Q.Show that $|\vec{a}|\vec{b}+|\vec{b}|\vec{a}$ is perpendicular to $|\vec{a}|\vec{b}-|\vec{b}|\vec{a},$ for any two nonzero vectors $\vec{a}$ and $\vec{b}.$

Showing the 12 most recent of 19 on this concept.

1. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.Let $\vec{a} = 2\vec{i} - \vec{j} + \vec{k}$, $\vec{b} = 2\vec{j} - 3\vec{k}$. If $\vec{b} = \vec{c} - \vec{d}$, $\vec{a}$ is parallel to $\vec{c}$ and perpendicular to $\vec{d}$, then $\vec{c} + \vec{d} =$ (A) $-\dfrac{1}{6}(2\vec{a} + 5\vec{b})$ (B) $\dfrac{1}{3}(3\vec{a} + 5\vec{b})$ (C) $\dfrac{1}{6}(5\vec{a} + 2\vec{b})$ (D) $-\dfrac{1}{3}(5\vec{a} + 3\vec{b})$

   ›Reveal solutionSolution

   We use the conditions for parallel and perpendicular vectors to express the unknown vectors $\vec{c}$ and $\vec{d}$ in terms of $\vec{a}$ and $\vec{b}$, then sum them. The result is $\boxed{-\dfrac{1}{3}(5\vec{a} + 3\vec{b})}$.

   The problem asks us to find the sum of two unknown vectors, $\vec{c}$ and $\vec{d}$, given their relationship with known vectors $\vec{a}$ and $\vec{b}$, and specific conditions about their orientation. The core idea is to translate the geometric conditions (parallelism and perpendicularity) into algebraic equations using scalar multiplication and the dot product. This allows us to express $\vec{c}$ and $\vec{d}$ in terms of $\vec{a}$ and $\vec{b}$ and then find their sum.

   Here's how we approach this:

   1. Understand Parallel Vectors: If two non-zero vectors $\vec{u}$ and $\vec{v}$ are parallel, it means they point in the same or opposite direction. Mathematically, this is expressed as $\vec{u} = k\vec{v}$ for some non-zero scalar $k$.
   2. Understand Perpendicular Vectors: If two non-zero vectors $\vec{u}$ and $\vec{v}$ are perpendicular (orthogonal), their dot product is zero. Mathematically, this is expressed as $\vec{u} \cdot \vec{v} = 0$.
   3. Use the given relationships: We are given $\vec{b} = \vec{c} - \vec{d}$. This equation connects $\vec{c}$ and $\vec{d}$ to $\vec{b}$.
   4. Combine conditions: We will use the parallel condition to express $\vec{c}$ in terms of $\vec{a}$ and an unknown scalar. Then, we'll use the given vector equation to express $\vec{d}$ in terms of $\vec{a}$, $\vec{b}$, and the same unknown scalar. Finally, the perpendicular condition will allow us to solve for this scalar.

   Let's work through the steps:

   1. Express $\vec{c}$ using the parallel condition: We are given that $\vec{a}$ is parallel to $\vec{c}$. This means $\vec{c}$ must be a scalar multiple of $\vec{a}$. Let this scalar be $k$.

   $\vec{c} = k\vec{a}$

   Here, $k$ is an unknown scalar that we need to determine.
   

   2. Express $\vec{d}$ in terms of $\vec{a}$, $\vec{b}$, and $k$:

   We are given the relation $\vec{b} = \vec{c} - \vec{d}$.

   We can rearrange this to find $\vec{d}$:

   $\vec{d} = \vec{c} - \vec{b}$

   Now, substitute the expression for $\vec{c}$ from Step 1:
   

   $\vec{d} = k\vec{a} - \vec{b}$

   3. Use the perpendicular condition to find $k$: We are given that $\vec{a}$ is perpendicular to $\vec{d}$. This means their dot product is zero:

   $\vec{a} \cdot \vec{d} = 0$

   Substitute the expression for $\vec{d}$ from Step 2:
   

   $\vec{a} \cdot (k\vec{a} - \vec{b}) = 0$

   Using the distributive property of the dot product:
   

   $k(\vec{a} \cdot \vec{a}) - (\vec{a} \cdot \vec{b}) = 0$

   4. Calculate the necessary dot products: We are given $\vec{a} = 2\vec{i} - \vec{j} + \vec{k}$ and $\vec{b} = 2\vec{j} - 3\vec{k}$. First, calculate $\vec{a} \cdot \vec{a}$:

   $\vec{a} \cdot \vec{a} = (2)(2) + (-1)(-1) + (1)(1) = 4 + 1 + 1 = 6$

   Next, calculate $\vec{a} \cdot \vec{b}$: …
   

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2. 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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3. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If $\vec{r} = 2\vec{i} - \vec{j} + 2\vec{k}$, $\vec{s} = 3\vec{i} - 3\vec{j} + 3\vec{k}$, $\vec{t} = \vec{i} + 2\vec{j} + \vec{k}$ are three vectors and $\vec{a}$ is a vector such that $\vec{s} \times \vec{a} = \vec{r} \times \vec{a}$ and $|\vec{t} \times \vec{a}| = \sqrt{128}$, then $|\vec{t} \cdot \vec{a}| =$ (A) 3 (B) 6 (C) 4 (D) 8

   ›Reveal solutionSolution

   $\vec a$ is parallel to $\vec s-\vec r$; this gives $|\vec t\cdot\vec a| = 4$ (C).

   The condition $\vec s\times\vec a=\vec r\times\vec a$ gives $(\vec s-\vec r)\times\vec a=\vec 0$, so $\vec a$ is parallel to

   $\vec s-\vec r=(3-2,\,-3+1,\,3-2)=(1,-2,1).$

   Write $\vec a=k\,(1,-2,1)$. With $\vec t=(1,2,1)$:

   $\vec t\times(1,-2,1)=\big(2\cdot1-1\cdot(-2),\ -(1\cdot1-1\cdot1),\ 1\cdot(-2)-2\cdot1\big)=(4,0,-4).$

   So $|\vec t\times\vec a|=|k|\,\sqrt{16+16}=|k|\sqrt{32}$. Given $|\vec t\times\vec a|=\sqrt{128}$: …

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4. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.$\vec{a}, \vec{b}, \vec{c}$ are three non-coplanar and mutually perpendicular vectors of same magnitude K. $\vec{r}$ is any vector satisfying $\vec{a} \times ((\vec{r} - \vec{b}) \times \vec{a}) + \vec{b} \times ((\vec{r} - \vec{c}) \times \vec{b}) + \vec{c} \times ((\vec{r} - \vec{a}) \times \vec{c}) = \vec{0}$, then $\vec{r} =$ (A) $\dfrac{\vec{a} + \vec{b} + \vec{c}}{K^2 + 1}$ (B) $\dfrac{K^2 (\vec{a} + \vec{b} + \vec{c})}{3K^2 - 1}$ (C) $\dfrac{K(\vec{a} + \vec{b} + \vec{c})}{K + 1}$ (D) $\dfrac{\vec{a} + \vec{b} + \vec{c}}{2}$

   ›Reveal solutionSolution

   Expand each vector triple product with the BAC-CAB rule; orthogonality of $\vec{a},\vec{b},\vec{c}$ (each of magnitude $K$) reduces the equation to $2K^2\vec{r}=K^2(\vec{a}+\vec{b}+\vec{c})$, so $\vec{r}=\frac{1}{2}(\vec{a}+\vec{b}+\vec{c})$.

   Concept — vector triple product. For any vectors, $\vec{a} \times (\vec{u} \times \vec{a}) = (\vec{a}\cdot\vec{a})\,\vec{u} - (\vec{a}\cdot\vec{u})\,\vec{a}$.

   Step 1 — expand each term. With $|\vec{a}|=|\vec{b}|=|\vec{c}|=K$:

   $\vec{a}\times((\vec{r}-\vec{b})\times\vec{a}) = K^2(\vec{r}-\vec{b}) - (\vec{a}\cdot(\vec{r}-\vec{b}))\vec{a}$

   and similarly for the $\vec{b}$ and $\vec{c}$ terms.

   Step 2 — add the three terms. Since $\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{c}=\vec{c}\cdot\vec{a}=0$, the dot products of one vector with another vanish, leaving

   $K^2\big[3\vec{r}-(\vec{a}+\vec{b}+\vec{c})\big] - \big[(\vec{a}\cdot\vec{r})\vec{a}+(\vec{b}\cdot\vec{r})\vec{b}+(\vec{c}\cdot\vec{r})\vec{c}\big] = \vec{0}.$ …

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5. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.$\vec{a}, \vec{b}, \vec{c}$ are three non-coplanar and mutually perpendicular vectors of same magnitude K. $\vec{r}$ is any vector satisfying $\vec{a} \times ((\vec{r} - \vec{b}) \times \vec{a}) + \vec{b} \times ((\vec{r} - \vec{c}) \times \vec{b}) + \vec{c} \times ((\vec{r} - \vec{a}) \times \vec{c}) = \vec{0}$, then $\vec{r} =$ (A) $\dfrac{K^2 (\vec{a} + \vec{b} + \vec{c})}{3K^2 - 1}$ (B) $\dfrac{\vec{a} + \vec{b} + \vec{c}}{2}$ (C) $\dfrac{K(\vec{a} + \vec{b} + \vec{c})}{K+1}$ (D) $\dfrac{\vec{a} + \vec{b} + \vec{c}}{K^2 + 1}$

   ›Reveal solutionSolution

   BAC–CAB expansion + orthogonality collapses the equation to $2K^2\vec{r} = K^2(\vec a + \vec b + \vec c)$, so $\vec r = \dfrac{\vec a + \vec b + \vec c}{2}$ — option (B).

   Concept. Use the vector triple-product identity $\vec{A}\times(\vec{B}\times\vec{C}) = \vec{B}(\vec{A}\cdot\vec{C}) - \vec{C}(\vec{A}\cdot\vec{B})$, plus the facts that $\vec a, \vec b, \vec c$ are mutually perpendicular ($\vec a\cdot\vec b = \vec b\cdot\vec c = \vec c\cdot\vec a = 0$) with $|\vec a| = |\vec b| = |\vec c| = K$, and that they form an orthogonal basis: any $\vec r$ satisfies $\vec a(\vec a\cdot\vec r) + \vec b(\vec b\cdot\vec r) + \vec c(\vec c\cdot\vec r) = K^2\vec r$.

   Step 1 — expand one term.

   $\vec a \times ((\vec r - \vec b)\times\vec a) = (\vec r - \vec b)(\vec a\cdot\vec a) - \vec a\,(\vec a\cdot(\vec r - \vec b)) = K^2\vec r - K^2\vec b - \vec a(\vec a\cdot\vec r),$

   using $\vec a\cdot\vec b = 0$.

   Step 2 — the cyclic sum. Similarly,

   $\vec b\times((\vec r-\vec c)\times\vec b) = K^2\vec r - K^2\vec c - \vec b(\vec b\cdot\vec r), \qquad \vec c\times((\vec r-\vec a)\times\vec c) = K^2\vec r - K^2\vec a - \vec c(\vec c\cdot\vec r).$

   Adding all three and setting the sum to $\vec 0$: …

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6. 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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7. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
   Q.The vector in the direction of the sum of the vectors $\vec{a} = 2\hat{i} - 2\hat{j} + 5\hat{k}$ and $\vec{b} = -2\hat{i} + 5\hat{j} - 3\hat{k}$ is (A) Perpendicular to ZX - plane (B) Parallel to ZX - plane (C) Parallel to YZ - plane (D) Perpendicular to YZ - plane

   ›Reveal solutionSolution

   The sum is $3\hat{j}+2\hat{k}$ (no $\hat{i}$ term), so it lies in and is parallel to the YZ-plane — option (C).

   Add the vectors component-wise:

   $\vec{a}+\vec{b}=(2-2)\hat{i}+(-2+5)\hat{j}+(5-3)\hat{k}=0\,\hat{i}+3\,\hat{j}+2\,\hat{k}.$ …

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8. 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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9. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.$\vec{a}, \vec{b}, \vec{c}$ are three vectors such that $|\vec{a}| = 3, |\vec{b}| = 2\sqrt{2}, |\vec{c}| = 5$ and $\vec{c}$ is perpendicular to the plane of $\vec{a}$ and $\vec{b}$. If the angle between the vectors $\vec{a}$ and $\vec{b}$ is $\dfrac{\pi}{4}$ then

   [!FORMULA] $|\vec{a} + \vec{b} + \vec{c}| =$

   (A) $5\sqrt{3}$ (B) $2\sqrt{5}$ (C) $10$ (D) $3\sqrt{6}$

   ›Reveal solutionSolution

   Since $\vec c\perp$ plane of $\vec a,\vec b$, the cross terms with $\vec c$ vanish and $\vec a\cdot\vec b = 6$. Then $|\vec a+\vec b+\vec c|^2 = 9+8+25+2(6)=54$, so the magnitude is $3\sqrt6$, option (D).

   Expand the squared magnitude

   $|\vec a+\vec b+\vec c|^2 = |\vec a|^2+|\vec b|^2+|\vec c|^2 + 2\left(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a\right).$

   Evaluate the pieces

   Magnitudes:

   $|\vec a|^2 = 9,\qquad |\vec b|^2 = (2\sqrt2)^2 = 8,\qquad |\vec c|^2 = 25.$

   Because $\vec c$ is perpendicular to the plane of $\vec a$ and $\vec b$, it is perpendicular to both:

   $\vec b\cdot\vec c = 0,\qquad \vec c\cdot\vec a = 0.$

   The angle between $\vec a$ and $\vec b$ is $\dfrac{\pi}{4}$: …

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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 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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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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