} \times \vec{b}$:

   $\vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 4 & -1 & 2 \\ 1 & 3 & -2 \end{vmatrix}$

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

   $= \vec{i}(2 - 6) - \vec{j}(-8 - 2) + \vec{k}(12 + 1)$

   $= -4\vec{i} + 10\vec{j} + 13\vec{k}$

   Magnitude:

   $|\vec{a} \times \vec{b}| = \sqrt{(-4)^2 + 10^2 + 13^2} = \sqrt{16 + 100 + 169} = \sqrt{285}$

   Hence,

   $$$$

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9. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If $\vec{a} = 2\vec{i} + 3\mu \vec{j} - \vec{k}$, $\vec{b} = \mu \vec{i} - 2\vec{j} + 3\vec{k}$ and $\vec{c} = \vec{i} + 3\vec{j} - 2\mu \vec{k}$ are three vectors such that $\alpha \vec{a} + \beta \vec{b} + \gamma \vec{c} = \vec{0}$ only when $\alpha = \beta = \gamma = 0$, then the set of all real values of $\mu$ is (A) $\mathbb{R} - \left\{ 9, 1, -\dfrac{7}{6} \right\}$ (B) $\mathbb{R} - \left\{ 1 \right\}$ (C) $\mathbb{R} - \left\{ 1, -\dfrac{5}{3} \right\}$ (D) $\mathbb{R} - \left\{ 0 \right\}$

   ›Reveal solutionSolution

   The three vectors are linearly independent exactly when their scalar triple product $\Delta\ne0$. Here $\Delta=2(\mu-1)(3\mu^2+3\mu+10)$, whose only real root is $\mu=1$, so the set is $\mathbb{R}-\{1\}$ — option (B).

   Condition. "$\alpha\vec a+\beta\vec b+\gamma\vec c=\vec 0$ only when $\alpha=\beta=\gamma=0$" means $\vec a,\vec b,\vec c$ are linearly independent, i.e. their determinant (scalar triple product) is non-zero.

   Set up the determinant. With $\vec a=(2,3\mu,-1)$, $\vec b=(\mu,-2,3)$, $\vec c=(1,3,-2\mu)$,

   $\Delta=\begin{vmatrix}2 & 3\mu & -1\\\mu & -2 & 3\\1 & 3 & -2\mu\end{vmatrix}.$

   Expand along the first row.

   $\Delta=2\big[(-2)(-2\mu)-9\big]-3\mu\big[\mu(-2\mu)-3\big]+(-1)\big[3\mu+2\big]$

   $=2(4\mu-9)-3\mu(-2\mu^2-3)-(3\mu+2)$

   $=8\mu-18+6\mu^3+9\mu-3\mu-2=6\mu^3+14\mu-20.$

   Factor.

   $\Delta=2(3\mu^3+7\mu-10).$

   Testing $\mu=1$: $3+7-10=0$, so $(\mu-1)$ is a factor: …

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10. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
    Q.Let $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{b} = 2\hat{i} - 3\hat{j} + \hat{k}$ and $\vec{c} = 3\hat{i} + \hat{j} - 2\hat{k}$ be three vectors. If $\vec{r}$ is a vector such that $\vec{r}.\vec{a} = 0$, $\vec{r}.\vec{b} = -2$ and $\vec{r}.\vec{c} = 6$ then $\vec{r}.(3\hat{i} + \hat{j} + \hat{k}) =$ (A) $1$ (B) $0$ (C) $3$ (D) $2$

    ›Reveal solutionSolution

    $\vec{r}\cdot(3\hat{i}+\hat{j}+\hat{k}) = 3$ — option (C).

    Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. The conditions give:

    $\vec{r}\cdot\vec{a}=0:\quad x + 2y + 3z = 0,$

    $\vec{r}\cdot\vec{b}=-2:\quad 2x - 3y + z = -2,$

    $\vec{r}\cdot\vec{c}=6:\quad 3x + y - 2z = 6.$

    From the first equation $x = -2y - 3z$. Substituting:

    $2(-2y-3z) - 3y + z = -2 \;\Rightarrow\; -7y - 5z = -2 \;\Rightarrow\; 7y + 5z = 2,$

    $3(-2y-3z) + y - 2z = 6 \;\Rightarrow\; -5y - 11z = 6 \;\Rightarrow\; 5y + 11z = -6.$ …

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11. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
    Q.Consider the vectors $\vec{a} = 3\hat{i} + 5\hat{j} + 2\hat{k}$, $\vec{b} = 2\hat{i} - 3\hat{j} - 5\hat{k}$ and $\vec{c} = -5\hat{i} - 2\hat{j} + 3\hat{k}$. If $l, m$ and $n$ are length of projections of $a$ on $b$, $b$ on $c$ and $c$ on $a$ respectively, then (A) $l + m - n = 0$ (B) $l = m = n$ (C) $l - m + n = 0$ (D) $m + n - l = 0$

    ›Reveal solutionSolution

    We calculate the length of the projection of $\vec{a}$ on $\vec{b}$, $\vec{b}$ on $\vec{c}$, and $\vec{c}$ on $\vec{a}$ using the scalar projection formula. All three lengths turn out to be equal, so the correct option is (B).

    The length of the projection of one vector onto another is a fundamental concept in vector algebra. It represents the magnitude of the component of the first vector that lies along the direction of the second vector.

    To find the length of the projection of a vector $\vec{P}$ onto another vector $\vec{Q}$, we use the scalar projection formula. This formula essentially tells us how much of $\vec{P}$ "points in the direction of" $\vec{Q}$. The dot product $\vec{P} \cdot \vec{Q}$ gives us a measure of how much the vectors align, and dividing by the magnitude of $\vec{Q}$ normalizes this to give the component along $\vec{Q}$. Since we are looking for a "length", we take the absolute value of this scalar projection to ensure it's non-negative.

    The length of the projection of vector $\vec{P}$ onto vector $\vec{Q}$ is given by:

    $\text{Length of projection} = \frac{|\vec{P} \cdot \vec{Q}|}{|\vec{Q}|}$

    Let's apply this concept to find $l, m,$ and $n$.

    1. Identify the given vectors:

       We are given the three vectors:

       $\vec{a} = 3\hat{i} + 5\hat{j} + 2\hat{k}$

       $\vec{b} = 2\hat{i} - 3\hat{j} - 5\hat{k}$

       $\vec{c} = -5\hat{i} - 2\hat{j} + 3\hat{k}$

    2. Calculate $l$, the length of the projection of $\vec{a}$ on $\vec{b}$:

       First, calculate the dot product $\vec{a} \cdot \vec{b}$:

       $\vec{a} \cdot \vec{b} = (3)(2) + (5)(-3) + (2)(-5)$

       $\vec{a} \cdot \vec{b} = 6 - 15 - 10 = -19$

       Next, calculate the magnitude of $\vec{b}$:

       $|\vec{b}| = \sqrt{2^2 + (-3)^2 + (-5)^2}$

       $|\vec{b}| = \sqrt{4 + 9 + 25} = \sqrt{38}$

       Now, use the projection formula for $l$:

       $l = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{b}|} = \frac{|-19|}{\sqrt{38}} = \frac{19}{\sqrt{38}}$

    3. Calculate $m$, the length of the projection of $\vec{b}$ on $\vec{c}$:

       First, calculate the dot product $\vec{b} \cdot \vec{c}$:

       $\vec{b} \cdot \vec{c} = (2)(-5) + (-3)(-2) + (-5)(3)$

       $\vec{b} \cdot \vec{c} = -10 + 6 - 15 = -19$

       Next, calculate the magnitude of $\vec{c}$:

       $|\vec{c}| = \sqrt{(-5)^2 + (-2)^2 + 3^2}$

       $|\vec{c}| = \sqrt{25 + 4 + 9} = \sqrt{38}$

       Now, use the projection formula for $m$:

       $m = \frac{|\vec{b} \cdot \vec{c}|}{|\vec{c}|} = \frac{|-19|}{\sqrt{38}} = \frac{19}{\sqrt{38}}$

    4. Calculate $n$, the length of the projection of $\vec{c}$ on $\vec{a}$:

       First, calculate the dot product $\vec{c} \cdot \vec{a}$: …

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12. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
    Q.If the equation of the plane passing through the points $(2,1,2),\ (1,2,1)$ and perpendicular to the plane $2x-y+2z=1$ is $ax+by+cz+d=0$ then $\dfrac{a+b}{c+d}=$ (A) $0$ (B) $1$ (C) $-1$ (D) $2$

    ›Reveal solutionSolution

    The required plane is $x-z=0$, so $\dfrac{a+b}{c+d}=\dfrac{1}{-1}=-1$ — option (C).

    Direction lying in the plane. With $P(2,1,2),\ Q(1,2,1)$, the vector $\overrightarrow{PQ}=(-1,1,-1)$ lies in the plane.

    Normal of the given plane. $2x-y+2z=1\Rightarrow \mathbf{n}_1=(2,-1,2)$.

    Normal of the required plane. It must be perpendicular to both $\overrightarrow{PQ}$ (lies in the plane) and $\mathbf{n}_1$ (perpendicular planes have perpendicular normals):

    $\mathbf{n}=\mathbf{n}_1\times\overrightarrow{PQ}=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\2&-1&2\\-1&1&-1\end{vmatrix}=(-1,\,0,\,1).$ …

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