Q.The scalar product of the vector $\hat{i} + \hat{j} + \hat{k}$ with a unit vector along the sum of vectors $2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\lambda\hat{i} + 2\hat{j} + 3\hat{k}$ is equal to one. Find the value of $\lambda$.

Showing the 12 most recent of 46 on this concept.

1. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
   Q.The dot product of unit vectors $\hat{n}_1$ and $\hat{n}_2$ that are parallel to $5\hat{i} + 12\hat{j}$ and $3\hat{i} + 4\hat{j}$ respectively is (A) $\frac{63}{65}$ (B) $63$ (C) $\frac{63}{4225}$ (D) $\frac{63}{845}$

   ›Reveal solutionSolution

   This tests unit-vector construction and the dot product formula $\hat a\cdot\hat b=\cos\theta$; the answer is $\frac{63}{65}$.

   Concept and Intuition

   A unit vector just points in the same direction as the original vector but has magnitude 1 — you get

   it by dividing each component by the vector's own magnitude. Once both vectors are unit vectors,

   their dot product is simply $\cos\theta$ between them, computed the usual way: sum of the products

   of corresponding components.

   Step-by-Step Solution

   1. Magnitude of $5\hat i+12\hat j$: $\sqrt{5^2+12^2}=\sqrt{25+144}=\sqrt{169}=13$. So $\hat n_1 = \dfrac{5}{13}\hat i + \dfrac{12}{13}\hat j$.
   2. Magnitude of $3\hat i+4\hat j$: $\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$. So $\hat n_2 = \dfrac{3}{5}\hat i + \dfrac{4}{5}\hat j$. …

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2. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.Let $(\bar{a}, \bar{b})$ denote the angle between vectors $\bar{a}$ and $\bar{b}$. If $\bar{a} = 2\bar{i} + 3\bar{j} + 6\bar{k}$, $\bar{a}.\bar{b} = 4$ and $(\bar{a}, \bar{b}) = \cos^{-1}\left(\dfrac{4}{21}\right)$, then $\bar{a} + \bar{b} =$ (A) $3\bar{i} + \bar{j} + 8\bar{k}$ (B) $3\bar{i} + 5\bar{j} + 4\bar{k}$ (C) $3\bar{i} + 5\bar{j} + 8\bar{k}$ (D) $\bar{i} + \bar{j} + 8\bar{k}$

   ›Reveal solutionSolution

   Using $|\bar a+\bar b|^2=|\bar a|^2+2\bar a\cdot\bar b+|\bar b|^2$ pins down the magnitude of $\bar a+\bar b$, which uniquely identifies the matching option: $\bar i+\bar j+8\bar k$.

   Concept and Intuition

   Even without knowing $\bar b$ explicitly, the magnitude of $\bar a+\bar b$ can be computed purely from $|\bar a|$, $|\bar b|$, and $\bar a\cdot\bar b$ — this is the vector analogue of the law of cosines. Since only one answer option has that exact magnitude, it must be the correct vector sum (and it can be verified to be consistent with an actual vector $\bar b$).

   Step-by-Step Solution

   1. $\bar a=2\bar i+3\bar j+6\bar k\Rightarrow|\bar a|=\sqrt{4+9+36}=\sqrt{49}=7$.
   2. $\cos(\bar a,\bar b)=\dfrac{\bar a\cdot\bar b}{|\bar a||\bar b|}=\dfrac{4}{21}$. Substituting $\bar a\cdot\bar b=4$: $\dfrac{4}{7|\bar b|}=\dfrac{4}{21}\Rightarrow|\bar b|=3$.
   3. $|\bar a+\bar b|^2=|\bar a|^2+2(\bar a\cdot\bar b)+|\bar b|^2=49+2(4)+9=66$.
   4. Compute $|\cdot|^2$ for each option: (A) $9+1+64=74$; (B) $9+25+16=50$; (C) $9+25+64=98$; (D) $1+1+64=66$.
   5. Only (D) matches 66. …

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3. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.Let $\vec{a} = 4\vec{i} - \vec{j} + \alpha\vec{k}$ and $\vec{b} = \vec{i} + \alpha\vec{j} - 4\vec{k}$ be two vectors. If $\alpha_1, \alpha_2$ ($\alpha_1 < \alpha_2$) are two different values of $\alpha$ such that $(\vec{a}, \vec{b}) = \cos^{-1}\left(-\dfrac{2}{7}\right)$, then $\alpha_1 + 2\alpha_2 =$ (A) $15$ (B) $24$ (C) $33$ (D) $52$

   ›Reveal solutionSolution

   Setting up $\cos\theta=\vec a\cdot\vec b/(|\vec a||\vec b|)=-2/7$ gives a quadratic in $\alpha$ with roots $2$ and $15.5$; then $\alpha_1+2\alpha_2=33$.

   Concept and Intuition

   Both vectors have the same magnitude expression in $\alpha$ (a nice simplification to notice first), which keeps the resulting equation a clean single-variable quadratic instead of something messier.

   Step-by-Step Solution

   1. $\vec a\cdot\vec b=4(1)+(-1)(\alpha)+\alpha(-4)=4-\alpha-4\alpha=4-5\alpha$.
   2. $|\vec a|=\sqrt{16+1+\alpha^2}=\sqrt{17+\alpha^2}$ and $|\vec b|=\sqrt{1+\alpha^2+16}=\sqrt{17+\alpha^2}$ — identical.
   3. So $\cos\theta=\dfrac{4-5\alpha}{17+\alpha^2}=-\dfrac27$.
   4. Cross-multiply: $7(4-5\alpha)=-2(17+\alpha^2)\Rightarrow28-35\alpha=-34-2\alpha^2\Rightarrow2\alpha^2-35\alpha+62=0$. …

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4. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.Let L be the line passing through the points $\bar{i}-9\bar{k}$ and $7\bar{j}+\bar{k}$ and $\pi$ be the plane passing through the point $6\bar{i}+\bar{j}$ and perpendicular to the vector $\bar{i}+\bar{j}+\bar{k}$. If $\theta$ is the angle between L and $\pi$, then $\sin\theta =$ (A) $\dfrac{8\sqrt{2}}{15}$ (B) $\dfrac{3\sqrt{3}}{8}$ (C) $\dfrac{7}{13}$ (D) $\dfrac{24}{25}$

   ›Reveal solutionSolution

   This tests the line–plane angle formula $\sin\theta=\dfrac{|\vec d\cdot\bar n|}{|\vec d||\bar n|}$ using $L$'s direction vector and $\pi$'s normal; the answer is $\dfrac{8\sqrt2}{15}$.

   Concept and Intuition

   The angle between a line and a plane is measured from the line to its projection on the plane, so it uses sine, not cosine — because the plane's normal is perpendicular to the plane itself. If $\phi$ is the angle between the line's direction $\vec d$ and the normal $\bar n$, then $\theta = 90^\circ-\phi$, so $\sin\theta=\cos\phi=\dfrac{|\vec d\cdot\bar n|}{|\vec d||\bar n|}$.

   Step-by-Step Solution

   1. Direction of $L$: $\vec d = (7\bar j+\bar k)-(\bar i-9\bar k) = -\bar i+7\bar j+10\bar k$.
   2. The plane is perpendicular to $\bar i+\bar j+\bar k$, so this vector IS the plane's normal $\bar n$ — the point $6\bar i+\bar j$ is not needed for the angle.
   3. $\vec d\cdot\bar n = (-1)(1)+(7)(1)+(10)(1)=16$.
   4. $|\vec d|=\sqrt{(-1)^2+7^2+10^2}=\sqrt{150}=5\sqrt6$, and $|\bar n|=\sqrt{3}$. …

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5. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.Let $\vec{a},\vec{b},\vec{c}$ be three vectors such that $\vec{a}$ is perpendicular to $\vec{b}$ and $\vec{b}$ is perpendicular to $\vec{c}$. If $|\vec{a}|=2, |\vec{b}|=3, |\vec{c}|=5$ and $|\vec{a}+\vec{b}+\vec{c}|=4\sqrt3$, then the angle between $\vec{a}$ and $\vec{c}$ is (A) $\cos^{-1}\left(\dfrac{2}{5}\right)$ (B) $\dfrac{\pi}{3}$ (C) $\cos^{-1}\left(\dfrac{2}{3}\right)$ (D) $\dfrac{\pi}{6}$

   ›Reveal solutionSolution

   Expand $|\vec a+\vec b+\vec c|^2$; the perpendicularity conditions kill two of the three cross terms, leaving $\vec a\cdot\vec c$ to solve for. Answer: (B).

   Concept and Intuition

   Squaring a vector sum brings out all pairwise dot products; when some pairs are given as perpendicular, those dot-product terms vanish, isolating the one unknown dot product — here $\vec a\cdot\vec c$, which directly gives the angle between $\vec a$ and $\vec c$.

   Step-by-Step Solution

   1. $|\vec a+\vec b+\vec c|^2=|\vec a|^2+|\vec b|^2+|\vec c|^2+2\vec a\cdot\vec b+2\vec b\cdot\vec c+2\vec a\cdot\vec c$.
   2. Since $\vec a\perp\vec b$, $\vec a\cdot\vec b=0$; since $\vec b\perp\vec c$, $\vec b\cdot\vec c=0$.
   3. So $|\vec a+\vec b+\vec c|^2=4+9+25+2\vec a\cdot\vec c=38+2\vec a\cdot\vec c$.
   4. Given $|\vec a+\vec b+\vec c|=4\sqrt3\Rightarrow|\vec a+\vec b+\vec c|^2=48$. So $38+2\vec a\cdot\vec c=48\Rightarrow\vec a\cdot\vec c=5$. …

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6. AP EAPCET 2023Set ap-2023-05-22-FN1 markMCQ
   Q.The angle between two vectors $(\hat{i}+\hat{j})$ and $(\hat{j}+\hat{k})$ is (A) $60^{\circ}$ (B) $30^{\circ}$ (C) $45^{\circ}$ (D) $90^{\circ}$

   ›Reveal solutionSolution

   A direct application of the dot-product formula for the angle between two vectors. Answer: (A) $60^{\circ}$.

   Concept and Intuition

   The angle between two vectors can be found from $\vec{A}\cdot\vec{B} = |\vec{A}||\vec{B}|\cos\theta$. Writing each vector in component form and computing the dot product and magnitudes directly gives $\cos\theta$, from which $\theta$ follows.

   Step-by-Step Solution

   1. Write $\vec{A} = (1,1,0)$ and $\vec{B} = (0,1,1)$.
   2. Compute the dot product: $\vec{A}\cdot\vec{B} = (1)(0)+(1)(1)+(0)(1) = 1$.
   3. Compute magnitudes: $|\vec{A}| = \sqrt{1^{2}+1^{2}+0^{2}} = \sqrt{2}$, similarly $|\vec{B}| = \sqrt{2}$.
   4. Apply the formula: $\cos\theta = \dfrac{1}{\sqrt{2}\cdot\sqrt{2}} = \dfrac{1}{2}$. …

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7. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
   Q.The value of $\dfrac{(\vec{a} \times \vec{b})^2 + (\vec{a} \cdot \vec{b})^2}{2(\vec{a})^2(\vec{b})^2}$ is (A) $0$ (B) $1$ (C) $\dfrac{1}{2}$ (D) $\dfrac{1}{4}$

   ›Reveal solutionSolution

   The identity $|\vec a\times\vec b|^2+(\vec a\cdot\vec b)^2 = |\vec a|^2|\vec b|^2$ makes the given ratio collapse instantly to $\frac12$, independent of the actual vectors.

   Concept and Intuition

   The cross-product magnitude captures the $\sin\theta$ part of the angle between two vectors, while the dot product captures the $\cos\theta$ part. Squaring and adding them recovers $a^2b^2(\sin^2\theta+\cos^2\theta)=a^2b^2$ — a clean Pythagorean-style identity that eliminates the angle entirely.

   Step-by-Step Solution

   1. Recall $|\vec a\times\vec b| = |\vec a||\vec b|\sin\theta$ and $\vec a\cdot\vec b = |\vec a||\vec b|\cos\theta$, where $\theta$ is the angle between $\vec a$ and $\vec b$.
   2. Square both: $(\vec a\times\vec b)^2 = a^2b^2\sin^2\theta$ and $(\vec a\cdot\vec b)^2=a^2b^2\cos^2\theta$. …

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8. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.If $\theta$ is the angle between $\overline{f} = i + 2j - 3k$ and $\overline{g} = 2i - 3j + ak$ and $\sin\theta = \sqrt{\dfrac{24}{28}}$ then $7a^2 + 24a =$ (A) $10$ (B) $12$ (C) $36$ (D) $15$

   ›Reveal solutionSolution

   This tests using $\cos\theta=\dfrac{\overline f\cdot\overline g}{|\overline f||\overline g|}$ together with $\sin^2\theta+\cos^2\theta=1$ to build an equation in the unknown $a$. Answer: $7a^2+24a=10$.

   Concept and Intuition

   Given $\sin\theta$ between two vectors, first get $\cos^2\theta$ from the Pythagorean identity, then equate $\cos^2\theta$ to $\left(\dfrac{\overline f\cdot\overline g}{|\overline f||\overline g|}\right)^2$ — squaring avoids sign ambiguity and leaves a clean polynomial equation in $a$.

   Step-by-Step Solution

   1. $\overline f=i+2j-3k$, $\overline g=2i-3j+ak$.
   2. $\overline f\cdot\overline g=(1)(2)+(2)(-3)+(-3)(a)=2-6-3a=-(4+3a)$.
   3. $|\overline f|^2=1+4+9=14$. $|\overline g|^2=4+9+a^2=13+a^2$.
   4. Given $\sin^2\theta=\dfrac{24}{28}=\dfrac67$, so $\cos^2\theta=1-\dfrac67=\dfrac17$. …

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9. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.Let $\bar{a}, \bar{b}$ be two unit vector. If $\bar{c} = \bar{a} + 2\bar{b}$ and $\bar{d} = 5\bar{a} - 4\bar{b}$ are perpendicular to each other, then the angle between $\bar{a}$ and $\bar{b}$ is (A) $\dfrac{\pi}{6}$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{\pi}{8}$

   ›Reveal solutionSolution

   Expand the perpendicularity condition $\bar c\cdot\bar d=0$ to isolate $\bar a\cdot\bar b$.

   Concept and Intuition

   Two vectors are perpendicular exactly when their dot product is zero. Expanding the dot product of linear combinations of unit vectors reduces everything to the single unknown $\bar a\cdot\bar b=\cos\theta$.

   Step-by-Step Solution

   1. $\bar c\cdot\bar d = (\bar a+2\bar b)\cdot(5\bar a-4\bar b) = 5(\bar a\cdot\bar a) -4(\bar a\cdot\bar b) + 10(\bar b\cdot\bar a) -8(\bar b\cdot\bar b)$.
   2. Since $|\bar a|=|\bar b|=1$: $=5(1) + 6(\bar a\cdot\bar b) - 8(1) = 6(\bar a\cdot\bar b) - 3$.
   3. Set to zero: $6(\bar a\cdot\bar b)=3\Rightarrow \bar a\cdot\bar b = \dfrac12$. …

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10. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
    Q.If the points A, B, C, D with position vectors $\bar{i}+\bar{j}-\bar{k}$, $\bar{i}-\bar{j}+2\bar{k}$, $\bar{i}-2\bar{j}+\bar{k}$, $2\bar{i}+\bar{j}+\bar{k}$ respectively form a tetrahedron, then the angle between the faces ABC and ABD of the tetrahedron is (A) $\mathrm{Cos}^{-1}\left(\dfrac{-4}{\sqrt{29}}\right)$ (B) $\mathrm{Cos}^{-1}\left(\dfrac{-4}{5}\right)$ (C) $\mathrm{Cos}^{-1}\left(\dfrac{3}{5}\right)$ (D) $\mathrm{Cos}^{-1}\left(\dfrac{\sqrt{29}}{\sqrt{33}}\right)$

    ›Reveal solutionSolution

    The angle between the two triangular faces sharing edge AB equals the angle between their normal vectors, computed via a pair of cross products as $\cos^{-1}\left(\dfrac{-4}{\sqrt{29}}\right)$.

    Concept and Intuition

    The dihedral angle between two planes meeting along a common edge can be found from the angle between their normal vectors (normals are perpendicular to their respective planes, so the angle between normals directly reflects the angle between the planes, up to sign conventions).

    Step-by-Step Solution

    1. Position vectors: $A=(1,1,-1)$, $B=(1,-1,2)$, $C=(1,-2,1)$, $D=(2,1,1)$.
    2. Compute edge vectors from $A$: $\overrightarrow{AB}=B-A=(0,-2,3)$, $\overrightarrow{AC}=C-A=(0,-3,2)$, $\overrightarrow{AD}=D-A=(1,0,2)$.
    3. Normal to face $ABC$: $\bar n_1=\overrightarrow{AB}\times\overrightarrow{AC} = \begin{vmatrix}\bar i&\bar j&\bar k\\0&-2&3\\0&-3&2\end{vmatrix} = \bar i[(-2)(2)-(3)(-3)] - \bar j[(0)(2)-(3)(0)]+\bar k[(0)(-3)-(-2)(0)] = \bar i(-4+9) - \bar j(0) + \bar k(0) = (5,0,0)$.
    4. Normal to face $ABD$: $\bar n_2=\overrightarrow{AB}\times\overrightarrow{AD} = \begin{vmatrix}\bar i&\bar j&\bar k\\0&-2&3\\1&0&2\end{vmatrix} = \bar i[(-2)(2)-(3)(0)] - \bar j[(0)(2)-(3)(1)]+\bar k[(0)(0)-(-2)(1)] = \bar i(-4) - \bar j(-3)+\bar k(2) = (-4,3,2)$. …

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11. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
    Q.If $\bar{a}, \bar{b}, \bar{c}$ are 3 vectors such that $|\bar{a}|=5, |\bar{b}|=8, |\bar{c}|=11$ and $\bar{a}+\bar{b}+\bar{c}=\bar{0}$ then the angle between the vectors $\bar{a}$ and $\bar{b}$ is (A) $\cos^{-1}\dfrac{2}{5}$ (B) $\cos^{-1}\dfrac{10}{11}$ (C) $\cos^{-1}\dfrac{41}{55}$ (D) $\dfrac{\pi}{3}$

    ›Reveal solutionSolution

    From $\bar c=-(\bar a+\bar b)$, $|\bar c|^2=|\bar a|^2+|\bar b|^2+2\,\bar a\!\cdot\!\bar b$ gives $\cos\theta=\tfrac25$.

    Since $\bar a+\bar b+\bar c=\bar 0$, we have $\bar c=-(\bar a+\bar b)$, so

    $|\bar c|^2=|\bar a+\bar b|^2=|\bar a|^2+|\bar b|^2+2|\bar a||\bar b|\cos\theta,$

    where $\theta$ is the angle between $\bar a$ and $\bar b$.

    Substituting $|\bar a|=5,\ |\bar b|=8,\ |\bar c|=11$:

    $121 = 25 + 64 + 2(5)(8)\cos\theta = 89 + 80\cos\theta.$ …

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12. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
    Q.The values of x for which the angle between the vectors $x^2\bar i+2x\bar j+\bar k$ and $\bar i-2\bar j+x\bar k$ is obtuse lie in the interval (A) $(-\infty,0)\cup(3,\infty)$ (B) $(0,3)$ (C) $[0,3]$ (D) $(-\infty,0)\cup[3,\infty)$

    ›Reveal solutionSolution

    An angle between two vectors is obtuse exactly when their dot product is negative; solving the resulting quadratic inequality gives $x\in(0,3)$.

    Concept and Intuition

    $\cos\theta = \dfrac{\vec u\cdot\vec v}{|\vec u||\vec v|}$ is negative iff $\theta$ is obtuse (strictly between $90^\circ$ and $180^\circ$), provided the vectors are not anti-parallel (which would make $\theta=180^\circ$, not obtuse).

    Step-by-Step Solution

    1. Dot product: $(x^2)(1)+(2x)(-2)+(1)(x) = x^2-4x+x = x^2-3x$.
    2. Require $x^2-3x<0 \Rightarrow x(x-3)<0 \Rightarrow 0<x<3$. …

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