Q.Find the angle between the vectors $\hat{i}-2\hat{j}+3\hat{k}$ and $3\hat{i}-2\hat{j}+\hat{k}$.

Showing the 12 most recent of 46 on this concept.

1. 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: $\cos\theta = \dfrac{1}{\sqrt{2}\cdot\sqrt{2}} = \dfrac{1}{2}$.
   5. So $\theta = \cos^{-1}(1/2) = 60^{\circ}$.

   Common Mistakes

   • Miscomputing the dot product by forgetting that $\hat{i},\hat{j},\hat{k}$ are mutually orthogonal (only matching components contribute).
   • Arithmetic slip converting $\cos\theta = 1/2$ into an angle other than $60^{\circ}$.

   ✓Final answer

   The correct option is (A) — $60^{\circ}$.

   ANSWER: A

2. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.Let $\pi_1$ be the plane determined by the vectors $\bar{i}+2\bar{j}$ and $3\bar{j}-2\bar{k}$. Let $\pi_2$ be the plane determined by the vectors $\bar{j}+2\bar{k}$ and $3\bar{k}-2\bar{i}$. If $\theta$ is the angle between $\pi_1$ and $\pi_2$, then $\cos\theta=$ (A) $\dfrac{7}{26}$ (B) $-\dfrac{14}{29}$ (C) $-\dfrac{32}{5\sqrt{2}}$ (D) $\dfrac{23}{38}$

   ›Reveal solutionSolution

   The angle between two planes equals the angle between their normal vectors, found here via cross products of the given spanning vectors, giving $\cos\theta=-\dfrac{14}{29}$.

   Concept and Intuition

   A plane spanned by two vectors has a normal vector equal to their cross product. Once both planes' normals are known, the angle between the planes is the angle between these normals (up to a sign ambiguity, which the options resolve for us).

   Step-by-Step Solution

   1. $\pi_1$ is spanned by $\bar i+2\bar j=(1,2,0)$ and $3\bar j-2\bar k=(0,3,-2)$. Normal $n_1 = (1,2,0)\times(0,3,-2)$: $n_1 = \left(2(-2)-0(3),\ -(1(-2)-0(0)),\ 1(3)-2(0)\right) = (-4,\ 2,\ 3)$.
   2. $\pi_2$ is spanned by $\bar j+2\bar k=(0,1,2)$ and $3\bar k-2\bar i=(-2,0,3)$. Normal $n_2 = (0,1,2)\times(-2,0,3)$: $n_2 = \left(1(3)-2(0),\ -(0(3)-2(-2)),\ 0(0)-1(-2)\right) = (3,\ -4,\ 2)$.
   3. Dot product: $n_1\cdot n_2 = (-4)(3)+(2)(-4)+(3)(2) = -12-8+6 = -14$.
   4. Magnitudes: $|n_1|=\sqrt{16+4+9}=\sqrt{29}$, $|n_2|=\sqrt{9+16+4}=\sqrt{29}$.
   5. $\cos\theta = \dfrac{-14}{\sqrt{29}\cdot\sqrt{29}} = \dfrac{-14}{29}$.

   Common Mistakes

   • Sign or component errors in the cross-product determinant expansion (a very common source of error in these vector-geometry problems).
   • Forgetting that the angle between planes uses the angle between normals directly (with the sign convention matching the given options).

   ✓Final answer

   The correct option is (B) — $-\dfrac{14}{29}$.

   ANSWER: B

3. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
   Q.Find the angle between the vectors $\vec{A} = 2\hat{i} + 4\hat{j} + 4\hat{k}$ and $\vec{B} = 4\hat{i} + 2\hat{j} - 4\hat{k}$. (A) $0^\circ$ (B) $45^\circ$ (C) $60^\circ$ (D) $90^\circ$

   ›Reveal solutionSolution

   The dot product of the two vectors is exactly zero, so the angle between them is $90^\circ$.

   Concept and Intuition

   The angle between two vectors is found from $\cos\theta=\dfrac{\vec A\cdot\vec B}{|\vec A||\vec B|}$; a zero dot product directly signals perpendicularity without needing the magnitudes.

   Step-by-Step Solution

   1. $\vec A\cdot\vec B=(2)(4)+(4)(2)+(4)(-4)=8+8-16=0$.
   2. Since $|\vec A|,|\vec B|\ne0$, $\cos\theta=0\Rightarrow\theta=90^\circ$.

   Common Mistakes

   • Sign slip while multiplying the $k$-components ($4\times(-4)=-16$, not $+16$).

   ✓Final answer

   The correct option is (D) — $90^\circ$.

   ANSWER: D

4. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.If $\bar{a} = -4\bar{i} + 2\bar{j} + 4\bar{k}$, $\bar{b} = \sqrt{2}\bar{i} - \sqrt{2}\bar{j}$ are two vectors then angle between the vectors $2\bar{a}$ and $\dfrac{\bar{b}}{2}$ is (A) $30^\circ$ (B) $135^\circ$ (C) $90^\circ$ (D) $0^\circ$

   ›Reveal solutionSolution

   The angle between $2\bar a$ and $\bar b/2$ equals the angle between $\bar a$ and $\bar b$ (scalar multiples by positive numbers don't change direction); computing that angle gives $135^\circ$.

   Concept and Intuition

   Multiplying a vector by a positive scalar only changes its magnitude, not its direction. So $\theta(2\bar a,\ \bar b/2)=\theta(\bar a,\ \bar b)$, and we can use the original vectors directly in the cosine formula.

   Step-by-Step Solution

   1. $\bar a\cdot\bar b=(-4)(\sqrt2)+(2)(-\sqrt2)+(4)(0)=-4\sqrt2-2\sqrt2+0=-6\sqrt2$.
   2. $|\bar a|=\sqrt{(-4)^2+2^2+4^2}=\sqrt{16+4+16}=\sqrt{36}=6$.
   3. $|\bar b|=\sqrt{(\sqrt2)^2+(-\sqrt2)^2+0^2}=\sqrt{2+2}=\sqrt4=2$.
   4. $\cos\theta=\dfrac{\bar a\cdot\bar b}{|\bar a||\bar b|}=\dfrac{-6\sqrt2}{6\times2}=\dfrac{-6\sqrt2}{12}=-\dfrac{\sqrt2}{2}=-\dfrac1{\sqrt2}$.
   5. $\theta=\cos^{-1}\left(-\dfrac1{\sqrt2}\right)=135^\circ$.

   Common Mistakes

   • Wasting time actually computing $2\bar a$ and $\bar b/2$ component-wise instead of recognizing the angle is scale-invariant.
   • Sign slip in the dot product from the negative components.

   ✓Final answer

   The correct option is (B) — $135^\circ$.

   ANSWER: B

5. 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 $ABD$: $\bar n_2=\overrightarrow{AB}\times\overrightarrow{AD} = \begin{vmatrix}\bar i&\bar j&\bar k\\0&-2&3\\1&0&2\end{vmatrix} = \bar i[(-2)(2)-(3)(0)] - \bar j[(0)(2)-(3)(1)]+\bar k[(0)(0)-(-2)(1)] = \bar i(-4) - \bar j(-3)+\bar k(2) = (-4,3,2)$.
   5. $|\bar n_1|=5$, $|\bar n_2|=\sqrt{16+9+4}=\sqrt{29}$.
   6. $\bar n_1.\bar n_2 = 5(-4)+0(3)+0(2)=-20$.
   7. $\cos\theta=\dfrac{-20}{5\sqrt{29}}=\dfrac{-4}{\sqrt{29}}$, so $\theta=\cos^{-1}\left(\dfrac{-4}{\sqrt{29}}\right)$.

   Common Mistakes

   • Sign errors in the cross-product cofactor expansion (especially the middle term's negative sign).
   • Using edge vectors not sharing a common vertex — always build both normals from vectors emanating from the shared edge's endpoint (here $A$) to keep the computation clean.

   ✓Final answer

   The correct option is (A) — $\mathrm{Cos}^{-1}\left(\dfrac{-4}{\sqrt{29}}\right)$.

   ANSWER: A

6. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.The angle between the planes $2x - y + z = 6$ and $x + y + 2z = 3$ is ______ (A) $\dfrac{\pi}{3}$ (B) $\cos^{-1}\left(\dfrac{1}{6}\right)$ (C) $\dfrac{\pi}{4}$ (D) $\dfrac{\pi}{6}$

   ›Reveal solutionSolution

   Tests finding the angle between two planes via the angle between their normal vectors.

   Concept and Intuition

   The angle between two planes equals the angle between their normal vectors (up to supplementary ambiguity, resolved by taking the acute angle). If a plane is $Ax+By+Cz=D$, its normal vector is $(A,B,C)$, and the angle between two normals is found using the dot-product formula.

   Step-by-Step Solution

   1. Plane 1: $2x-y+z=6$, normal $n_1=(2,-1,1)$.
   2. Plane 2: $x+y+2z=3$, normal $n_2=(1,1,2)$.
   3. $n_1\cdot n_2 = 2(1)+(-1)(1)+1(2)=2-1+2=3$.
   4. $|n_1|=\sqrt{4+1+1}=\sqrt6$, $|n_2|=\sqrt{1+1+4}=\sqrt6$.
   5. $\cos\theta=\dfrac{3}{\sqrt6\cdot\sqrt6}=\dfrac{3}{6}=\dfrac12$.
   6. So $\theta=\cos^{-1}\left(\dfrac12\right)=\dfrac{\pi}{3}$.

   Common Mistakes

   • Forgetting to normalize (divide by the magnitudes), landing on the wrong cosine value.
   • Not recognizing $\cos^{-1}(1/2)=\pi/3$ is a standard angle, and instead leaving the answer in an unsimplified inverse-cosine form (which happens to also appear as a distractor option (B) computed for different vectors).

   ✓Final answer

   The correct option is (A) — $\dfrac{\pi}{3}$.

   ANSWER: A

7. 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 $x^2-3x<0 \Rightarrow x(x-3)<0 \Rightarrow 0<x<3$.
   3. Check for anti-parallel case: if $x^2\vec i+2x\vec j+\vec k = t(\vec i-2\vec j+x\vec k)$ for some $t<0$, matching the $\vec k$-component gives $t=1/x$, and the $\vec i$-component gives $x^2=t=1/x\Rightarrow x^3=1\Rightarrow x=1$; but then the $\vec j$-component requires $2(1)=-2/1=-2$, false. So no real $x$ makes the vectors anti-parallel — the strict inequality region is exactly the obtuse-angle region.
   4. Hence the interval is $(0,3)$, open at both ends (endpoints give a right angle, not obtuse).

   Common Mistakes

   • Forgetting to exclude the anti-parallel (straight-angle) case, though here it doesn't actually occur.
   • Including the endpoints where the angle is exactly $90°$.

   ✓Final answer

   The correct option is (B) — $(0,3)$.

   ANSWER: B

8. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.Angle between the planes $\bar{r}.(12\bar{i}+4\bar{j}-3\bar{k}) = 5$ and $\bar{r}.(5\bar{i}+3\bar{j}+4\bar{k}) = 7$ is (A) $\cos^{-1}\left(\dfrac{12}{13}\right)$ (B) $\cos^{-1}\left(\dfrac{6\sqrt{2}}{13}\right)$ (C) $\cos^{-1}\left(\dfrac{3\sqrt{2}}{13}\right)$ (D) $\cos^{-1}\left(\dfrac{6}{13}\right)$

   ›Reveal solutionSolution

   The angle between two planes equals the angle between their normal vectors.

   Using the dot product formula, the cosine of the angle is $\frac{6\sqrt{2}}{13}$, so the correct option is (B).

   The key idea: the angle between two planes is defined as the angle between their normal vectors.

   Given plane equations in vector form $\bar{r} \cdot \bar{n} = d$, the normal vectors are simply the coefficients of $\bar{i}, \bar{j}, \bar{k}$.

   1. Identify the normal vectors

      For the first plane: $\bar{n}_1 = 12\bar{i} + 4\bar{j} - 3\bar{k}$

      For the second plane: $\bar{n}_2 = 5\bar{i} + 3\bar{j} + 4\bar{k}$

   2. Compute the dot product

      $\bar{n}_1 \cdot \bar{n}_2 = (12)(5) + (4)(3) + (-3)(4) = 60 + 12 - 12 = 60$

   3. Compute the magnitudes

      $|\bar{n}_1| = \sqrt{12^2 + 4^2 + (-3)^2} = \sqrt{144 + 16 + 9} = \sqrt{169} = 13$

      $|\bar{n}_2| = \sqrt{5^2 + 3^2 + 4^2} = \sqrt{25 + 9 + 16} = \sqrt{50} = 5\sqrt{2}$

   4. Apply the dot product formula for the angle

      $\cos\theta = \frac{\bar{n}_1 \cdot \bar{n}_2}{|\bar{n}_1|\,|\bar{n}_2|} = \frac{60}{13 \cdot 5\sqrt{2}} = \frac{60}{65\sqrt{2}} = \frac{12}{13\sqrt{2}}$

   5. Rationalize

      $\frac{12}{13\sqrt{2}} = \frac{12\sqrt{2}}{13 \cdot 2} = \frac{6\sqrt{2}}{13}$

   Thus $\theta = \cos^{-1}\left(\frac{6\sqrt{2}}{13}\right)$.

   Watch out

   A common mistake is to forget that the angle between planes is the acute angle between their normals — but here the dot product is positive, so no adjustment is needed.

   Tip

   Always simplify $\frac{60}{65\sqrt{2}}$ by cancelling 5 first: $\frac{12}{13\sqrt{2}}$, then rationalize.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

9. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.If $P = (0, 1, 2), Q = (4, -2, 1)$ and $O = (0, 0, 0)$ then $\angle POQ =$ (A) $\dfrac{\pi}{6}$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{\pi}{2}$

   ›Reveal solutionSolution

   Compute the dot product of the position vectors of $P$ and $Q$ from the origin; a zero dot product means the angle between them is a right angle.

   Concept and Intuition

   For any two vectors $\vec u,\vec v$, $\cos\theta=\dfrac{\vec u\cdot\vec v}{|\vec u||\vec v|}$. If the dot product is zero, $\cos\theta=0$ regardless of the magnitudes, so $\theta=\dfrac{\pi}{2}$ — no need to even compute the lengths.

   Step-by-Step Solution

   1. Since $O$ is the origin, $\vec{OP}=P-O=(0,1,2)$ and $\vec{OQ}=Q-O=(4,-2,1)$.
   2. Dot product: $\vec{OP}\cdot\vec{OQ}=(0)(4)+(1)(-2)+(2)(1)=0-2+2=0$.
   3. A zero dot product means $\vec{OP}\perp\vec{OQ}$, so $\angle POQ=\dfrac{\pi}{2}$.

   Common Mistakes

   • Sign slip while multiplying corresponding components.
   • Unnecessarily computing $|\vec{OP}|,|\vec{OQ}|$ when the zero dot product alone already settles the angle.

   ✓Final answer

   The correct option is (D) — $\dfrac{\pi}{2}$.

   ANSWER: D

10. 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 $38+2\vec a\cdot\vec c=48\Rightarrow\vec a\cdot\vec c=5$.
    5. $\cos\theta=\dfrac{\vec a\cdot\vec c}{|\vec a||\vec c|}=\dfrac{5}{2\times5}=\dfrac12\Rightarrow\theta=\dfrac{\pi}{3}$.

    Common Mistakes

    • Forgetting to double the cross terms when expanding $|\vec a+\vec b+\vec c|^2$.
    • Assuming $\vec a\perp\vec c$ as well just because both are perpendicular to $\vec b$ — that's not implied in 3D.

    ✓Final answer

    The correct option is (B) — $\dfrac{\pi}{3}$.

    ANSWER: B

11. 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: $7(4-5\alpha)=-2(17+\alpha^2)\Rightarrow28-35\alpha=-34-2\alpha^2\Rightarrow2\alpha^2-35\alpha+62=0$.
    5. Discriminant $=35^2-4(2)(62)=1225-496=729=27^2$. Roots: $\alpha=\dfrac{35\pm27}{4}$, i.e. $\alpha=2$ or $\alpha=15.5$.
    6. Since $\alpha_1<\alpha_2$: $\alpha_1=2,\ \alpha_2=15.5$. Then $\alpha_1+2\alpha_2=2+2(15.5)=2+31=33$.

    Common Mistakes

    • Not noticing $|\vec a|=|\vec b|$ and computing both magnitudes the long way (more error-prone but same result).
    • Sign error cross-multiplying the negative cosine value.

    ✓Final answer

    The correct option is (C) — $33$.

    ANSWER: C

12. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
    Q.The angle made by the resultant vector of two vectors $2\bar{i}+3\bar{j}+4\bar{k}$ and $2\bar{i}-7\bar{j}-4\bar{k}$ with x-axis is (A) $60°$ (B) $45°$ (C) $90°$ (D) $120°$

    ›Reveal solutionSolution

    Add the vectors component-wise, then use the direction-cosine formula with the x-axis. Answer: $45°$.

    Concept and Intuition

    The angle a vector makes with the x-axis is found from its direction cosine $\cos\theta = \dfrac{R_x}{|\vec R|}$, where $R_x$ is the x-component of the resultant vector and $|\vec R|$ is its magnitude.

    Step-by-Step Solution

    1. Add the vectors: $(2\bar i+3\bar j+4\bar k) + (2\bar i-7\bar j-4\bar k) = 4\bar i - 4\bar j + 0\bar k$.
    2. Magnitude: $|\vec R| = \sqrt{4^2+(-4)^2+0^2} = \sqrt{32} = 4\sqrt2$.
    3. Direction cosine with x-axis: $\cos\theta = \dfrac{R_x}{|\vec R|} = \dfrac{4}{4\sqrt2} = \dfrac{1}{\sqrt2}$.
    4. So $\theta = 45°$.

    Common Mistakes

    • Adding the vectors incorrectly (sign errors on the $j$ or $k$ components).
    • Confusing the angle with the x-axis with the angle with another axis, or using the wrong component in the cosine formula.

    ✓Final answer

    The correct option is (B) — $45°$.

    ANSWER: B