Q.Show that the straight lines whose direction cosines are given by $2l + 2m - n = 0$ and $mn + nl + lm = 0$ are at right angles.

Showing the 12 most recent of 16 on this concept.

1. CBSE 2026Set 65/3/11 markMCQ
   Q.If $l_1, m_1, n_1$ and $l_2, m_2, n_2$ are direction cosines of lines $L_1$ and $L_2$ respectively and $\theta$ is the acute angle between them, then: (A) $\cos \theta = l_1 l_2 + m_1 m_2 + n_1 n_2$ (B) $\sin \theta = l_1 l_2 + m_1 m_2 + n_1 n_2$ (C) $\tan \theta = \frac{l_1}{l_2} + \frac{m_1}{m_2} + \frac{n_1}{n_2}$ (D) $\cos \theta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$

   ›Reveal solutionSolution

   The angle between two lines is found using the dot product of their direction vectors. For direction cosines, the cosine of the angle is simply the sum of the products of corresponding cosines. Since the angle is acute, we take the absolute value to ensure a non-negative cosine. The correct choice is (D).

   The core idea here is beautifully simple: direction cosines are the components of a unit vector along each axis. So if you have two lines, their direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ are just the coordinates of two unit vectors pointing along those lines.

   Now, what does the dot product of two unit vectors give you? Exactly the cosine of the angle between them. That’s the geometric meaning of the dot product. So:

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

   But there’s a subtlety the question is testing: the angle between two lines is always taken as the acute angle (between $0^\circ$ and $90^\circ$). The dot product formula above can give a negative value if the angle is obtuse (greater than $90^\circ$). To get the acute angle, we take the absolute value.

   Let’s walk through the options one by one.

   1. Option (A): $\cos \theta = l_1 l_2 + m_1 m_2 + n_1 n_2$

      This is almost correct, but it doesn’t account for the acute angle condition. If the lines make an obtuse angle, this sum is negative, and $\cos \theta$ for the acute angle should be positive. So (A) is not fully correct for the acute angle.

   2. Option (B): $\sin \theta = l_1 l_2 + m_1 m_2 + n_1 n_2$

      This is simply wrong. The sum of products of direction cosines gives cosine, not sine. No further discussion needed — discard.

   3. Option (C): $\tan \theta = \frac{l_1}{l_2} + \frac{m_1}{m_2} + \frac{n_1}{n_2}$

      This is nonsense. Division by a direction cosine is not defined if that cosine is zero, and even when defined, it has no relation to the tangent of the angle between lines. Discard. …

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2. CBSE 20241 markMCQ
   Q.The angle between the lines $\frac{x+1}{2} = \frac{2-y}{-5} = \frac{z}{4}$ and $\frac{x-3}{1} = \frac{y-7}{2} = \frac{5-z}{3}$ is: (A) $\frac{\pi}{4}$ (B) $\frac{\pi}{2}$ (C) $\frac{\pi}{3}$ (D) $\frac{\pi}{6}$

   ›Reveal solutionSolution

   The angle between two lines in space is found using the dot product of their direction vectors. After extracting the direction ratios and computing the cosine, the angle is $\frac{\pi}{2}$, so the correct option is (B).

   The key idea is simple: in 3D geometry, the angle between two lines is defined as the acute angle between their direction vectors. We don’t care about where the lines are placed — only their orientation matters. That’s why we can ignore the points they pass through and focus entirely on the direction ratios.

   Let’s extract those direction vectors carefully.

   1. First line: $\frac{x+1}{2} = \frac{2-y}{-5} = \frac{z}{4}$ The standard symmetric form is $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$, where $(a,b,c)$ are the direction ratios. Here, the $y$-term is $\frac{2-y}{-5}$. Rewrite it as $\frac{y-2}{5}$ (multiply numerator and denominator by $-1$). So the line becomes:

   $\frac{x+1}{2} = \frac{y-2}{5} = \frac{z}{4}$

   Hence, direction ratios for the first line are $(2, 5, 4)$.

   2. Second line: $\frac{x-3}{1} = \frac{y-7}{2} = \frac{5-z}{3}$ The $z$-term is $\frac{5-z}{3}$. Rewrite as $\frac{z-5}{-3}$. So the line is:

   $\frac{x-3}{1} = \frac{y-7}{2} = \frac{z-5}{-3}$

   Hence, direction ratios for the second line are $(1, 2, -3)$.

   Watch out

   A common mistake is to take the $z$-direction ratio as $3$ instead of $-3$ from $\frac{5-z}{3}$. Always rewrite in the form $\frac{z - z_1}{c}$ — the sign of $c$ matters.

   3. Compute the angle using the dot product formula: …

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3. CBSE 20201 markMCQ
   Q.The plane $2x - 3y + 6z - 11 = 0$ makes an angle $\sin^{-1}(\alpha)$ with the $x$-axis. The value of $\alpha$ is (A) $\frac{\sqrt{3}}{2}$ (B) $\frac{\sqrt{2}}{3}$ (C) $\frac{2}{7}$ (D) $\frac{3}{7}$

   ›Reveal solutionSolution

   The angle between a plane and a line is the complement of the angle between the plane’s normal and the line. Using the direction ratios of the $x$-axis and the normal vector of the plane, we find $\sin\theta = \frac{2}{7}$, so $\alpha = \frac{2}{7}$.

   The problem asks for the sine of the angle between a given plane and the $x$-axis. A common mistake is to directly use the angle between the normal and the line — but that gives the complement. Let’s build the correct idea from scratch.

   Concept first: When we talk about the angle between a line and a plane, we mean the smallest angle between the line and its projection onto the plane. This is not the angle between the line and the normal to the plane. In fact, if $\theta$ is the angle between the line and the plane, and $\phi$ is the angle between the line and the normal, then $\theta + \phi = 90^\circ$. So $\sin\theta = \cos\phi$.

   Watch out

   Never confuse the angle between a line and a plane with the angle between the line and the normal. They are complementary. If you compute the angle with the normal directly, you get $\phi$, not $\theta$.

   The plan: Find $\phi$ (angle between the $x$-axis and the normal vector), then use $\sin\theta = \cos\phi$.

   ---

   1. Identify the direction vectors

   The $x$-axis has direction ratios $(1, 0, 0)$.

   The plane $2x - 3y + 6z - 11 = 0$ has normal vector $\vec{n} = (2, -3, 6)$.

   2. Find $\cos\phi$, where $\phi$ is the angle between the $x$-axis and $\vec{n}$

   The formula for the cosine of the angle between two vectors $\vec{a}$ and $\vec{b}$ is:

   $\cos\phi = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{a}| \, |\vec{b}|}$

   We take the absolute value because the angle between a line and a plane is always taken as acute (between $0^\circ$ and $90^\circ$).

   For $\vec{a} = (1, 0, 0)$ and $\vec{b} = (2, -3, 6)$:

   $\vec{a} \cdot \vec{b} = 1(2) + 0(-3) + 0(6) = 2$

   $|\vec{a}| = 1, \quad |\vec{b}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$

   So: …

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4. CBSE 2026Set A1 markMCQ
   Q.The angle between the straight lines $\frac{x-2}{2} = \frac{y-1}{7} = \frac{z+3}{-3}$ and $\frac{x+2}{-1} = \frac{y-4}{2} = \frac{z-5}{4}$ is

   (a) $\frac{\pi}{2}$

   (b) $0$

   (c) $\frac{\pi}{6}$

   (d) $\frac{\pi}{4}$

   ›Reveal solutionSolution

   Dot product of the direction ratios is zero $\Rightarrow$ angle $=\tfrac{\pi}{2}$.

   Direction ratios of the two lines are $(2,7,-3)$ and $(-1,2,4)$. Their dot product:

   $2(-1) + 7(2) + (-3)(4) = -2 + 14 - 12 = 0.$ …

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5. CBSE 2026Set A1 markMCQ
   Q.If the direction cosines of two straight lines are $l_1, m_1, n_1$ and $l_2, m_2, n_2$ then the cosine of the angle between the lines will be

   (a) $(l_1 + m_1 + n_1)(l_2 + m_2 + n_2)$

   (b) $\frac{l_1}{l_2} + \frac{m_1}{m_2} + \frac{n_1}{n_2}$

   (c) $l_1 l_2 + m_1 m_2 + n_1 n_2$

   (d) none of these

   ›Reveal solutionSolution

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

   Since direction cosines are already normalized ($l^2+m^2+n^2=1$), the cosine of the angle between two lines is just the dot product of their direction-cosine tri …

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6. CBSE 2026Set ANNUAL1 markMCQ
   Q.Write the angle between the lines through the points $(4,7,8)$, $(2,3,4)$ and $(-1,-2,1)$, $(1,2,5)$.

   (a) $\frac{\pi}{2}$

   (b) $\frac{\pi}{4}$

   (c) 0

   (d) $\frac{\pi}{6}$

   ›Reveal solutionSolution

   Both lines have the same direction ratios (up to a scalar), so they are parallel and the angle between them is $0$.

   Direction ratios of line 1 through $(4,7,8)$ and $(2,3,4)$:

   $(2-4,\ 3-7,\ 4-8)=(-2,-4,-4)\ \propto\ (1,2,2)$

   Direction ratios of line 2 through $(-1,-2,1)$ and $(1,2,5)$:

   $(1-(-1),\ 2-(-2),\ 5-1)=(2,4,4)\ \propto\ (1,2,2)$

   …

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7. CBSE 2025Set ANNUAL1 mark
   Q.Assertion (A): The line r⃗ = a⃗₁ + λb⃗₁ and r⃗ = a⃗₂ + μb⃗₂ are perpendicular when b⃗₁ − b⃗₂ = 0. Reason (R): The angle 'θ' between the line r⃗ = a⃗₁ + λb⃗₁ and r⃗ = a⃗₂ + μb⃗₂ is given by cosθ = (b⃗₁ − b⃗₂)/(|b⃗₁||b⃗₂|).

   ›Reveal solutionSolution

   Perpendicularity of two lines requires the dot product of their direction vectors to vanish ($\vec b_1\cdot\vec b_2=0$), not their difference; the angle formula also uses the dot product, not the difference — so both statements as given are false.

   Checking Assertion (A): Two lines $\vec r=\vec a_1+\lambda\vec b_1$ and $\vec r=\vec a_2+\mu\vec b_2$ are perpendicular when the angle between their direction vectors is $90^\circ$, i.e. when $\vec b_1\cdot\vec b_2=0$. The condition $\vec b_1-\vec b_2=0$ actually means $\vec b_1=\vec b_2$, which makes the lines parallel, not perpendicular. So Assertion (A) is false.

   …

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8. CBSE 2024Set A1 mark
   Q.Write True or False: If $l_1, m_1, n_1$ and $l_2, m_2, n_2$ are the direction cosines of two lines and $\theta$ is the acute angle between the two lines, then $\sin\theta = |l_1l_2 + m_1m_2 + n_1n_2|$.

   ›Reveal solutionSolution

   The correct formula uses $\cos\theta$, not $\sin\theta$: $\cos\theta=|l_1l_2+m_1m_2+n_1n_2|$.

   For two lines with direction cosines $l_1,m_1,n_1$ and $l_2,m_2,n_2$, the acute angle $\theta$ between them satisfies $\cos\theta = |l_1l_2+m_1m_2+n_1n_2|$ (this follows from $\vec u\cdot\vec v=|\vec u||\vec v|\cos\theta$ with $\vec u,\vec v$ unit vecto …

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9. CBSE 2024Set ANNUAL1 markMCQ
   Q.Assertion (A): The angle between the straight lines $\dfrac{x+1}{2} = \dfrac{y-2}{5} = \dfrac{z+3}{4}$ and $\dfrac{x-1}{1} = \dfrac{y+2}{2} = \dfrac{z-3}{-3}$ is $90°$. Reason (R): Skew lines are lines in different planes which are parallel and intersecting.

   (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

   (b) Both Assertion (A) and Reason (R) are true and Reason (R) is not the correct explanation of Assertion (A).

   (c) Assertion (A) is true, but Reason (R) is false.

   (d) Assertion (A) is false, but Reason (R) is true.

   ›Reveal solutionSolution

   Assertion (A) is true, but Reason (R) is false — option (c).

   Checking Assertion: direction ratios are $\vec{d_1}=(2,5,4)$ and $\vec{d_2}=(1,2,-3)$.

   $\vec{d_1}\cdot\vec{d_2} = 2(1)+5(2)+4(-3) = 2+10-12 = 0$

   Since the dot product is zero, the lines are perpendicular, i.e. the angle between them is $90^\circ$. Assertion (A) is true.

   …

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10. CBSE 2023Set 65/3/11 markMCQ
    Q.The value of $\lambda$ for which the angle between the lines $\vec{r} = \hat{i} + \hat{j} + \hat{k} + p(2\hat{i} + \hat{j} + 2\hat{k})$ and $\vec{r} = (1+q)\hat{i} + (1+q\lambda)\hat{j} + (1+q)\hat{k}$ is $\frac{\pi}{2}$ is :

    (a) $-4$

    (b) $4$

    (c) $2$

    (d) $-2$

    ›Reveal solutionSolution

    The angle between two lines is $\frac{\pi}{2}$ when their direction vectors are perpendicular (dot product = 0). For the given lines, this gives $\lambda = -4$, so the correct option is (a).

    Concept & Intuition

    When two lines in space are perpendicular, the angle between them is $90^\circ$ ($\pi/2$ radians). The key idea is that the direction vectors of the lines determine this angle — not their position vectors. The constant terms ($\hat{i} + \hat{j} + \hat{k}$ in the first line, and the $q$-dependent position in the second) only tell us where the lines are located, not which way they point.

    For two lines with direction vectors $\vec{d}_1$ and $\vec{d}_2$, the angle $\theta$ between them satisfies:

    $\cos\theta = \frac{\vec{d}_1 \cdot \vec{d}_2}{|\vec{d}_1|\,|\vec{d}_2|}$

    When $\theta = \frac{\pi}{2}$, $\cos\theta = 0$, so the numerator must be zero: $\vec{d}_1 \cdot \vec{d}_2 = 0$. That's the entire condition — no need to compute magnitudes or worry about the constant terms.

    Watch out

    A common mistake is to include the constant position vectors ($\hat{i} + \hat{j} + \hat{k}$ etc.) in the dot product. Those are just points on the line, not directions. Only the coefficients of $p$ and $q$ matter.

    Step-by-step solution

    1. Extract the direction vectors from each line. The first line is $\vec{r} = \hat{i} + \hat{j} + \hat{k} + p(2\hat{i} + \hat{j} + 2\hat{k})$. The coefficient of $p$ is the direction vector:

    $\vec{d}_1 = 2\hat{i} + \hat{j} + 2\hat{k}$

    The second line is $\vec{r} = (1+q)\hat{i} + (1+q\lambda)\hat{j} + (1+q)\hat{k}$.

    Rewrite it in the standard form $\vec{r} = \text{(constant)} + q(\text{direction})$:

    $\vec{r} = (\hat{i} + \hat{j} + \hat{k}) + q(\hat{i} + \lambda\hat{j} + \hat{k})$

    So the direction vector is: …

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11. CBSE 2023Set ANNUAL1 mark
    Q.If the coordinates of the points A, B, C and D are $(1,2,3), (4,5,7), (-4,3,-6)$ and $(2,9,2)$ respectively, the acute angle between the lines AB and CD will be ______.

    ›Reveal solutionSolution

    Compute direction vectors of AB and CD; if one is a positive scalar multiple of the other, the lines are parallel and the angle between them is $0^\circ$.

    Direction of AB: $B-A=(4-1,5-2,7-3)=(3,3,4)$

    Direction of CD: $D-C=(2-(-4),9-3,2-(-6))=(6,6,8)$

    …

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12. CBSE 2022Set ANNUAL1 markMCQ
    Q.The slope of the line which makes an angle $45°$ with the line $3x-y=-5$ are:

    (a) $1, \frac{1}{2}$

    (b) $1, -1$

    (c) $2, \frac{-1}{2}$

    (d) $\frac{1}{2}, -2$

    ›Reveal solutionSolution

    The line $3x-y=-5$ has slope 3; solving $\tan45°=\left|\dfrac{m-3}{1+3m}\right|=1$ gives $m=\dfrac12$ or $m=-2$.

    Rewrite $3x-y=-5$ as $y=3x+5$, so its slope is $m_1=3$.

    The angle $\theta$ between two lines of slopes $m_1,m_2$ satisfies $\tan\theta = \left|\dfrac{m_2-m_1}{1+m_1m_2}\right|$. With $\theta=45°$, $\tan45°=1$:

    $\left|\dfrac{m-3}{1+3m}\right| = 1$

    Case 1: $\dfrac{m-3}{1+3m}=1 \Rightarrow m-3 = 1+3m \Rightarrow -2m=4 \Rightarrow m=-2$.

    …

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