Q.Find the angle between the pair of lines given by $\vec{r} = 3\hat{i} + 2\hat{j} - 4\hat{k} + \lambda(\hat{i} + 2\hat{j} + 2\hat{k})$ and $\vec{r} = 5\hat{i} - 2\hat{j} + \mu(3\hat{i} + 2\hat{j} + 6\hat{k})$.

Showing the 12 most recent of 67 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.The acute angle between the lines whose direction cosines satisfy the relations $l^2-5m^2+n^2=0$ and $l+m-n=0$ is (A) $\mathrm{Cos}^{-1}\left(\dfrac{\sqrt{3}}{4}\right)$ (B) $\dfrac{\pi}{3}$ (C) $\mathrm{Cos}^{-1}\left(\dfrac{2}{3}\right)$ (D) $\dfrac{\pi}{6}$

   ›Reveal solutionSolution

   The two relations on direction cosines actually describe a pair of lines; solving them simultaneously extracts both direction ratios, and the angle between them is $\pi/3$.

   Concept and Intuition

   A single homogeneous quadratic relation like $l^2-5m^2+n^2=0$ together with a linear relation like $l+m-n=0$ defines two lines through the origin (the linear relation is a plane, and the quadratic relation restricted to that plane factors into two linear factors — i.e. two direction ratios). Once we have both direction ratio triples, the angle between the lines is just the standard angle-between-vectors formula.

   Step-by-Step Solution

   1. From $l+m-n=0$: $n=l+m$.
   2. Substitute into $l^2-5m^2+n^2=0$: $l^2-5m^2+(l+m)^2=0$.
   3. Expand: $l^2-5m^2+l^2+2lm+m^2=0\Rightarrow 2l^2+2lm-4m^2=0$.
   4. Divide by 2: $l^2+lm-2m^2=0$.
   5. Factor: $(l+2m)(l-m)=0$, so $l=-2m$ or $l=m$.
   6. Case $l=m$: take $m=1\Rightarrow l=1,\ n=l+m=2$. Direction ratios $(1,1,2)$.
   7. Case $l=-2m$: take $m=1\Rightarrow l=-2,\ n=l+m=-1$. Direction ratios $(-2,1,-1)$.
   8. Angle between $(1,1,2)$ and $(-2,1,-1)$: dot product $=1(-2)+1(1)+2(-1)=-2+1-2=-3$.
   9. Magnitudes: $\sqrt{1+1+4}=\sqrt6$ and $\sqrt{4+1+1}=\sqrt6$.
   10. $\cos\theta=\dfrac{|-3|}{\sqrt6\cdot\sqrt6}=\dfrac{3}{6}=\dfrac12$ (taking absolute value since we want the acute angle).
   11. $\theta=\cos^{-1}(1/2)=\pi/3$.

   Common Mistakes

   • Forgetting to take the absolute value of the dot product when the problem asks for the acute angle specifically.
   • Arithmetic slip while expanding $(l+m)^2$.

   ✓Final answer

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

   ANSWER: B

2. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.If the direction ratios of two lines are given by $3lm - 4ln + mn = 0$ and $l + 2m + 3n = 0$, then the angle between the lines is ________ (A) $\frac{\pi}{2}$ (B) $\frac{\pi}{3}$ (C) $\frac{\pi}{4}$ (D) $\frac{\pi}{6}$

   ›Reveal solutionSolution

   The linear relation combined with the quadratic relation gives two explicit direction-ratio triples; their dot product is zero. Answer: $\theta=\pi/2$.

   Concept and Intuition

   A pair of homogeneous-degree-2 relation and a linear relation in $(l,m,n)$ together represent two actual lines through a point. Eliminating one variable from the linear relation and substituting into the quadratic relation gives a single-variable quadratic whose two roots correspond to the two lines' direction ratios.

   Step-by-Step Solution

   1. From $l+2m+3n=0$: $l=-2m-3n$.
   2. Substitute into $3lm-4ln+mn=0$: $3(-2m-3n)m - 4(-2m-3n)n + mn = -6m^2-9mn+8mn+12n^2+mn = -6m^2+12n^2=0$.
   3. So $m^2=2n^2 \Rightarrow m=\pm\sqrt2\,n$. Take $n=1$.
   4. Case 1: $m=\sqrt2$, $l=-2\sqrt2-3$. Case 2: $m=-\sqrt2$, $l=2\sqrt2-3$.
   5. Dot product: $l_1l_2+m_1m_2+n_1n_2 = (-2\sqrt2-3)(2\sqrt2-3) + (\sqrt2)(-\sqrt2) + 1\cdot1$. $(-2\sqrt2-3)(2\sqrt2-3) = -(8-9)=1$ (using $(a+b)(a-b)$ pattern with sign care), so sum $=1-2+1=0$.
   6. Since the dot product is zero, the lines are perpendicular: $\theta=\pi/2$.

   Common Mistakes

   • Sign errors expanding $(-2\sqrt2-3)(2\sqrt2-3)$ — treat it carefully term by term.
   • Forgetting that a homogeneous quadratic + linear pair represents two lines, not one.

   ✓Final answer

   The correct option is (A) — $\frac{\pi}{2}$.

   ANSWER: A

3. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If $(K, 3, 5), (2, -1, 2)$ are direction ratios of two lines and the angle between them is $45^\circ$, then a value of $K$ is (A) $2$ (B) $4$ (C) $6$ (D) $8$

   ›Reveal solutionSolution

   Applying the direction-cosine angle formula and solving the resulting quadratic in $K$ gives $K=4$ (the other root, $52$, isn't among the choices). Answer: (B).

   Concept and Intuition

   The angle between two lines with direction ratios $(l_1,m_1,n_1)$ and $(l_2,m_2,n_2)$ satisfies $\cos\theta=\dfrac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{l_1^2+m_1^2+n_1^2}\sqrt{l_2^2+m_2^2+n_2^2}}$. Setting this equal to $\cos45^\circ$ gives an equation purely in $K$.

   Step-by-Step Solution

   1. $d_1=(K,3,5)$, $d_2=(2,-1,2)$. Dot product: $2K-3+10=2K+7$.
   2. $|d_1|=\sqrt{K^2+9+25}=\sqrt{K^2+34}$, $|d_2|=\sqrt{4+1+4}=3$.
   3. $\cos45^\circ=\frac{2K+7}{3\sqrt{K^2+34}}=\frac{1}{\sqrt2}$
   4. Cross-multiplying: $2(2K+7)=3\sqrt2\sqrt{K^2+34} \Rightarrow 4K+14=3\sqrt2\sqrt{K^2+34}$.
   5. Squaring: $(4K+14)^2=18(K^2+34) \Rightarrow 16K^2+112K+196=18K^2+612$

   $\Rightarrow 2K^2-112K+416=0 \Rightarrow K^2-56K+208=0$

   6. Discriminant $=56^2-4(208)=3136-832=2304=48^2$. $K=\dfrac{56\pm48}{2}=52\text{ or }4$.
   7. Only $K=4$ is among the options. Verify: $d_1=(4,3,5)$, dot with $(2,-1,2)$ is $8-3+10=15$; $|d_1|=\sqrt{50}=5\sqrt2$; $\cos\theta=\dfrac{15}{5\sqrt2\cdot3}=\dfrac{1}{\sqrt2}$, i.e., $\theta=45^\circ$. ✓

   Common Mistakes

   • Dropping one of the two quadratic roots without checking which matches the given options.
   • Sign error in the dot product (the $-1$ in $d_2$ must multiply the $3$, giving $-3$, not $+3$).

   ✓Final answer

   The correct option is (B) — $4$.

   ANSWER: B

4. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.Let A = (2, 0, -1), B = (1, -2, 0), C = (1, 2, -1) and D = (0, -1, -2) be four points. If $\theta$ is the acute angle between the plane determined by A, B, C and the plane determined by A, C, D, then $\tan\theta =$ (A) $\sqrt{\dfrac{14}{5}}$ (B) $\dfrac{3}{\sqrt{14}}$ (C) $\dfrac{3}{\sqrt{5}}$ (D) $\dfrac{\sqrt{5}}{3}$

   ›Reveal solutionSolution

   This tests finding the angle between two planes via their normal vectors (cross products of edge vectors) and converting cosine to tangent. Answer: $\tan\theta=\dfrac{3}{\sqrt5}$.

   Concept and Intuition

   The angle between two planes equals the angle between their normal vectors (taking the acute value). A plane through three points has its normal given by the cross product of any two vectors lying in it (e.g. $\vec{AB}\times\vec{AC}$). Once both normals are known, $\cos\theta$ follows from the dot product formula, and $\tan\theta$ from $\sin\theta/\cos\theta$ using $\sin\theta=\sqrt{1-\cos^2\theta}$.

   Step-by-Step Solution

   1. $A=(2,0,-1),\,B=(1,-2,0),\,C=(1,2,-1),\,D=(0,-1,-2)$.
   2. Plane $ABC$: $\vec{AB}=B-A=(-1,-2,1)$, $\vec{AC}=C-A=(-1,2,0)$. Normal $N_1=\vec{AB}\times\vec{AC}=\big((-2)(0)-(1)(2),\ (1)(-1)-(-1)(0),\ (-1)(2)-(-2)(-1)\big)=(-2,-1,-4)$.
   3. Plane $ACD$: $\vec{AD}=D-A=(-2,-1,-1)$. Normal $N_2=\vec{AC}\times\vec{AD}=\big((2)(-1)-(0)(-1),\ (0)(-2)-(-1)(-1),\ (-1)(-1)-(2)(-2)\big)=(-2,-1,5)$.
   4. $N_1\cdot N_2=(-2)(-2)+(-1)(-1)+(-4)(5)=4+1-20=-15$. $|N_1|=\sqrt{4+1+16}=\sqrt{21}$, $|N_2|=\sqrt{4+1+25}=\sqrt{30}$.
   5. $\cos\theta=\dfrac{|-15|}{\sqrt{21}\sqrt{30}}=\dfrac{15}{\sqrt{630}}=\dfrac{15}{3\sqrt{70}}=\dfrac{5}{\sqrt{70}}$ (taking the acute angle).
   6. $\sin\theta=\sqrt{1-\dfrac{25}{70}}=\sqrt{\dfrac{45}{70}}=\sqrt{\dfrac{9}{14}}=\dfrac{3}{\sqrt{14}}$.
   7. $\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{3/\sqrt{14}}{5/\sqrt{70}}=\dfrac{3}{\sqrt{14}}\cdot\dfrac{\sqrt{70}}{5}=\dfrac{3\sqrt{70/14}}{5}=\dfrac{3\sqrt5}{5}=\dfrac{3}{\sqrt5}$.

   Common Mistakes

   • Cross-product component sign errors (easy to mix up the $j$-component sign, which is negated relative to $i,k$).
   • Forgetting to take the absolute value of the dot product when the angle between planes is required to be acute.

   ✓Final answer

   The correct option is (C) — $\dfrac{3}{\sqrt5}$.

   ANSWER: C

5. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.The angle between the straight lines $3x + 4y + 9 = 0$ and $x - 7y - 22 = 0$ is ______ (A) $\dfrac{\pi}{4}$ (B) $\dfrac{\pi}{6}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{\pi}{8}$

   ›Reveal solutionSolution

   This tests the standard formula for the angle between two lines given their slopes. The answer is $\dfrac{\pi}{4}$.

   Concept and Intuition

   The angle $\theta$ between two lines with slopes $m_1,m_2$ satisfies $\tan\theta=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|$. We extract each slope from its line equation and substitute.

   Step-by-Step Solution

   1. $3x+4y+9=0 \Rightarrow y=-\dfrac34x-\dfrac94$, so $m_1=-\dfrac34$.
   2. $x-7y-22=0 \Rightarrow y=\dfrac17x-\dfrac{22}{7}$, so $m_2=\dfrac17$.
   3. $\tan\theta=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|=\left|\dfrac{-\frac34-\frac17}{1+\left(-\frac34\right)\left(\frac17\right)}\right|$.
   4. Numerator: $-\dfrac{21}{28}-\dfrac{4}{28}=-\dfrac{25}{28}$. Denominator: $1-\dfrac{3}{28}=\dfrac{25}{28}$.
   5. $\tan\theta=\left|\dfrac{-25/28}{25/28}\right|=1 \Rightarrow \theta=\dfrac{\pi}{4}$.

   Common Mistakes

   • Slope sign error when rearranging $3x+4y+9=0$ to slope-intercept form.
   • Forgetting the absolute value, which could otherwise give a negative or obtuse-angle tangent.

   ✓Final answer

   The correct option is (A) — $\dfrac{\pi}{4}$.

   ANSWER: A

6. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.If the direction cosines of two lines are given by $l + m + n = 0$ and $mn - 2lm - 2nl = 0$, then the acute angle between those lines is (A) $2\pi/5$ (B) $\pi/3$ (C) $\pi/4$ (D) $\pi/60$

   ›Reveal solutionSolution

   Eliminate $n$ using the linear relation, factor the resulting quadratic in $l,m$ to get two sets of direction ratios, then use the cosine formula between two lines.

   Concept and Intuition

   When direction cosines satisfy one linear and one quadratic (or bilinear) relation, substituting the linear relation into the quadratic one reduces it to a single quadratic in the ratio $l:m$, whose two roots give the direction ratios of the two lines being described.

   Step-by-Step Solution

   1. From $l+m+n=0$: $n=-(l+m)$.
   2. Substitute into $mn-2lm-2nl=0$:

   $m(-(l+m)) - 2lm - 2(-(l+m))l = -lm-m^2-2lm+2l^2+2lm = 2l^2-lm-m^2=0$

   3. Solve for $l$ in terms of $m$: $2l^2-lm-m^2=0 \Rightarrow l=\dfrac{m\pm\sqrt{m^2+8m^2}}{4}=\dfrac{m\pm3m}{4}$, giving $l=m$ or $l=-\dfrac{m}{2}$.
   4. Case 1: $l=m=1\Rightarrow n=-(1+1)=-2$. Direction ratios: $(1,1,-2)$.
   5. Case 2: $l=-1,m=2\Rightarrow n=-(-1+2)=-1$. Direction ratios: $(-1,2,-1)\propto(1,-2,1)$.
   6. Angle between the lines: $\cos\theta = \dfrac{|(1)(1)+(1)(-2)+(-2)(1)|}{\sqrt{1+1+4}\sqrt{1+4+1}}=\dfrac{|1-2-2|}{6}=\dfrac{3}{6}=\dfrac12$.
   7. $\theta=\dfrac{\pi}{3}$ (acute angle, taking absolute value of cosine).

   Common Mistakes

   • Not taking the absolute value in the cosine formula (would give an obtuse angle instead of the requested acute one).
   • Losing one of the two roots of the quadratic in $l,m$.

   ✓Final answer

   The correct option is (B) — $\pi/3$.

   ANSWER: B

7. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.If a line L makes angles $\frac{\pi}{3}$ and $\frac{\pi}{4}$ with Y-axis and Z-axis respectively, then the angle between L and another line having direction ratios 1, 1, 1 is (A) $\text{Cos}^{-1}\left(\frac{2}{\sqrt{6}}\right)$ (B) $\text{Cos}^{-1}\left(\frac{\sqrt{2}+1}{3\sqrt{3}}\right)$ (C) $\text{Cos}^{-1}\left(\frac{\sqrt{2}-1}{3}\right)$ (D) $\text{Cos}^{-1}\left(\frac{\sqrt{2}+1}{\sqrt{6}}\right)$

   ›Reveal solutionSolution

   Find the missing direction cosine from $l^2+m^2+n^2=1$, then use $\cos\theta=l l_2+mm_2+nn_2$ against $(1,1,1)$; the answer is $\text{Cos}^{-1}\!\left(\dfrac{\sqrt2+1}{\sqrt6}\right)$.

   Concept and Intuition

   The direction cosines of a line satisfy $l^2+m^2+n^2=1$ where $l=\cos\alpha$, $m=\cos\beta$, $n=\cos\gamma$ are the cosines of the angles the line makes with the $X$, $Y$, $Z$ axes respectively. Once all three are known, the angle between two lines is $\cos\theta=l_1l_2+m_1m_2+n_1n_2$.

   Step-by-Step Solution

   1. Given angle with $Y$-axis is $\dfrac{\pi}{3}$: $m=\cos\dfrac{\pi}{3}=\dfrac12$.
   2. Given angle with $Z$-axis is $\dfrac{\pi}{4}$: $n=\cos\dfrac{\pi}{4}=\dfrac{1}{\sqrt2}$.
   3. From $l^2+m^2+n^2=1$: $l^2=1-\dfrac14-\dfrac12=\dfrac14 \Rightarrow l=\dfrac12$ (taking the positive root).
   4. Direction cosines of the second line with ratios $(1,1,1)$: $\left(\dfrac{1}{\sqrt3},\dfrac{1}{\sqrt3},\dfrac{1}{\sqrt3}\right)$.
   5. $\cos\theta=l\cdot\dfrac{1}{\sqrt3}+m\cdot\dfrac{1}{\sqrt3}+n\cdot\dfrac{1}{\sqrt3}=\dfrac{l+m+n}{\sqrt3}=\dfrac{\frac12+\frac12+\frac{1}{\sqrt2}}{\sqrt3}=\dfrac{1+\frac{1}{\sqrt2}}{\sqrt3}=\dfrac{\sqrt2+1}{\sqrt2\cdot\sqrt3}=\dfrac{\sqrt2+1}{\sqrt6}$.
   6. So $\theta=\text{Cos}^{-1}\!\left(\dfrac{\sqrt2+1}{\sqrt6}\right)$.

   Common Mistakes

   • Forgetting the third direction cosine must be solved from $l^2+m^2+n^2=1$, not assumed.
   • Not converting $(1,1,1)$ (direction ratios) into direction cosines by dividing by $\sqrt3$ before using the dot-product formula.

   ✓Final answer

   The correct option is (D) — $\text{Cos}^{-1}\left(\dfrac{\sqrt2+1}{\sqrt6}\right)$.

   ANSWER: D

8. AP EAPCET 2021Set eng-2021-08-24-FN1 markMCQ
   Q.The angle between the lines whose direction cosines are given by the equations $l^2 + m^2 - n^2 = 0$, $l + m + n = 0$ is ____ (A) $\dfrac{\pi}{6}$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{\pi}{3}$ (D) $\dfrac{\pi}{2}$

   ›Reveal solutionSolution

   Eliminate $n$ between the two given relations to find the two actual sets of direction ratios, then compute the angle between them directly.

   Concept and Intuition

   The two given equations jointly define (generically) two lines through the origin whose direction cosines satisfy both. Eliminating one variable reduces the quadratic relation to a simple product-equals-zero form, revealing the two explicit direction-ratio triples.

   Step-by-Step Solution

   1. From $l+m+n=0$: $n=-(l+m)$.
   2. Substitute into $l^2+m^2-n^2=0$: $l^2+m^2-(l+m)^2 = l^2+m^2-l^2-2lm-m^2 = -2lm=0$.
   3. So $lm=0$, meaning $l=0$ or $m=0$.
   4. If $l=0$: $n=-m$, giving direction ratios $(0,1,-1)$.
   5. If $m=0$: $n=-l$, giving direction ratios $(1,0,-1)$.
   6. Angle between $(0,1,-1)$ and $(1,0,-1)$: $\cos\theta = \dfrac{(0)(1)+(1)(0)+(-1)(-1)}{\sqrt{0+1+1}\sqrt{1+0+1}} = \dfrac{1}{\sqrt2\cdot\sqrt2} = \dfrac12$.
   7. So $\theta = \cos^{-1}(1/2) = \dfrac{\pi}{3}$.

   Common Mistakes

   • Trying to use the general angle-between-lines formula for pairs of direction-cosine equations without first simplifying — direct elimination is far cleaner here.
   • Sign slips substituting $n=-(l+m)$.

   ✓Final answer

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

   ANSWER: C

9. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If the angle $\theta$ between the line $\dfrac{x+1}{1} = \dfrac{y-1}{2} = \dfrac{z-2}{2}$ and the plane $2x - y + \sqrt{\lambda}z + 4 = 0$ is such that $\sin\theta = \dfrac{1}{3}$, then the value of $\lambda$ is (A) $\dfrac{5}{3}$ (B) $-\dfrac{3}{5}$ (C) $\dfrac{3}{4}$ (D) $-\dfrac{4}{3}$

   ›Reveal solutionSolution

   Using $\sin\theta=\dfrac{|d\cdot n|}{|d||n|}$ for the line-plane angle and solving for $\lambda$ gives $\lambda=\dfrac53$. Answer: (A).

   Concept and Intuition

   The angle $\theta$ between a line (direction vector $d$) and a plane (normal vector $n$) is the complement of the angle between $d$ and $n$, so $\sin\theta=\cos(\text{angle between }d,n)=\dfrac{|d\cdot n|}{|d||n|}$. This directly converts the given $\sin\theta$ value into an equation for the unknown in the normal vector.

   Step-by-Step Solution

   1. Line: $\dfrac{x+1}{1}=\dfrac{y-1}{2}=\dfrac{z-2}{2}$ has direction $d=(1,2,2)$, $|d|=\sqrt{1+4+4}=3$.
   2. Plane: $2x-y+\sqrt\lambda z+4=0$ has normal $n=(2,-1,\sqrt\lambda)$, $|n|=\sqrt{4+1+\lambda}=\sqrt{5+\lambda}$.
   3. $d\cdot n = 1(2)+2(-1)+2(\sqrt\lambda)=2-2+2\sqrt\lambda=2\sqrt\lambda$
   4. $\sin\theta=\frac{|2\sqrt\lambda|}{3\sqrt{5+\lambda}}=\frac13$
   5. $\Rightarrow 2\sqrt\lambda=\sqrt{5+\lambda} \Rightarrow 4\lambda=5+\lambda \Rightarrow 3\lambda=5 \Rightarrow \lambda=\frac53$

   Common Mistakes

   • Using $\cos\theta$ instead of $\sin\theta$ for the line-plane angle formula (it's the complementary relation, not the same as line-line angle).
   • Sign/arithmetic slip in the dot product cancelling the first two terms to zero.

   ✓Final answer

   The correct option is (A) — $\dfrac53$.

   ANSWER: A

10. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
    Q.If the angle between the lines having direction ratios $(3, 1, 2)$ and $(1, -1, 2)$ is $\theta$, then $\cos 2\theta =$ (A) $\dfrac{1}{7}$ (B) $\dfrac{-1}{7}$ (C) $\dfrac{\sqrt{3}}{\sqrt{7}}$ (D) $\dfrac{-3}{\sqrt{7}}$

    ›Reveal solutionSolution

    Computing $\cos\theta$ from the direction-ratio dot-product formula and applying the double-angle identity gives $\cos2\theta=-\dfrac17$.

    Concept and Intuition

    The angle between two lines with direction ratios $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ satisfies

    $\cos\theta=\dfrac{a_1a_2+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\,\sqrt{a_2^2+b_2^2+c_2^2}}.$

    Once $\cos\theta$ is known, $\cos2\theta=2\cos^2\theta-1$ follows directly from the double angle formula — no need to find $\theta$ itself.

    Step-by-Step Solution

    1. Direction ratios: $(3,1,2)$ and $(1,-1,2)$.
    2. Dot product: $3(1)+1(-1)+2(2)=3-1+4=6$.
    3. Magnitudes: $\sqrt{9+1+4}=\sqrt{14}$, $\sqrt{1+1+4}=\sqrt6$.
    4. $\cos\theta=\dfrac{6}{\sqrt{14}\cdot\sqrt6}=\dfrac{6}{\sqrt{84}}=\dfrac{6}{2\sqrt{21}}=\dfrac{3}{\sqrt{21}}$.
    5. $\cos^2\theta = \dfrac{9}{21}=\dfrac37$.
    6. $\cos2\theta = 2\cdot\dfrac37-1 = \dfrac67-1=-\dfrac17$.

    Common Mistakes

    • Forgetting the identity needs $\cos^2\theta$, not $\cos\theta$ itself, before doubling.
    • Sign error in the dot product (missing that $1\times(-1)=-1$).

    ✓Final answer

    The correct option is (B) — $\dfrac{-1}{7}$.

    ANSWER: B

11. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
    Q.Let $\theta$ be the angle between the line $\frac{x+1}{1} = \frac{y-1}{2} = \frac{z-2}{2}$ and the plane $2x - y + \sqrt{\lambda}z + 4 = 0$. If $\sin\theta = \frac{1}{3}$, then $\lambda =$ (A) $\frac{4}{3}$ (B) $\frac{5}{3}$ (C) $\frac{2}{3}$ (D) $\frac{7}{3}$

    ›Reveal solutionSolution

    This tests the line-plane angle formula. Setting up $\sin\theta$ in terms of $\lambda$ and solving the resulting equation gives $\lambda=\frac53$.

    Concept and Intuition

    The angle $\theta$ between a line with direction $\vec d$ and a plane with normal $\vec n$ satisfies $\sin\theta=\frac{|\vec d\cdot\vec n|}{|\vec d||\vec n|}$ (since the angle between the line and the normal is $90^\circ-\theta$). Plugging in the components and the given $\sin\theta$ gives a single equation in $\lambda$.

    Step-by-Step Solution

    1. Direction of the line: $\vec d=(1,2,2)$, so $|\vec d|=\sqrt{1+4+4}=3$.
    2. Normal of the plane: $\vec n=(2,-1,\sqrt\lambda)$, so $|\vec n|=\sqrt{4+1+\lambda}=\sqrt{5+\lambda}$.
    3. $\vec d\cdot\vec n=1(2)+2(-1)+2(\sqrt\lambda)=2-2+2\sqrt\lambda=2\sqrt\lambda$.
    4. $\sin\theta=\frac{2\sqrt\lambda}{3\sqrt{5+\lambda}}=\frac13$.
    5. Cross-multiplying: $2\sqrt\lambda=\sqrt{5+\lambda}$.
    6. Squaring: $4\lambda=5+\lambda\Rightarrow3\lambda=5\Rightarrow\lambda=\frac53$.

    Common Mistakes

    • Using $\cos\theta$ instead of $\sin\theta$ for the line-plane angle formula (it is reversed compared to the line-line or plane-plane angle formulas).
    • Sign/arithmetic slip in the dot product $1(2)+2(-1)+2\sqrt\lambda=2\sqrt\lambda$ (the first two terms cancel).

    ✓Final answer

    The correct option is (B) — $\frac53$.

    ANSWER: B

12. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
    Q.If the slope of one of the pair of lines represented by $2x^2 + 3xy + Ky^2 = 0$ is $2$, then the angle between the pair of lines is (A) $\dfrac{\pi}{2}$ (B) $\dfrac{\pi}{3}$ (C) $\dfrac{\pi}{6}$ (D) $\dfrac{\pi}{4}$

    ›Reveal solutionSolution

    The homogeneous pair $2x^2+3xy+Ky^2=0$ gives slopes satisfying a quadratic in $m=y/x$; using the given slope $2$ we find $K$, then both slopes, and their product tells us the angle.

    Concept and Intuition

    A pair of straight lines through the origin, $ax^2+2hxy+by^2=0$, can be seen as $y=m_1x$ and $y=m_2x$. Substituting $y=mx$ turns the equation into a quadratic in $m$, whose two roots are the two slopes.

    Step-by-Step Solution

    1. Substitute $y=mx$ into $2x^2+3xy+Ky^2=0$: dividing by $x^2$, $K m^2+3m+2=0$.
    2. One slope is given as $m=2$: $K(4)+3(2)+2=0 \Rightarrow 4K+8=0 \Rightarrow K=-2$.
    3. The slope equation becomes $-2m^2+3m+2=0$, i.e. $2m^2-3m-2=0$.
    4. Factor: $(2m+1)(m-2)=0 \Rightarrow m=2$ or $m=-\dfrac12$.
    5. Product of slopes $m_1m_2 = 2\times\left(-\dfrac12\right) = -1$.
    6. Since $m_1 m_2=-1$, the two lines are perpendicular, so the angle between them is $\dfrac{\pi}{2}$.

    Common Mistakes

    • Forgetting that $K$ must first be determined from the given slope before finding the other slope.
    • Confusing "sum of slopes" and "product of slopes" formulas from the quadratic.

    ✓Final answer

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

    ANSWER: A