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 44 on this concept.

1. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
   Q.If the d.r.'s of two lines are connected by the relations $a - b + c = 0$, $a^2 - b^2 + 2c^2 = 0$ and $\theta$ is the angle between these lines then $\cos \theta =$ (A) $\dfrac{2}{\sqrt{7}}$ (B) $\dfrac{3}{2\sqrt{7}}$ (C) $\dfrac{3}{4\sqrt{2}}$ (D) $\dfrac{1}{3\sqrt{2}}$

   ›Reveal solutionSolution

   The relations give the two direction ratios $(1,1,0)$ and $(1,3,2)$; the angle between them has $\cos\theta=\dfrac{2}{\sqrt7}$.

   From $a-b+c=0$ we get $b=a+c$. Substitute into $a^2-b^2+2c^2=0$:

   $a^2-(a+c)^2+2c^2=0 \;\Rightarrow\; -2ac+c^2=0 \;\Rightarrow\; c(c-2a)=0.$

   So $c=0$ or $c=2a$, giving the two lines:

   • $c=0 \Rightarrow b=a$: direction ratios $(a,a,0)\propto(1,1,0)$.
   • $c=2a \Rightarrow b=3a$: direction ratios $(a,3a,2a)\propto(1,3,2)$.

   Angle between them:

   $\cos\theta=\frac{(1)(1)+(1)(3)+(0)(2)}{\sqrt{1^2+1^2+0^2}\,\sqrt{1^2+3^2+2^2}}=\frac{4}{\sqrt2\,\sqrt{14}}=\frac{4}{\sqrt{28}}=\frac{4}{2\sqrt7}=\frac{2}{\sqrt7}.$

   ✓Final answer

   $\cos\theta=\dfrac{2}{\sqrt7}$ — option (A).

2. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.The direction cosines of two lines are connected by the relations $l - m + n = 0$ and $2l - 3m + nl = 0$. If $\theta$ is the angle between these two lines, then $\cos \theta =$ (A) $\dfrac{1}{4}$ (B) $\dfrac{1}{\sqrt{19}}$ (C) $\dfrac{1}{3}$ (D) $\dfrac{1}{3\sqrt{2}}$

   ›Reveal solutionSolution

   Eliminating $m$ gives $2l^2=3n^2$, so the two lines have direction ratios $\left(\pm\sqrt3,\ \pm\sqrt3+\sqrt2,\ \sqrt2\right)$. Their dot product is $-2$ and the product of the magnitudes is $2\sqrt{19}$, giving $\cos\theta=\dfrac{1}{\sqrt{19}}$ — option (B).

   The concept first

   When two direction cosines relations are given — one linear and one homogeneous quadratic — the standard recipe is:

   1. use the linear relation to express one variable in terms of the other two;
   2. substitute into the quadratic, which becomes a homogeneous quadratic in the two remaining variables — hence an equation for a ratio;
   3. its two roots give the direction ratios of the two lines;
   4. finally

   $\cos\theta=\frac{l_1l_2+m_1m_2+n_1n_2}{\sqrt{l_1^2+m_1^2+n_1^2}\ \sqrt{l_2^2+m_2^2+n_2^2}} ,$

   where we may use direction ratios (not necessarily normalised) provided we divide by the magnitudes.

   Step-by-step

   1. Eliminate $m$. From $l-m+n=0$,

   $m=l+n .$

   2. Substitute into $2lm-3mn+nl=0$:

   $2l(l+n)-3(l+n)n+nl=2l^2+2ln-3ln-3n^2+nl$

   $=2l^2+(2-3+1)log-3n^2=2l^2-3n^2=0 .$

   The $log$ terms cancel exactly — that is what makes this problem tractable.

   3. Solve for the ratio.

   $2l^2=3n^2\ \Longrightarrow\ \frac{l}{n}=\pm\sqrt{\frac32}.$

   Choose the convenient scaling $n=\sqrt2$, so $l=\pm\sqrt3$, and $m=l+n$:

   $\text{Line 1: } \left(\sqrt3,\ \sqrt3+\sqrt2,\ \sqrt2\right),\qquad \text{Line 2: } \left(-\sqrt3,\ \sqrt2-\sqrt3,\ \sqrt2\right).$

   4. Dot product.

   $\left(\sqrt3\right)\left(-\sqrt3\right)+\left(\sqrt3+\sqrt2\right)\left(\sqrt2-\sqrt3\right)+\left(\sqrt2\right)\left(\sqrt2\right)$

   $=-3+\left(2-3\right)+2=-3-1+2=-2 .$

   5. Magnitudes.

   $|{\bf u}_1|^2=3+\left(\sqrt3+\sqrt2\right)^2+2=3+\left(5+2\sqrt6\right)+2=10+2\sqrt6,$

   $|{\bf u}_2|^2=3+\left(\sqrt2-\sqrt3\right)^2+2=10-2\sqrt6 .$

   So

   $|{\bf u}_1||{\bf u}_2|=\sqrt{\left(10\right)^2-\left(2\sqrt6\right)^2}=\sqrt{100-24}=\sqrt{76}=2\sqrt{19}.$

   6. Angle.

   $\cos\theta=\frac{-2}{2\sqrt{19}}=-\frac{1}{\sqrt{19}}\ \Longrightarrow\ |\cos\theta|=\frac{1}{\sqrt{19}} .$

   (The sign only records which of the two opposite directions we chose along each line; the angle between the lines is taken as the acute one, so $\cos\theta=\dfrac{1}{\sqrt{19}}$.)

   ✓Final answer

   $\cos\theta=\dfrac{1}{\sqrt{19}}$.

   ANSWER: B

3. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
   Q.If $\theta$ is the acute angle between the two lines whose direction cosines are connected by the relations $l + m + n = 0$ and $2lm + 2nl - mn = 0$, then $\cos \theta =$ (A) $\dfrac{1}{2}$ (B) $\dfrac{\sqrt{3}}{2}$ (C) $\dfrac{5}{6}$ (D) $\dfrac{3}{5}$

   ›Reveal solutionSolution

   The acute angle between the two lines is found by solving the given constraints for direction cosines, then using the dot product formula; the result is $\cos\theta = \frac{1}{2}$, so the correct option is (A).

   We are given two lines whose direction cosines $(l, m, n)$ satisfy two conditions:

   1. $l + m + n = 0$
   2. $2lm + 2nl - mn = 0$

   We need the cosine of the acute angle between these two lines.

   ---

   Concept and Intuition

   Direction cosines of a line satisfy $l^2 + m^2 + n^2 = 1$. For two lines with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$, the cosine of the angle between them is

   $\cos\theta = l_1 l_2 + m_1 m_2 + n_1 n_2.$

   Here, both lines share the same pair of equations, so we must find two distinct sets $(l, m, n)$ that satisfy both constraints. The trick: treat the equations as a system that yields a relation between the ratios of $l, m, n$, then find two independent direction vectors.

   ---

   Step-by-step solution

   1. Eliminate one variable using $l + m + n = 0$

      From $l + m + n = 0$, we have $n = -l - m$.

   2. Substitute into the second equation

      The second condition is $2lm + 2nl - mn = 0$. Substitute $n$:

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

   Simplify:

   $2lm - 2l^2 - 2lm + ml + m^2 = 0.$

   The $2lm$ and $-2lm$ cancel. We get:

   $-2l^2 + ml + m^2 = 0.$

   Multiply by $-1$:

   $2l^2 - ml - m^2 = 0.$

   3. Solve the quadratic in $l$ and $m$ Treat this as a quadratic in $l$:

   $2l^2 - m l - m^2 = 0.$

   Using the quadratic formula:

   $l = \frac{m \pm \sqrt{m^2 + 8m^2}}{4} = \frac{m \pm 3m}{4}.$

   So the two possibilities are:

   $l = \frac{m + 3m}{4} = m \quad \text{or} \quad l = \frac{m - 3m}{4} = -\frac{m}{2}.$

   4. Find the corresponding direction ratios

      • Case 1: $l = m$. Then $n = -l - m = -2l$. So direction ratios are $(l, l, -2l)$, i.e., proportional to $(1, 1, -2)$.
      • Case 2: $l = -\frac{m}{2}$. Then $m = -2l$, and $n = -l - (-2l) = l$. So direction ratios are $(l, -2l, l)$, i.e., proportional to $(1, -2, 1)$.

      These are two distinct lines.

   5. Compute $\cos\theta$ using the dot product

      For vectors $\mathbf{a} = (1, 1, -2)$ and $\mathbf{b} = (1, -2, 1)$:

   $\mathbf{a} \cdot \mathbf{b} = 1\cdot1 + 1\cdot(-2) + (-2)\cdot1 = 1 - 2 - 2 = -3.$

   Magnitudes:

   $|\mathbf{a}| = \sqrt{1^2 + 1^2 + (-2)^2} = \sqrt{6}, \quad |\mathbf{b}| = \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6}.$

   So

   $\cos\theta = \frac{-3}{\sqrt{6}\cdot\sqrt{6}} = \frac{-3}{6} = -\frac{1}{2}.$

   6. Take the acute angle The acute angle between lines is given by the absolute value of cosine, so $\cos\theta = \frac{1}{2}$.

   Tip

   The negative sign in the dot product only tells us the angle is obtuse; the acute angle between lines is always taken as the smaller one, so we take the absolute value.

   Watch out

   A common mistake is to forget that direction cosines are not unique up to sign — but here we used ratios, so the absolute value is correct for the acute angle.

   ---

   ✓Final answer

   The correct option is (A).

   ANSWER: A

4. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
   Q.If the direction cosines $(l, m, n)$ of two lines are connected by the relations $l + m + n = 0$ and $lm = 0$, then the angle between those lines is (A) $\dfrac{\pi}{3}$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{\pi}{2}$ (D) $\dfrac{\pi}{6}$

   ›Reveal solutionSolution

   The condition $l+m+n=0$ and $lm=0$ forces each line’s direction cosines to be a permutation of $(1,-1,0)/\sqrt{2}$, so the angle between them is $\pi/3$, making option (A) correct.

   ---

   We are given two lines whose direction cosines $(l, m, n)$ satisfy:

   $l + m + n = 0 \quad\text{and}\quad lm = 0.$

   We need the angle between these two lines.

   Concept & Intuition

   Direction cosines satisfy $l^2 + m^2 + n^2 = 1$. The conditions $l+m+n=0$ and $lm=0$ are symmetric but not fully symmetric — they force one of $l$ or $m$ to be zero. That gives us a family of possible triples, but the angle between two distinct lines from this family is fixed. The trick is to find two distinct triples that satisfy both conditions, then compute the dot product to get the cosine of the angle between them.

   ---

   Step-by-step reasoning

   1. Use the normalization condition Since $(l,m,n)$ are direction cosines, we have:

   $l^2 + m^2 + n^2 = 1.$

   Together with $l+m+n=0$, we can eliminate $n$: $n = -l - m$.

   2. Apply $lm = 0$

      This means either $l = 0$ or $m = 0$ (or both, but both zero would force $n=0$ from $l+m+n=0$, which is impossible because then $l^2+m^2+n^2=0 \neq 1$). So we have two cases:

      • Case 1: $l = 0$. Then $m + n = 0 \Rightarrow n = -m$. Normalization: $0^2 + m^2 + (-m)^2 = 2m^2 = 1 \Rightarrow m = \pm \frac{1}{\sqrt{2}}$. So one line has direction cosines $(0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}})$ or $(0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}})$. These are essentially the same line (opposite direction), so pick one representative:

   $\text{Line } A: \left(0, \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right).$

   • Case 2: $m = 0$. Then $l + n = 0 \Rightarrow n = -l$. Normalization: $l^2 + 0^2 + (-l)^2 = 2l^2 = 1 \Rightarrow l = \pm \frac{1}{\sqrt{2}}$. Pick the representative:

   $\text{Line } B: \left(\frac{1}{\sqrt{2}}, 0, -\frac{1}{\sqrt{2}}\right).$

   These are two distinct lines satisfying the given relations.

   3. Compute the angle between them The cosine of the angle $\theta$ between two lines with direction cosines $(l_1,m_1,n_1)$ and $(l_2,m_2,n_2)$ is:

   $\cos\theta = l_1 l_2 + m_1 m_2 + n_1 n_2.$

   For our lines:

   $\cos\theta = 0\cdot\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\cdot 0 + \left(-\frac{1}{\sqrt{2}}\right)\left(-\frac{1}{\sqrt{2}}\right) = \frac{1}{2}.$

   Hence $\theta = \cos^{-1}\left(\frac12\right) = \frac{\pi}{3}$.

   Tip

   Notice that the two lines we found are just permutations of the triple $(\pm 1, \pm 1, 0)/\sqrt{2}$ with the sum zero. Any other pair from the same family will also give $\cos\theta = 1/2$ because the dot product always picks the product of the two non-zero components of opposite sign.

   Watch out

   A common mistake is to think $lm=0$ means both $l$ and $m$ are zero — but that would make $n=0$ from the sum condition, violating normalization. Always check that direction cosines satisfy $l^2+m^2+n^2=1$.

   ---

   ✓Final answer

   The correct option is (A).

   ANSWER: A

5. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If the direction cosines of two lines satisfy the equations $2l + m - n = 0$, $l^2 - 2m^2 + n^2 = 0$ and $\theta$ is the angle between the lines then $\cos \theta =$ (A) $\dfrac{1}{5}$ (B) $\dfrac{\pi}{4}$ (C) $\dfrac{2}{3}$ (D) $\dfrac{\pi}{3}$

   ›Reveal solutionSolution

   The direction cosines of each line satisfy two given equations; solving them yields two distinct direction vectors, and the cosine of the angle between them is found via dot product, giving $\cos\theta = \frac{1}{5}$.

   We are given that the direction cosines $(l, m, n)$ of two lines satisfy

   $2l + m - n = 0 \quad\text{and}\quad l^2 - 2m^2 + n^2 = 0.$

   The angle $\theta$ between the lines is the angle between their direction vectors. Since direction cosines satisfy $l^2 + m^2 + n^2 = 1$, each line’s $(l, m, n)$ is a unit vector. The two equations above must hold for both lines, but they define a set of possible unit vectors; the two distinct solutions give the two lines.

   Why this approach works:

   We treat the equations as a system in $l, m, n$ with the constraint $l^2+m^2+n^2=1$. Solving gives two unit vectors. Their dot product is $\cos\theta$.

   ---

   1. Express one variable in terms of another

      From $2l + m - n = 0$, we have $n = 2l + m$.

   2. Substitute into the second equation

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

   Expand:

   $l^2 - 2m^2 + 4l^2 + 4lm + m^2 = 0 \quad\Rightarrow\quad 5l^2 + 4lm - m^2 = 0.$

   3. Solve the quadratic relation between $l$ and $m$ Treat $5l^2 + 4lm - m^2 = 0$ as quadratic in $l$:

   $5l^2 + 4m\,l - m^2 = 0.$

   Using the quadratic formula:

   $l = \frac{-4m \pm \sqrt{16m^2 + 20m^2}}{10} = \frac{-4m \pm 6m}{10}.$

   So the two possibilities are:

   $l = \frac{2m}{10} = \frac{m}{5} \quad\text{or}\quad l = \frac{-10m}{10} = -m.$

   4. Find corresponding $(l, m, n)$ for each case

      • Case 1: $l = \frac{m}{5}$ Then $n = 2l + m = \frac{2m}{5} + m = \frac{7m}{5}$. The unit vector condition $l^2 + m^2 + n^2 = 1$ gives:

   $\left(\frac{m}{5}\right)^2 + m^2 + \left(\frac{7m}{5}\right)^2 = 1 \quad\Rightarrow\quad \frac{m^2}{25} + m^2 + \frac{49m^2}{25} = 1.$

    Combine: $\frac{1+25+49}{25}m^2 = \frac{75}{25}m^2 = 3m^2 = 1$, so $m^2 = \frac{1}{3}$.  
    Choose $m = \frac{1}{\sqrt{3}}$ (sign doesn’t matter for direction). Then  
   

   $l = \frac{1}{5\sqrt{3}},\quad n = \frac{7}{5\sqrt{3}}.$

    So one direction vector is  
   

   $\mathbf{v}_1 = \left(\frac{1}{5\sqrt{3}},\; \frac{1}{\sqrt{3}},\; \frac{7}{5\sqrt{3}}\right).$

   • Case 2: $l = -m$ Then $n = 2(-m) + m = -m$. Unit condition: $(-m)^2 + m^2 + (-m)^2 = 3m^2 = 1$, so $m^2 = \frac{1}{3}$. Take $m = \frac{1}{\sqrt{3}}$, then $l = -\frac{1}{\sqrt{3}}$, $n = -\frac{1}{\sqrt{3}}$. So the second direction vector is

   $\mathbf{v}_2 = \left(-\frac{1}{\sqrt{3}},\; \frac{1}{\sqrt{3}},\; -\frac{1}{\sqrt{3}}\right).$

   5. Compute $\cos\theta$ via dot product

   $\mathbf{v}_1 \cdot \mathbf{v}_2 = \frac{1}{5\sqrt{3}}\left(-\frac{1}{\sqrt{3}}\right) + \frac{1}{\sqrt{3}}\cdot\frac{1}{\sqrt{3}} + \frac{7}{5\sqrt{3}}\left(-\frac{1}{\sqrt{3}}\right).$

   Simplify each term:

   $= -\frac{1}{15} + \frac{1}{3} - \frac{7}{15} = \frac{-1 + 5 - 7}{15} = \frac{-3}{15} = -\frac{1}{5}.$

   The cosine of the angle between lines is the absolute value of the dot product (since direction cosines can be taken with either sign), so

   $\cos\theta = \frac{1}{5}.$

   Tip

   The negative dot product means the angle is obtuse; the acute angle between lines is taken, so we use the absolute value.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

6. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.If the angle between the straight lines whose direction cosines satisfy the equations $l - 2m + n = 0$ and $2l^2 - 3m^2 + n^2 = 0$ is $\theta$, then $\cos \theta =$ (A) $\dfrac{9}{\sqrt{105}}$ (B) $\dfrac{3\sqrt{3}}{\sqrt{7}}$ (C) $\dfrac{\pi}{2}$ (D) $\dfrac{\pi}{4}$

   ›Reveal solutionSolution

   The problem asks for the cosine of the angle between two lines whose direction cosines satisfy two given equations. By solving the system for possible direction ratios and using the dot product formula, we find $\cos\theta = \frac{9}{\sqrt{105}}$, which corresponds to option (A).

   We are given two conditions that the direction cosines $(l, m, n)$ of each line must satisfy:

   $l - 2m + n = 0 \quad \text{and} \quad 2l^2 - 3m^2 + n^2 = 0.$

   Since direction cosines also satisfy $l^2 + m^2 + n^2 = 1$, but we don’t need that directly — we only need the ratios of direction cosines to find the angle between the lines. The key idea: each line corresponds to a set $(l, m, n)$ (up to a common factor) that satisfies both equations. The angle between two such lines is found from the dot product of their direction vectors.

   ---

   1. Eliminate one variable using the linear equation

   From $l - 2m + n = 0$, we have

   $n = 2m - l.$

   2. Substitute into the quadratic equation

   Plug into $2l^2 - 3m^2 + n^2 = 0$:

   $2l^2 - 3m^2 + (2m - l)^2 = 0.$

   Expand $(2m - l)^2 = 4m^2 - 4lm + l^2$, so:

   $2l^2 - 3m^2 + 4m^2 - 4lm + l^2 = 0,$

   $3l^2 + m^2 - 4lm = 0.$

   3. Treat as a quadratic in $l/m$

   Divide through by $m^2$ (assuming $m \neq 0$; we’ll check later):

   $3\left(\frac{l}{m}\right)^2 - 4\left(\frac{l}{m}\right) + 1 = 0.$

   Let $t = l/m$. Then:

   $3t^2 - 4t + 1 = 0.$

   Solve:

   $t = \frac{4 \pm \sqrt{16 - 12}}{6} = \frac{4 \pm 2}{6}.$

   So $t = 1$ or $t = \frac{1}{3}$.

   4. Find direction ratios for each case

   Case 1: $l/m = 1 \Rightarrow l = m$.

   From $n = 2m - l = 2m - m = m$.

   So direction ratios are $(l, m, n) = (1, 1, 1)$.

   Case 2: $l/m = 1/3 \Rightarrow l = m/3$.

   Then $n = 2m - l = 2m - m/3 = \frac{5m}{3}$.

   So direction ratios are $(1/3, 1, 5/3)$, or multiply by 3: $(1, 3, 5)$.

   Thus the two lines have direction vectors $\mathbf{a} = (1, 1, 1)$ and $\mathbf{b} = (1, 3, 5)$.

   Tip

   We didn’t need to normalize to unit vectors because the cosine formula uses ratios:

   $\cos\theta = \frac{|\mathbf{a} \cdot \mathbf{b}|}{\|\mathbf{a}\| \|\mathbf{b}\|}$ works with any proportional vectors.

   5. Compute the cosine

   Dot product:

   $\mathbf{a} \cdot \mathbf{b} = 1\cdot1 + 1\cdot3 + 1\cdot5 = 9.$

   Magnitudes:

   $\|\mathbf{a}\| = \sqrt{1^2+1^2+1^2} = \sqrt{3},$

   $\|\mathbf{b}\| = \sqrt{1^2+3^2+5^2} = \sqrt{1+9+25} = \sqrt{35}.$

   Thus:

   $\cos\theta = \frac{9}{\sqrt{3} \cdot \sqrt{35}} = \frac{9}{\sqrt{105}}.$

   Watch out

   A common mistake is to forget that the two equations define two distinct lines, not just one. Solving the quadratic gives two possible ratios, each corresponding to a different line. The angle is then between these two lines.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

7. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.Let $\overrightarrow{OA}$, $\overrightarrow{OB}$, $\overrightarrow{OC}$ lying along X, Y, Z-axes respectively represent the coterminous edges of a rectangular parallelepiped. If $OA = 1$, $OB = 2$, $OC = 3$ then the angle between a pair of diagonals of the parallelepiped drawn through the vertices O and A is (A) $\dfrac{\pi}{3}$ (B) $\cos^{-1}\left(\dfrac{5}{7}\right)$ (C) $\cos^{-1}\left(\dfrac{6}{7}\right)$ (D) $\dfrac{\pi}{4}$

   ›Reveal solutionSolution

   The angle between the space diagonals through O and A of a rectangular box is found using the dot product of their direction vectors. The correct answer is $\cos^{-1}\left(\frac{6}{7}\right)$, option (C).

   The problem gives us a rectangular parallelepiped (a box) with edges along the coordinate axes. The three coterminous edges from O are $\overrightarrow{OA}$ along X, $\overrightarrow{OB}$ along Y, and $\overrightarrow{OC}$ along Z, with lengths 1, 2, and 3 respectively.

   The key idea: a rectangular box has four space diagonals. Two of them pass through O and A (opposite vertices). The angle between any pair of space diagonals can be found by writing their direction vectors and using the dot product formula.

   1. Set up coordinates. Place O at the origin $(0,0,0)$. Then:

      • A is at $(1,0,0)$ (since OA = 1 along X)
      • B is at $(0,2,0)$
      • C is at $(0,0,3)$ The opposite vertex to O is the one with all three coordinates: $(1,2,3)$. Call it D.

   2. Identify the two diagonals through O and A. The diagonal through O goes from O to D: vector $\overrightarrow{OD} = (1,2,3)$. The diagonal through A goes from A to the vertex opposite A, which is the vertex with coordinates $(0,2,3)$ — call it E. So the diagonal through A is $\overrightarrow{AE} = (0-1, 2-0, 3-0) = (-1,2,3)$.

   3. Find the angle between these two diagonals. Use the dot product:

   $\overrightarrow{OD} \cdot \overrightarrow{AE} = (1)(-1) + (2)(2) + (3)(3) = -1 + 4 + 9 = 12$

   Magnitudes:

   $|\overrightarrow{OD}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$

   $|\overrightarrow{AE}| = \sqrt{(-1)^2 + 2^2 + 3^2} = \sqrt{14}$

   So:

   $\cos\theta = \frac{12}{\sqrt{14} \cdot \sqrt{14}} = \frac{12}{14} = \frac{6}{7}$

   Hence $\theta = \cos^{-1}\left(\frac{6}{7}\right)$.

   Watch out

   A common mistake is to take the diagonal through A as going from A to O instead of from A to the opposite vertex. That would give a different vector and a wrong angle. Always identify the correct opposite vertex.

   Tip

   For any rectangular box with edges $a, b, c$, the angle between two space diagonals through opposite vertices is $\cos^{-1}\left(\frac{a^2 + b^2 + c^2 - 2(\text{product of the two differing coordinates})}{a^2+b^2+c^2}\right)$ — but it's faster to just set coordinates and compute.

   ✓Final answer

   The angle is $\cos^{-1}\left(\dfrac{6}{7}\right)$, which corresponds to option (C).

8. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
   Q.If $\theta$ is the acute angle between the lines $\dfrac{x}{a} + \dfrac{y}{b} = 1$, $\dfrac{x}{b} + \dfrac{y}{a} = 1$ then $\sin \theta =$ (A) $\dfrac{2ab}{a^2 + b^2}$ (B) $\left| \dfrac{a-b}{a+b} \right|$ (C) $\left| \dfrac{a^2 - b^2}{2ab} \right|$ (D) $\left| \dfrac{a^2 - b^2}{a^2 + b^2} \right|$

   ›Reveal solutionSolution

   The acute angle between two lines depends only on their slopes. Rewriting each line in slope-intercept form, computing the slopes, and applying the tangent formula for the angle between lines leads to $\sin\theta = \left| \frac{a^2 - b^2}{a^2 + b^2} \right|$, which matches option (D).

   We are given two lines:

   $\frac{x}{a} + \frac{y}{b} = 1 \quad \text{and} \quad \frac{x}{b} + \frac{y}{a} = 1$

   We want the acute angle $\theta$ between them. The classic approach: find the slopes of both lines, then use the formula for the tangent of the angle between two lines. From $\tan\theta$, we can find $\sin\theta$ using a right-triangle relation.

   ---

   1. Find the slopes

   Rewrite each line in the form $y = mx + c$.

   • For the first line:

   $\frac{x}{a} + \frac{y}{b} = 1 \quad\Rightarrow\quad \frac{y}{b} = 1 - \frac{x}{a}$

   Multiply by $b$:

   $y = b - \frac{b}{a}x$

   So slope $m_1 = -\dfrac{b}{a}$.

   • For the second line:

   $\frac{x}{b} + \frac{y}{a} = 1 \quad\Rightarrow\quad \frac{y}{a} = 1 - \frac{x}{b}$

   Multiply by $a$:

   $y = a - \frac{a}{b}x$

   So slope $m_2 = -\dfrac{a}{b}$.

   ---

   2. Use the angle formula

   If two lines have slopes $m_1$ and $m_2$, the acute angle $\theta$ between them satisfies:

   $\tan\theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$

   Substitute $m_1 = -\frac{b}{a}$ and $m_2 = -\frac{a}{b}$:

   $m_1 - m_2 = -\frac{b}{a} + \frac{a}{b} = \frac{-b^2 + a^2}{ab} = \frac{a^2 - b^2}{ab}$

   $1 + m_1 m_2 = 1 + \left(-\frac{b}{a}\right)\left(-\frac{a}{b}\right) = 1 + 1 = 2$

   Thus:

   $\tan\theta = \left| \frac{\frac{a^2 - b^2}{ab}}{2} \right| = \left| \frac{a^2 - b^2}{2ab} \right|$

   ---

   3. Convert $\tan\theta$ to $\sin\theta$

   We know $\tan\theta = \frac{\text{opposite}}{\text{adjacent}}$. Imagine a right triangle where the side opposite $\theta$ is $|a^2 - b^2|$ and the adjacent side is $2|ab|$. Then the hypotenuse is:

   $\sqrt{(a^2 - b^2)^2 + (2ab)^2} = \sqrt{a^4 - 2a^2b^2 + b^4 + 4a^2b^2} = \sqrt{a^4 + 2a^2b^2 + b^4} = \sqrt{(a^2 + b^2)^2} = a^2 + b^2$

   Since $\theta$ is acute, $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{|a^2 - b^2|}{a^2 + b^2}$.

   ---

   Watch out

   A common mistake is to stop at $\tan\theta$ and match it to option (C). But the question asks for $\sin\theta$, not $\tan\theta$. Option (C) is the tangent, not the sine.

   Tip

   Notice that the slopes are negative reciprocals only when $a = b$ (then lines are perpendicular). In general, the neat simplification $1 + m_1 m_2 = 2$ is what makes the final expression clean.

   ---

   ✓Final answer

   The correct option is (D).

   ANSWER: D

9. TG EAPCET 2021Set eng-2021-08-06-AN1 markMCQ
   Q.Let $\vec{a}, \vec{b}, \vec{c}$ be three unit vectors such that $\vec{a} \times (\vec{b} \times \vec{c}) = \frac{1}{2} \vec{b}$. If the angle between $\vec{a}, \vec{b}$ is $\theta_1$ and the angle between $\vec{a}, \vec{c}$ is $\theta_2$, then $\theta_1 + \theta_2 =$ (A) $150^\circ$ (B) $180^\circ$ (C) $120^\circ$ (D) $90^\circ$

   ›Reveal solutionSolution

   Using the vector triple product identity, the given condition forces $\vec{a}$ to be perpendicular to $\vec{c}$ and at $60^\circ$ to $\vec{b}$, so $\theta_1 + \theta_2 = 150^\circ$.

   The key is the vector triple product identity:

   $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$.

   This identity rewrites a cross of crosses into a combination of scalars and vectors — exactly what we need to turn the given equation into dot-product relations.

   Since $\vec{a}, \vec{b}, \vec{c}$ are unit vectors, their dot products are simply cosines of the angles between them:

   $\vec{a} \cdot \vec{b} = \cos\theta_1$, $\vec{a} \cdot \vec{c} = \cos\theta_2$, and $\vec{b} \cdot \vec{c} = \cos\theta_3$ (where $\theta_3$ is the angle between $\vec{b}$ and $\vec{c}$, which we may not need directly).

   1. Apply the identity to the given condition:

   $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \frac{1}{2}\vec{b}.$

   2. Bring terms together:

   $(\vec{a} \cdot \vec{c})\vec{b} - \frac{1}{2}\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = 0,$

   or

   $\left(\cos\theta_2 - \frac12\right)\vec{b} - (\cos\theta_1)\,\vec{c} = 0.$

   3. This is a linear combination of the two non-zero vectors $\vec{b}$ and $\vec{c}$ equaling zero. Since $\vec{b}$ and $\vec{c}$ are not necessarily parallel (they could be, but we must check), the only way a linear combination $p\vec{b} + q\vec{c} = \vec{0}$ holds without forcing $\vec{b}$ and $\vec{c}$ to be parallel is if both coefficients are zero. If $\vec{b}$ and $\vec{c}$ were parallel, then $\vec{b} \times \vec{c} = \vec{0}$, making the original left side $\vec{a} \times \vec{0} = \vec{0}$, which would give $\frac12\vec{b} = \vec{0}$ — impossible for a unit vector. So $\vec{b}$ and $\vec{c}$ are not parallel, and we must have:

   $\cos\theta_2 - \frac12 = 0 \quad\text{and}\quad \cos\theta_1 = 0.$

   4. From $\cos\theta_1 = 0$, with $\theta_1$ between $0^\circ$ and $180^\circ$ (the usual range for angles between vectors), we get:

   $\theta_1 = 90^\circ.$

   5. From $\cos\theta_2 = \frac12$, we get:

   $\theta_2 = 60^\circ.$

   6. Therefore:

   $\theta_1 + \theta_2 = 90^\circ + 60^\circ = 150^\circ.$

   Watch out

   A common mistake is to forget that $\vec{b}$ and $\vec{c}$ might be parallel and simply set both coefficients to zero without justification. Here, checking that case eliminates it cleanly.

   ✓Final answer

   The sum is $\boxed{150^\circ}$, which corresponds to option (A).

10. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
    Q.One of the pair of lines $x^2 - 3y^2 - 4x - 6\sqrt{3}y - 5 = 0$ is $x + by + c = 0$ $(b < 0)$. If the other line intersects the curve $x^2 - 5y^2 - 4x = 0$ at two points A and B, then $\angle AOB =$ (A) $\dfrac{\pi}{4}$ (B) $\dfrac{\pi}{3}$ (C) $\dfrac{\pi}{6}$ (D) $\dfrac{\pi}{2}$

    ›Reveal solutionSolution

    The given degenerate hyperbola splits into two lines; one is $x+by+c=0$ with $b<0$. The other line, when intersected with a second hyperbola, gives points $A$ and $B$ such that $OA \perp OB$, so the angle is $\frac{\pi}{2}$.

    ---

    Concept and Intuition

    The equation $x^2 - 3y^2 - 4x - 6\sqrt{3}y - 5 = 0$ is a degenerate conic — it represents a pair of straight lines. The trick is to factor it into two linear factors. One of them is given as $x + by + c = 0$ with $b<0$. Once we find $b$ and $c$, we can write the other line. That other line intersects the curve $x^2 - 5y^2 - 4x = 0$ (a hyperbola) at two points $A$ and $B$. The angle $\angle AOB$ is the angle subtended at the origin by the chord $AB$. For a hyperbola centered at the origin, if a chord passes through a fixed point and satisfies a certain condition, the angle at the origin can be constant. Here, we can find $A$ and $B$ explicitly and compute the dot product of their position vectors.

    ---

    Step-by-step solution

    1. Factor the degenerate conic.

    The given equation is

    $x^2 - 3y^2 - 4x - 6\sqrt{3}y - 5 = 0.$

    Complete the square in $x$ and $y$:

    $(x^2 - 4x) - 3(y^2 + 2\sqrt{3}y) = 5.$

    Add $4$ to complete $x^2-4x+4 = (x-2)^2$, and inside the $y$ part: $y^2+2\sqrt{3}y + 3 = (y+\sqrt{3})^2$, so we add $-3 \times 3 = -9$ to the left. Balance:

    $(x-2)^2 - 3(y+\sqrt{3})^2 = 5 + 4 - 9 = 0.$

    Thus

    $(x-2)^2 - 3(y+\sqrt{3})^2 = 0.$

    This factors as a difference of squares:

    $\left[(x-2) - \sqrt{3}(y+\sqrt{3})\right] \left[(x-2) + \sqrt{3}(y+\sqrt{3})\right] = 0.$

    So the two lines are:

    $x - \sqrt{3}y - 2 - 3 = 0 \quad\Rightarrow\quad x - \sqrt{3}y - 5 = 0,$

    $x + \sqrt{3}y - 2 + 3 = 0 \quad\Rightarrow\quad x + \sqrt{3}y + 1 = 0.$

    2. Identify which line matches $x+by+c=0$ with $b<0$.

    The second line is $x + \sqrt{3}y + 1 = 0$. Here $b = \sqrt{3} > 0$, not allowed.

    The first line is $x - \sqrt{3}y - 5 = 0$, which can be written as $x + (-\sqrt{3})y + (-5) = 0$. So $b = -\sqrt{3} < 0$, $c = -5$. This matches the given condition.

    Thus the other line (the one not given) is

    $x + \sqrt{3}y + 1 = 0.$

    3. Intersect this other line with the second curve.

    The second curve is

    $x^2 - 5y^2 - 4x = 0.$

    From the line, $x = -\sqrt{3}y - 1$. Substitute:

    $(-\sqrt{3}y - 1)^2 - 5y^2 - 4(-\sqrt{3}y - 1) = 0.$

    Expand:

    $3y^2 + 2\sqrt{3}y + 1 - 5y^2 + 4\sqrt{3}y + 4 = 0.$

    Simplify:

    $-2y^2 + 6\sqrt{3}y + 5 = 0 \quad\Rightarrow\quad 2y^2 - 6\sqrt{3}y - 5 = 0.$

    4. Find coordinates of $A$ and $B$.

    Let the roots be $y_1, y_2$. Then

    $y_1 + y_2 = 3\sqrt{3}, \quad y_1 y_2 = -\frac{5}{2}.$

    Corresponding $x$: $x_i = -\sqrt{3}y_i - 1$. So

    $x_1 + x_2 = -\sqrt{3}(y_1+y_2) - 2 = -\sqrt{3}(3\sqrt{3}) - 2 = -9 - 2 = -11,$

    $x_1 x_2 = (-\sqrt{3}y_1 - 1)(-\sqrt{3}y_2 - 1) = 3y_1y_2 + \sqrt{3}(y_1+y_2) + 1.$

    Substitute:

    $x_1 x_2 = 3\left(-\frac{5}{2}\right) + \sqrt{3}(3\sqrt{3}) + 1 = -\frac{15}{2} + 9 + 1 = -\frac{15}{2} + 10 = \frac{5}{2}.$

    5. Compute $\angle AOB$.

    Vectors $OA$ and $OB$ are $(x_1, y_1)$ and $(x_2, y_2)$. The cosine of the angle between them is

    $\cos\theta = \frac{x_1 x_2 + y_1 y_2}{\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}}.$

    We have

    $x_1 x_2 + y_1 y_2 = \frac{5}{2} + \left(-\frac{5}{2}\right) = 0.$

    Thus $\cos\theta = 0$, so $\theta = \frac{\pi}{2}$.

    Tip

    The numerator of the dot product vanished immediately — no need to compute the denominator. This is a classic sign that the chord subtends a right angle at the origin.

    ---

    ✓Final answer

    The correct option is (D).

    ANSWER: D

11. TG EAPCET 2021Set eng-2021-08-04-AN1 markMCQ
    Q.The angle between the pair of straight lines $3y^2 - 8xy - 3x^2 - 29x + 3y - 18 = 0$ is (A) $90^\circ$ (B) $35^\circ$ (C) $45^\circ$ (D) $30^\circ$

    ›Reveal solutionSolution

    The angle between a pair of straight lines given by a general second-degree equation is determined by the coefficients of the $x^2$, $xy$, and $y^2$ terms. When the sum of the coefficients of $x^2$ and $y^2$ is zero, the lines are perpendicular. For the given equation, this sum is zero, so the angle is $\boxed{90^\circ}$.

    The general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ represents a pair of straight lines if a specific condition (the determinant of the associated matrix being zero) is met. The problem statement confirms that the given equation represents a pair of straight lines, so we don't need to verify this condition.

    The angle between these two lines is determined solely by the coefficients of the homogeneous part of the equation, which is $ax^2 + 2hxy + by^2 = 0$. The linear terms ($2gx + 2fy$) and the constant term ($c$) only affect the position of the intersection point of the lines, not the angle between them.

    The angle $\theta$ between the pair of straight lines represented by $ax^2 + 2hxy + by^2 = 0$ is given by:

    $\tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} \right|$

    A crucial special case arises when $a+b=0$. If $a+b=0$, it implies $a = -b$. In this situation, the formula for $\tan \theta$ becomes undefined, which indicates that the angle $\theta$ is $90^\circ$. This means the lines are perpendicular. We can see this intuitively: if $a = -b$, the homogeneous part becomes $-bx^2 + 2hxy + by^2 = 0$, or $by^2 + 2hxy - bx^2 = 0$. If we divide by $bx^2$ (assuming $x \neq 0$), we get $(y/x)^2 + (2h/b)(y/x) - 1 = 0$. If $m_1$ and $m_2$ are the slopes of the lines, then $m_1 m_2 = -1$, which is the condition for perpendicular lines.

    Let's apply this understanding to the given problem.

    1. Identify the coefficients:

       The given equation is $3y^2 - 8xy - 3x^2 - 29x + 3y - 18 = 0$.

       To compare it with the standard form $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$, we rearrange the terms:

       $-3x^2 - 8xy + 3y^2 - 29x + 3y - 18 = 0$

       Comparing the coefficients:

       • Coefficient of $x^2$, $a = -3$
       • Coefficient of $xy$, $2h = -8 \implies h = -4$
       • Coefficient of $y^2$, $b = 3$
       • Coefficient of $x$, $2g = -29 \implies g = -29/2$
       • Coefficient of $y$, $2f = 3 \implies f = 3/2$
       • Constant term, $c = -18$

    2. Check the sum of coefficients $a+b$:

       We need to calculate $a+b$:

       $a+b = (-3) + (3) = 0$

    3. Determine the angle:

       Since $a+b=0$, the lines represented by the equation are perpendicular to each other.

       Therefore, the angle between them is $90^\circ$.

    Watch out

    Do not directly substitute $a+b=0$ into the denominator of the $\tan \theta$ formula without understanding the implication. A zero denominator indicates a special case where the angle is $90^\circ$, not that the angle is undefined in a physical sense.

    ›Proof

    We can optionally verify that the given equation indeed represents a pair of straight lines using the condition $abc + 2fgh - af^2 - bg^2 - ch^2 = 0$.

    Given $a = -3, b = 3, h = -4, g = -29/2, f = 3/2, c = -18$.

    $abc = (-3)(3)(-18) = 162$

    $2fgh = 2(3/2)(-29/2)(-4) = 3(-29/2)(-4) = 3(58) = 174$

    $af^2 = (-3)(3/2)^2 = (-3)(9/4) = -27/4$

    $bg^2 = (3)(-29/2)^2 = (3)(841/4) = 2523/4$

    $ch^2 = (-18)(-4)^2 = (-18)(16) = -288$

    Substituting these values into the condition:

    $162 + 174 - (-27/4) - (2523/4) - (-288)$

    $= 162 + 174 + 27/4 - 2523/4 + 288$

    $= (162 + 174 + 288) + (27/4 - 2523/4)$

    $= 624 + (-2496/4)$

    $= 624 - 624 = 0$

    Since the condition is satisfied, the equation indeed represents a pair of straight lines.

    The correct option is (A).

    ✓Final answer

    The angle between the pair of straight lines is $\boxed{90^\circ}$.

12. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
    Q.A line makes angles $60^\circ$, $45^\circ$, $\theta$ with positive X, Y, Z-axes respectively. If $\theta$ is an acute angle, then $\tan \theta =$ (A) $\sqrt{3}$ (B) $\dfrac{1}{\sqrt{3}}$ (C) $1$ (D) $2$

    ›Reveal solutionSolution

    The direction cosines of a line satisfy $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$. Given $\alpha = 60^\circ$, $\beta = 45^\circ$, and $\gamma = \theta$ acute, we find $\cos\theta = \frac12$, so $\tan\theta = \sqrt{3}$. The correct option is (A).

    The key idea is that any line in 3D space has direction cosines — the cosines of the angles it makes with the positive coordinate axes. These three cosines are not independent; they satisfy a fundamental Pythagorean-like identity. Once we know two angles, we can solve for the third, and then compute its tangent.

    1. Recall the direction cosine identity If a line makes angles $\alpha$, $\beta$, $\gamma$ with the positive X, Y, Z axes, then

    $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1.$

    This holds because the direction vector’s components are proportional to these cosines, and its squared length is the sum of squares of those components.

    2. Plug in the given angles We have $\alpha = 60^\circ$, $\beta = 45^\circ$, $\gamma = \theta$ (acute).

    $\cos 60^\circ = \frac12, \quad \cos 45^\circ = \frac{\sqrt{2}}{2}.$

    So

    $\left(\frac12\right)^2 + \left(\frac{\sqrt{2}}{2}\right)^2 + \cos^2\theta = 1.$

    3. Simplify to find $\cos\theta$

    $\frac14 + \frac{2}{4} + \cos^2\theta = 1 \quad\Rightarrow\quad \frac34 + \cos^2\theta = 1.$

    Hence

    $\cos^2\theta = \frac14 \quad\Rightarrow\quad \cos\theta = \frac12 \quad (\text{since } \theta \text{ is acute, cosine positive}).$

    4. Find $\tan\theta$ If $\cos\theta = \frac12$, then $\theta = 60^\circ$ (acute).

    $\tan 60^\circ = \sqrt{3}.$

    Watch out

    A common mistake is to forget the squares and write $\cos 60^\circ + \cos 45^\circ + \cos\theta = 1$. That would give a different (and wrong) angle. Always square the cosines.

    Tip

    Once you get $\cos\theta = \frac12$, you instantly recognize $\theta = 60^\circ$, so $\tan\theta = \sqrt{3}$ without any further calculation.

    ✓Final answer

    The correct option is (A).

    ANSWER: A