Q.State True or False: If $|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|$, then the vectors $\vec{a}$ and $\vec{b}$ are orthogonal.

Showing the 12 most recent of 19 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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6. 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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7. 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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8. 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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9. 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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10. 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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11. 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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12. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
    Q.$y = f(x)$ and $x = g(y)$ are two curves and P$(x, y)$ is a common point of the two curves. If at P, on the curve $y = f(x)$, $\dfrac{dy}{dx} = Q(x)$ and at the same point P on the curve $x = g(y)$, $\dfrac{dx}{dy} = -Q(x)$, then (A) the two curves have a common tangent (B) the angle between two curves is $45^\circ$ (C) tangent drawn at P to one curve is normal to the other curve at P (D) the two curves never intersect orthogonally

    ›Reveal solutionSolution

    The key idea is that the slopes of the two curves at the common point are negative reciprocals, which means the tangent of one is the normal of the other. The correct option is (C).

    We are given two curves: $y = f(x)$ and $x = g(y)$. At a common point $P(x, y)$, the derivative on the first curve is $\frac{dy}{dx} = Q(x)$. On the second curve, the derivative is $\frac{dx}{dy} = -Q(x)$.

    The concept here is the relationship between the slopes of tangents and normals. If two lines are perpendicular, their slopes multiply to $-1$. But here we have a twist: one derivative is with respect to $x$, the other with respect to $y$. We must interpret what $\frac{dx}{dy}$ means geometrically.

    ---

    1. Interpret the slopes at P On the curve $y = f(x)$, the slope of the tangent at P is

    $m_1 = \frac{dy}{dx} = Q(x).$

    On the curve $x = g(y)$, the derivative $\frac{dx}{dy}$ gives the slope of the tangent when $x$ is treated as a function of $y$. But the slope of the tangent in the usual $xy$-plane (rise over run) is $\frac{dy}{dx}$. For the second curve, we have

    $\frac{dx}{dy} = -Q(x).$

    Therefore, the slope of the tangent to the second curve at P is

    $m_2 = \frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{-Q(x)} = -\frac{1}{Q(x)}.$

    This is valid provided $Q(x) \neq 0$.

    2. Compare the two slopes We now have:

    $m_1 = Q(x), \quad m_2 = -\frac{1}{Q(x)}.$

    Their product is

    $m_1 \cdot m_2 = Q(x) \cdot \left(-\frac{1}{Q(x)}\right) = -1.$

    This is the condition for two lines to be perpendicular.

    3. Geometric meaning …

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