Q.Find the angle between the lines whose direction ratios are $a, b, c$ and $b-c, c-a, a-b$.

Showing the 12 most recent of 67 on this concept.

1. 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

2. 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

3. 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

4. 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

5. 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

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 2022Set eng-2022-07-07-FN1 markMCQ
   Q.Suppose $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are the directional cosines of two lines and $\theta$ is the angle between them and $\cos\theta = \pm(l_1 l_2 + m_1 m_2 + n_1 n_2)$. Let $A = (1, -2, 3)$, $B = (3, 1, -3)$ and $C = (-3, 1, 3)$ be the vertices of a triangle ABC. Then $\cos A =$ (A) $-\dfrac{1}{35}$ (B) $\dfrac{1}{7}$ (C) $-\dfrac{1}{7}$ (D) $\dfrac{1}{35}$

   ›Reveal solutionSolution

   The angle at vertex $A$ between sides $AB$ and $AC$ is found from the dot product formula $\cos A=\dfrac{\vec{AB}\cdot\vec{AC}}{|\vec{AB}||\vec{AC}|}=\dfrac{1}{35}$.

   Concept and Intuition

   The angle between two lines through a common vertex of a triangle equals the angle between the vectors from that vertex to the other two vertices. This is exactly the direction-cosine formula for the angle between two lines given in the problem, applied to vectors $\vec{AB}$ and $\vec{AC}$.

   Step-by-Step Solution

   1. $A=(1,-2,3),\,B=(3,1,-3),\,C=(-3,1,3)$.
   2. $\vec{AB}=B-A=(2,3,-6)$; $|\vec{AB}|=\sqrt{4+9+36}=\sqrt{49}=7$.
   3. $\vec{AC}=C-A=(-4,3,0)$; $|\vec{AC}|=\sqrt{16+9+0}=\sqrt{25}=5$.
   4. $\vec{AB}\cdot\vec{AC}=(2)(-4)+(3)(3)+(-6)(0)=-8+9+0=1$.
   5. $\cos A=\dfrac{\vec{AB}\cdot\vec{AC}}{|\vec{AB}||\vec{AC}|}=\dfrac{1}{7\times5}=\dfrac{1}{35}$.

   Common Mistakes

   • Using $\vec{BA}$ and $\vec{CA}$ inconsistently (mixing directions), which can flip the sign — although here the dot product is positive so the sign issue is not fatal, it is easy to make an error in general.
   • Forgetting that the angle "at $A$" always uses vectors emanating from $A$, not $B$ or $C$.

   ✓Final answer

   The correct option is (D) — $\dfrac{1}{35}$.

   ANSWER: D

8. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.The angle between the lines whose direction cosines are given by the equations $2l - m + n = 0$ and $lm + 2mn - 10nl = 0$ is $\theta$, then $\cos\theta =$ (A) $\dfrac{20}{3\sqrt{70}}$ (B) $\dfrac{10}{\sqrt{70}}$ (C) $\dfrac{8}{\sqrt{70}}$ (D) $\dfrac{16}{3\sqrt{70}}$

   ›Reveal solutionSolution

   Eliminating $n$ between the two given relations yields a quadratic in $m/l$ whose two roots are the direction ratios of the two lines; their angle has $\cos\theta=8/\sqrt{70}$.

   Concept and Intuition

   When direction cosines satisfy one linear relation and one homogeneous quadratic relation, eliminating one variable between them produces a quadratic whose two roots correspond to the direction ratios of the two lines being jointly described.

   Step-by-Step Solution

   1. From $2l-m+n=0$: $n=m-2l$.
   2. Substitute into $lm+2mn-10nl=0$: $lm+2m(m-2l)-10(m-2l)l=0$.
   3. Expand: $lm+2m^2-4ml-10ml+20l^2=0 \Rightarrow 2m^2+(1-4-10)lm+20l^2=0 \Rightarrow 2m^2-13lm+20l^2=0$.
   4. Divide by $l^2$, let $k=m/l$: $2k^2-13k+20=0\Rightarrow k=\dfrac{13\pm\sqrt{169-160}}{4}=\dfrac{13\pm3}{4}$, giving $k=4$ or $k=2.5$.
   5. Case $k=4$: $m=4l$, $n=m-2l=2l$ — direction ratios $(1,4,2)$ (taking $l=1$).
   6. Case $k=2.5$: taking $l=2$, $m=5$, $n=m-2l=1$ — direction ratios $(2,5,1)$.
   7. $\cos\theta=\dfrac{(1)(2)+(4)(5)+(2)(1)}{\sqrt{1+16+4}\sqrt{4+25+1}}=\dfrac{2+20+2}{\sqrt{21}\cdot\sqrt{30}}=\dfrac{24}{\sqrt{630}}=\dfrac{24}{3\sqrt{70}}=\dfrac{8}{\sqrt{70}}$.

   Common Mistakes

   • Sign error collecting the $lm$ coefficient ($1-4-10=-13$, easy to miscount).
   • Using an inconvenient value for $l$ that leaves fractional direction ratios (choosing $l=2$ for the second root avoids this).

   ✓Final answer

   The correct option is (C) — $\dfrac8{\sqrt{70}}$.

   ANSWER: C

9. 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

10. AP EAPCET 2021Set eng-2021-08-25-FN1 markMCQ
    Q.The angle between the lines $ab(x^2 - y^2) + (a^2 - b^2)xy = 0$ is ______ (A) $\dfrac{\pi}{2}$ (B) $\dfrac{\pi}{3}$ (C) $\dfrac{\pi}{4}$ (D) $\dfrac{\pi}{6}$

    ›Reveal solutionSolution

    The coefficients of $x^2$ and $y^2$ in this pair-of-lines equation are exact negatives of each other, which is exactly the condition for the two lines to be perpendicular.

    Concept and Intuition

    For a homogeneous pair of lines $Ax^2+2Hxy+By^2=0$, the lines are perpendicular precisely when the coefficient of $x^2$ plus the coefficient of $y^2$ is zero ($A+B=0$) — this is a standard, easily-checked shortcut, avoiding the need to compute the actual slopes.

    Step-by-Step Solution

    1. Expand the given equation: $ab\,x^2-ab\,y^2+(a^2-b^2)xy=0$, i.e. $ab\,x^2+(a^2-b^2)xy-ab\,y^2=0$.
    2. Compare to $Ax^2+2Hxy+By^2=0$: $A=ab$, $B=-ab$.
    3. Check $A+B=ab+(-ab)=0$.
    4. This satisfies the perpendicularity condition for a homogeneous pair of lines, so the angle between them is $\dfrac{\pi}{2}$, independent of the specific values of $a$ and $b$.

    Common Mistakes

    • Trying to compute the angle formula $\tan\theta=\dfrac{2\sqrt{H^2-AB}}{A+B}$ directly without noticing $A+B=0$ makes the denominator zero (which itself signals $\theta=\pi/2$, rather than being an error).
    • Missing the sign flip ($-ab$, not $+ab$) when reading off $B$ from the $y^2$ term.

    ✓Final answer

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

    ANSWER: A

11. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
    Q.The angle between the line with the direction ratios $(2, 5, 1)$ and the plane $8x + 2y - z = 4$ is (A) $\cos^{-1}\left(\dfrac{64}{\sqrt{9804}}\right)$ (B) $\sin^{-1}\left(\dfrac{64}{\sqrt{9804}}\right)$ (C) $\sin^{-1}\left(\dfrac{25}{\sqrt{2070}}\right)$ (D) $\cos^{-1}\left(\dfrac{25}{\sqrt{2070}}\right)$

    ›Reveal solutionSolution

    The angle between a line and a plane uses the sine formula with the plane's normal vector (not the cosine formula used for line–line or plane–plane angles).

    Concept and Intuition

    If $\phi$ is the angle between the line and the normal to the plane, then the angle $\theta$ between the line and the plane itself is the complement: $\theta = 90^\circ-\phi$, so $\sin\theta=\cos\phi = \dfrac{|\vec n\cdot\vec d|}{|\vec n||\vec d|}$.

    Step-by-Step Solution

    1. Plane $8x+2y-z=4$ has normal $\vec n=(8,2,-1)$.
    2. Line has direction ratios $\vec d=(2,5,1)$.
    3. $\vec n\cdot\vec d = 8(2)+2(5)+(-1)(1) = 16+10-1=25$.
    4. $|\vec n| = \sqrt{64+4+1}=\sqrt{69}$.
    5. $|\vec d| = \sqrt{4+25+1}=\sqrt{30}$.
    6. $|\vec n||\vec d| = \sqrt{69\times30}=\sqrt{2070}$.
    7. $\sin\theta = \dfrac{25}{\sqrt{2070}}$, so $\theta=\sin^{-1}\left(\dfrac{25}{\sqrt{2070}}\right)$.

    Common Mistakes

    • Using $\cos^{-1}$ instead of $\sin^{-1}$ (that formula is for the angle between the line and the plane's normal, not the plane itself).
    • Arithmetic slip computing $|\vec n|$ or $|\vec d|$.

    ✓Final answer

    The correct option is (C) — $\sin^{-1}\left(\dfrac{25}{\sqrt{2070}}\right)$.

    ANSWER: C

12. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
    Q.The angle between the pair of lines represented by the equation $(a^2 - 3b^2)x^2 + 8abxy + (b^2 - 3a^2)y^2 = 0$ is (A) $30°$ (B) $45°$ (C) $75°$ (D) $60°$

    ›Reveal solutionSolution

    Plug the coefficients into the pair-of-lines angle formula; the $(a^2+b^2)$ factors cancel neatly, leaving a constant angle independent of $a,b$. The answer is (D).

    Concept and Intuition

    Even though the coefficients depend on parameters $a,b$, the angle-between-lines formula often simplifies to a value independent of those parameters — a hallmark of this classic type.

    Step-by-Step Solution

    1. Here $A=a^2-3b^2$, $B=b^2-3a^2$, $2H=8ab\Rightarrow H=4ab$.
    2. $A+B = (a^2-3b^2)+(b^2-3a^2) = -2a^2-2b^2=-2(a^2+b^2)$.
    3. $AB = (a^2-3b^2)(b^2-3a^2) = a^2b^2-3a^4-3b^4+9a^2b^2 = 10a^2b^2-3a^4-3b^4$.
    4. $H^2-AB = 16a^2b^2 - (10a^2b^2-3a^4-3b^4) = 6a^2b^2+3a^4+3b^4 = 3(a^4+2a^2b^2+b^4)=3(a^2+b^2)^2$.
    5. $\tan\theta = \left|\dfrac{2\sqrt{H^2-AB}}{A+B}\right| = \left|\dfrac{2\sqrt3(a^2+b^2)}{-2(a^2+b^2)}\right| = \sqrt3$.
    6. $\theta=60^\circ$.

    Common Mistakes

    • Arithmetic slip expanding $(a^2-3b^2)(b^2-3a^2)$ — easy to mis-collect the $a^2b^2$ terms.
    • Forgetting to take the absolute value, which would otherwise suggest an obtuse angle.

    ✓Final answer

    The correct option is (D) — $60°$.

    ANSWER: D