Q.Direction cosines of the line $\frac{x-1}{2} = \frac{1-y}{3} = \frac{2z-1}{12}$ are:

Showing the 12 most recent of 15 on this concept.

1. CBSE 2026Set 65/3/11 markMCQ
   Q.Direction ratios of lines $L_1$ and $L_2$ are $\langle 12, -3, 9 \rangle$ and $\langle 4, q, -p \rangle$ respectively. The values of $p$ and $q$ for which $L_1$ and $L_2$ are parallel are respectively: (A) $-1, 3$ (B) $3, 1$ (C) $-3, -1$ (D) $-1, -3$

   ›Reveal solutionSolution

   Two lines are parallel when their direction vectors are scalar multiples of each other. For the given direction ratios, this condition gives $p = -3$ and $q = -1$, which corresponds to option (C).

   The key idea here is simple: two lines in space are parallel if and only if their direction vectors are proportional. That means one vector is a constant multiple of the other — every component must scale by the same factor.

   Let’s unpack what that means for the numbers we have.

   1. Write down the direction vectors.

      For $L_1$, the direction ratios are $\langle 12, -3, 9 \rangle$.

      For $L_2$, they are $\langle 4, q, -p \rangle$.

   2. Set up the proportionality condition.

      If $L_1 \parallel L_2$, then there exists some scalar $k$ such that:

   $\langle 12, -3, 9 \rangle = k \cdot \langle 4, q, -p \rangle$

   This gives us three equations:

   $12 = 4k, \quad -3 = kq, \quad 9 = k(-p)$

   3. Solve for $k$ from the first equation.

   $12 = 4k \quad\Rightarrow\quad k = 3$

   4. Use $k = 3$ to find $q$. From $-3 = kq$:

   $-3 = 3q \quad\Rightarrow\quad q = -1$

   5. Use $k = 3$ to find $p$. From $9 = k(-p)$: $9 = 3(-p) \quad\Rightarrow\quad 9 = -3p \quad\Rightarrow\quad p = -3$ …

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2. CBSE 2025Set E1 markMCQ
   Q.The direction ratios of the straight line $\frac{x - 19}{13} = \frac{y - 17}{11} = \frac{z - 15}{9}$ are

   (a) $19, 17, 15$

   (b) $13, 11, 9$

   (c) $19, 17, 9$

   (d) None of these

   ›Reveal solutionSolution

   In $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$, the direction ratios are $a,b,c$.

   …

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3. CBSE 2025Set ANNUAL1 markMCQ
   Q.The direction ratios of the line 6x − 2 = 3y + 1 = 2z − 2 is .................

   (a) 3, 2, 1

   (b) 1/3, 1/3, −1

   (c) 1, 2, 3

   (d) 1/6, 1/3, −1

   ›Reveal solutionSolution

   Rewrite the symmetric line equation in the standard form $\dfrac{x-x_1}{a} = \dfrac{y-y_1}{b} = \dfrac{z-z_1}{c}$; the denominators $a,b,c$ are the direction ratios.

   Given: $6x - 2 = 3y + 1 = 2z - 2$

   Step 1 — rewrite each part as (variable − constant) form by factoring the coefficient of the variable out of the linear expression:

   $6x - 2 = 6\left(x - \frac{1}{3}\right), \quad 3y+1 = 3\left(y + \frac{1}{3}\right), \quad 2z - 2 = 2(z - 1)$

   Step 2 — set the common parameter: Let each equal $6k$ (a convenient common value, using LCM of 6, 3, 2):

   $6\left(x - \tfrac13\right) = 6k \Rightarrow \frac{x - 1/3}{1} = k$ …

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4. CBSE 2025Set ANNUAL1 markMCQ
   Q.The fixed point which the line 3x + 1 = 6y − 2 = 1 − z passes through is –

   (i) (−1/3, 1/3, 1)

   (ii) (1/3, −1/3, 1)

   (iii) (1/3, 1/3, 1)

   (iv) (−1/3, −1/3, 1)

   ›Reveal solutionSolution

   Rewrite the equal-ratios form as $\dfrac{x-x_0}{l}=\dfrac{y-y_0}{m}=\dfrac{z-z_0}{n}$; the point $(x_0,y_0,z_0)$ is on the line.

   Given $3x+1=6y-2=1-z$. Set each equal to a parameter $t$:

   $3x+1=t \Rightarrow x=\dfrac{t-1}{3}=-\dfrac13+\dfrac{t}{3}$

   $6y-2=t \Rightarrow y=\dfrac{t+2}{6}=\dfrac13+\dfrac{t}{6}$

   $1-z=t \Rightarrow z=1-t$

   …

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5. CBSE 2025Set ANNUAL1 markMCQ
   Q.The straight line (x−2)/3 = (y−3)/1 = (z+1)/0 is –

   (i) parallel to the x-axis

   (ii) parallel to the y-axis

   (iii) parallel to the z-axis

   (iv) perpendicular to the z-axis

   ›Reveal solutionSolution

   The direction ratios are $(3,1,0)$; a zero z-component means the direction vector is perpendicular to the z-axis.

   The line $\dfrac{x-2}{3}=\dfrac{y-3}{1}=\dfrac{z+1}{0}$ has direction ratios $(3,1,0)$ (the '0' denominator is conventional notation meaning the z-coordinate is constant, $z=-1$, along the line).

   The z-axis has direction $(0,0,1)$. Dot product of the line's direction with the z-axis direction: $3(0)+1(0)+0(1)=0$.

   …

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6. CBSE 20241 markMCQ
   Q.The Cartesian equations of a line are given as $6x - 2 = 3y + 1 = 2z - 2$ The direction ratios of the line are : (A) $2, -1, 3$ (B) $1, -2, -3$ (C) $1, 2, 3$ (D) $3, 1, 2$

   ›Reveal solutionSolution

   Writing the line in standard symmetric form gives direction ratios $(1, 2, 3)$ — option (C).

   For a line in symmetric form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$, the denominators $(a, b, c)$ are the direction ratios — but only when each variable has coefficient $1$. The given equation has coefficients $6, 3, 2$, so make each variable's coefficient $1$ first.

   Rewrite each expression:

   $6x - 2 = 6\left(x - \tfrac{1}{3}\right), \qquad 3y + 1 = 3\left(y + \tfrac{1}{3}\right), \qquad 2z - 2 = 2(z - 1).$

   Equating them gives $6\left(x - \tfrac{1}{3}\right) = 3\left(y + \tfrac{1}{3}\right) = 2(z - 1)$. Dividing the whole chain by $6$ puts it in standard form:

   $\frac{x - \tfrac{1}{3}}{1} = \frac{y + \tfrac{1}{3}}{2} = \frac{z - 1}{3}.$

   The denominators $1, 2, 3$ are the direction ratios. …

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7. CBSE 2024Set 65/2/11 markMCQ
   Q.Direction ratios of a vector parallel to line $\dfrac{x - 1}{2} = -y = \dfrac{2z + 1}{6}$ are: (A) $2, -1, 6$ (B) $2, 1, 6$ (C) $2, 1, 3$ (D) $2, -1, 3$

   ›Reveal solutionSolution

   Rewrite the symmetric equation so all three parts equal the same parameter, then read off the denominators as direction ratios: $2, -1, 3$.

   The symmetric form of a line encodes its direction ratios directly in its structure. When a line is written as

   $\frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c},$

   the numbers $a, b, c$ are the direction ratios of any vector parallel to that line. The idea is simple: if we set each fraction equal to a parameter $t$, we get $x = x_0 + at$, $y = y_0 + bt$, $z = z_0 + ct$, which shows the line moves in the direction $\langle a, b, c \rangle$.

   The catch here is that the given equation isn't quite in standard form yet. We have

   $\frac{x - 1}{2} = -y = \frac{2z + 1}{6}.$

   Notice the middle term is $-y$, not $\frac{y - y_0}{b}$. We need to massage this into the proper shape.

   ---

   Step-by-step extraction:

   1. Rewrite the $y$-term with a denominator. The expression $-y$ can be written as $\frac{y - 0}{-1}$, because

   $-y = \frac{-y}{1} = \frac{y}{-1} = \frac{y - 0}{-1}.$

   2. Rewrite the $z$-term in standard form. We have $\frac{2z + 1}{6}$. Factor out the coefficient of $z$ from the numerator:

   $2z + 1 = 2\left(z + \frac{1}{2}\right),$

   so …

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8. CBSE 2024Set ANNUAL1 markMCQ
   Q.The direction ratio of the line $\frac{x}{1}=\frac{y}{1}=\frac{z}{1}$ is

   (a) $\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}$

   (b) $1, 1, 1$

   (c) $0, 0, 0$

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

   ›Reveal solutionSolution

   In the symmetric form x/a=y/b=z/c, the numbers a,b,c ARE the direction ratios of the line.

   The line $\dfrac{x}{1}=\dfrac{y}{1}=\dfrac{z}{1}$ is in standard symmetric form $\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}$ …

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9. CBSE 2023Set 65/1/11 markMCQ
   Q.Direction cosines of the line $\frac{x-1}{2} = \frac{1-y}{3} = \frac{2z-1}{12}$ are:

   (a) $\frac{2}{7}, \frac{3}{7}, \frac{6}{7}$

   (b) $\frac{2}{\sqrt{157}}, -\frac{3}{\sqrt{157}}, \frac{12}{\sqrt{157}}$

   (c) $\frac{2}{7}, -\frac{3}{7}, -\frac{6}{7}$

   (d) $\frac{2}{7}, -\frac{3}{7}, \frac{6}{7}$

   ›Reveal solutionSolution

   To find direction cosines, first convert the line equation to its standard symmetric form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ to correctly identify the direction ratios $(a, b, c)$. Then, divide each direction ratio by the magnitude $\sqrt{a^2+b^2+c^2}$ to get the direction cosines. The direction cosines are $\left(\frac{2}{7}, -\frac{3}{7}, \frac{6}{7}\right)$.

   When working with lines in 3D space, direction ratios and direction cosines are fundamental concepts that describe the orientation of the line.

   • Direction Ratios: These are any set of three numbers $(a, b, c)$ that are proportional to the components of a vector parallel to the line. There are infinitely many sets of direction ratios for a given line, as any scalar multiple $(ka, kb, kc)$ also represents valid direction ratios.
   • Direction Cosines: These are the cosines of the angles that the line makes with the positive $x, y,$ and $z$ axes, usually denoted as $(\ell, m, n)$. Unlike direction ratios, direction cosines are unique for a given direction (up to a sign, depending on which way along the line you consider). They have the property that $\ell^2 + m^2 + n^2 = 1$.

   The symmetric form of a line's equation is given by $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$. In this standard form, $(x_1, y_1, z_1)$ is a point on the line, and $(a, b, c)$ are the direction ratios of the line. The key here is that the coefficients of $x, y,$ and $z$ in the numerators must be $+1$. If they are not, we need to manipulate the equation to bring it into this standard form before we can correctly identify the direction ratios.

   Let's apply this understanding to the given problem.

   1. Convert the given equation to standard symmetric form.

      The given equation is $\frac{x-1}{2} = \frac{1-y}{3} = \frac{2z-1}{12}$.

      We need to ensure that the numerators are of the form $(x-x_1)$, $(y-y_1)$, and $(z-z_1)$.

      • The first term, $\frac{x-1}{2}$, is already in the correct form. Here, $x_1=1$ and the direction ratio component is $a=2$.

      • The second term is $\frac{1-y}{3}$. To get $(y-y_1)$, we factor out $-1$ from the numerator:

        $\frac{1-y}{3} = \frac{-(y-1)}{3} = \frac{y-1}{-3}$.

        Now it's in the correct form. Here, $y_1=1$ and the direction ratio component is $b=-3$.

      • The third term is $\frac{2z-1}{12}$. To get $(z-z_1)$, we factor out $2$ from the numerator:

        $\frac{2z-1}{12} = \frac{2(z - \frac{1}{2})}{12} = \frac{z - \frac{1}{2}}{6}$.

        Now it's in the correct form. Here, $z_1=\frac{1}{2}$ and the direction ratio component is $c=6$.

      So, the standard symmetric form of the line's equation is:

   $\frac{x-1}{2} = \frac{y-1}{-3} = \frac{z-\frac{1}{2}}{6}$

   > [!WARNING]
   > A common mistake is to directly take the denominators as direction ratios without ensuring the numerators are in the form $(x-x_1)$, $(y-y_1)$, and $(z-z_1)$. Always check the coefficients of $x, y, z$ in the numerator; they must be $+1$.
   

   2. Identify the direction ratios. …

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10. CBSE 2023Set E1 markMCQ
    Q.The direction ratios of the straight line $\dfrac{x+1}{3}=\dfrac{y-2}{3}=\dfrac{z-5}{6}$ are

    (a) $1,-2,5$

    (b) $3,2,5$

    (c) $3,3,6$

    (d) $1,3,5$

    ›Reveal solutionSolution

    The denominators $3,3,6$ are the line's direction ratios.

    A line in symmetric form $\dfrac{x-x_0}{a}=\dfrac{y-y_0}{b}=\dfrac{z-z_0}{c}$ has direction ratios $a,b,c$ (the denominators).

    …

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11. CBSE 2023Set ANNUAL1 markMCQ
    Q.Line $\dfrac{x-5}{3} = \dfrac{y+4}{7} = \dfrac{z-6}{2}$ passes through the point

    (a) $(5, -4, 6)$

    (b) $(5, 4, -6)$

    (c) $(4, 5, 6)$

    (d) none of these

    ›Reveal solutionSolution

    In the symmetric form $\dfrac{x-x_0}{a}=\dfrac{y-y_0}{b}=\dfrac{z-z_0}{c}$, the point $(x_0,y_0,z_0)$ lies on the line.

    Given $\dfrac{x-5}{3}=\dfrac{y+4}{7}=\dfrac{z-6}{2}$, rewrite the middle term as $\dfrac{y-(-4)}{7}$. So the fixed point is $x_0=5,\ y_0=-4,\ z_0=6$, i.e. $(5,-4,6)$.

    …

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12. CBSE 2023Set ANNUAL1 markMCQ
    Q.The line (x−3)/2 = (y−4)/−3 = (z−1)/5 crosses the xy plane at –

    (a) (13/5, 23/5, 0)

    (b) (13/5, 0, 23/5)

    (c) (0, 13/5, 23/5)

    (d) (13/5, 23/5, 33/5)

    ›Reveal solutionSolution

    Write the line in parametric form and set $z=0$ (the condition for the xy-plane) to solve for the parameter.

    Write the line $\dfrac{x-3}{2}=\dfrac{y-4}{-3}=\dfrac{z-1}{5}=t$ in parametric form:

    $x = 3+2t,\quad y = 4-3t,\quad z = 1+5t.$

    The xy-plane is where $z=0$:

    $1+5t=0 \implies t=-\frac15.$

    Substitute back: …

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