Q.State whether the following statement is True or False: The vector equation of the line $\dfrac{x-5}{3} = \dfrac{y+4}{7} = \dfrac{z-6}{2}$ is $\vec{r} = 5\hat{i} - 4\hat{j} + 6\hat{k} + \lambda(3\hat{i} + 7\hat{j} + 2\hat{k})$.

Showing the 12 most recent of 25 on this concept.

1. CBSE 2025Set 65/2/11 markMCQ
   Q.The equation of a line parallel to the vector $3\hat{i}+\hat{j}+2\hat{k}$ and passing through the point $(4,-3,7)$ is: (A) $x = 4t+3,\ y = -3t+1,\ z = 7t+2$ (B) $x = 3t+4,\ y = t+3,\ z = 2t+7$ (C) $x = 3t+4,\ y = t-3,\ z = 2t+7$ (D) $x = 3t+4,\ y = -t+3,\ z = 2t+7$

   ›Reveal solutionSolution

   The equation of a line is determined by a point it passes through and a vector parallel to it. We use the vector equation $\vec{r} = \vec{a} + t\vec{b}$ and convert it to parametric Cartesian form to find the correct option. The equation is $\boxed{x = 3t+4,\ y = t-3,\ z = 2t+7}$.

   To find the equation of a line in 3D space, we need two fundamental pieces of information:

   1. A point through which the line passes.
   2. A vector that is parallel to the line, which defines its direction.

   Imagine you are standing at a specific point in space. To define a unique line, you then need to know which way to walk. That "way to walk" is given by the direction vector. Any point on the line can be reached by starting at your initial point and moving some distance (which can be positive, negative, or zero) along the direction vector.

   Let $\vec{a}$ be the position vector of the known point $(x_1, y_1, z_1)$ through which the line passes. So, $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$.

   Let $\vec{b}$ be the vector parallel to the line, which is the direction vector. So, $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$.

   Let $\vec{r}$ be the position vector of any arbitrary point $(x, y, z)$ on the line. So, $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

   The vector equation of a line passing through a point with position vector $\vec{a}$ and parallel to a vector $\vec{b}$ is given by:

   $\vec{r} = \vec{a} + t\vec{b}$

   where $t$ is a scalar parameter.

   This equation states that to reach any point $\vec{r}$ on the line, you start at $\vec{a}$ and add a scalar multiple ($t$) of the direction vector $\vec{b}$. As $t$ varies over all real numbers, $\vec{r}$ traces out all points on the line.

   Let's apply this concept to the given problem.

   1. Identify the given information.

      The line passes through the point $(4, -3, 7)$. The position vector of this point is $\vec{a} = 4\hat{i} - 3\hat{j} + 7\hat{k}$.

      The line is parallel to the vector $3\hat{i} + \hat{j} + 2\hat{k}$. This is our direction vector, $\vec{b} = 3\hat{i} + \hat{j} + 2\hat{k}$.

   2. Formulate the vector equation of the line.

      Using the formula $\vec{r} = \vec{a} + t\vec{b}$, we substitute the identified vectors:

      $\vec{r} = (4\hat{i} - 3\hat{j} + 7\hat{k}) + t(3\hat{i} + \hat{j} + 2\hat{k})$

   3. Convert the vector equation to parametric Cartesian form.

      We know that $\vec{r}$ represents any point $(x, y, z)$ on the line, so $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$.

      Substitute this into the equation and group the $\hat{i}$, $\hat{j}$, and $\hat{k}$ components:

      $x\hat{i} + y\hat{j} + z\hat{k} = (4\hat{i} - 3\hat{j} + 7\hat{k}) + (3t\hat{i} + t\hat{j} + 2t\hat{k})$

      $x\hat{i} + y\hat{j} + z\hat{k} = (4 + 3t)\hat{i} + (-3 + t)\hat{j} + (7 + 2t)\hat{k}$ …

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2. CBSE 20231 mark
   Q.Assertion (A): The equation of the line passing through the points $(1, 2, 3)$ and $(3, -1, 3)$ is $\frac{x-3}{2} = \frac{y+1}{3} = \frac{z-3}{0}$. Reason (R): The equation of the line passing through the points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is $\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$.

   ›Reveal solutionSolution

   The key idea is that the equation of a line in 3D uses direction ratios from the difference of coordinates. Here, the given line has direction ratios $(2, -3, 0)$, but the assertion incorrectly writes the $y$-component as $+3$ instead of $-3$, making it false. The Reason (R) is the correct standard formula, so (A) is false but (R) is true.

   We need to check whether the line equation given in Assertion (A) actually passes through the two points, and whether Reason (R) correctly states the general formula.

   Concept first: In 3D geometry, the equation of a line through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is written using direction ratios — the differences $x_2 - x_1$, $y_2 - y_1$, $z_2 - z_1$. The symmetric form is:

   $\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}$

   provided none of the denominators is zero. If a denominator is zero, that coordinate is constant, and we write the numerator equal to zero (e.g., $z - z_1 = 0$).

   Now let’s apply this to the given points.

   1. Find the direction ratios.

      Points: $P(1, 2, 3)$ and $Q(3, -1, 3)$.

      Differences:

      $x_2 - x_1 = 3 - 1 = 2$

      $y_2 - y_1 = -1 - 2 = -3$

      $z_2 - z_1 = 3 - 3 = 0$

      So the direction ratios are $(2, -3, 0)$.

   2. Write the correct line equation using $Q$ as the base point.

      Using $(x_1, y_1, z_1) = (3, -1, 3)$, we get:

   $\frac{x - 3}{2} = \frac{y - (-1)}{-3} = \frac{z - 3}{0}$

   That simplifies to:

   $\frac{x - 3}{2} = \frac{y + 1}{-3} = \frac{z - 3}{0}$

   The $z$-coordinate is constant: $z = 3$, so the last part is written as $z - 3 = 0$.

   3. Compare with the assertion. Assertion (A) gives:

   $\frac{x-3}{2} = \frac{y+1}{3} = \frac{z-3}{0}$

   Notice the $y$-term: the denominator is $3$ instead of $-3$. That changes the sign of the direction ratio for $y$. The line with denominator $+3$ would have direction ratios $(2, 3, 0)$, which does not match the vector from $(1,2,3)$ to $(3,-1,3)$. So (A) is false. …

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3. CBSE 2019Set 65/1/11 mark
   Q.If a line makes angles $90^\circ, 135^\circ, 45^\circ$ with the $x$, $y$ and $z$ axes respectively, find its direction cosines.

   (OR)

   Find the vector equation of a line which passes through the point $(3, 4, 5)$ and is parallel to the vector $2\hat{i} + 2\hat{j} - 3\hat{k}$.

   ›Reveal solutionSolution

   1. Direction cosines $l=0,\ m=-\tfrac{1}{\sqrt2},\ n=\tfrac{1}{\sqrt2}$.
   2. Vector equation $\vec r=(3\hat i+4\hat j+5\hat k)+\lambda(2\hat i+2\hat j-3\hat k)$.

   Part (a)

   The direction cosines of a line are the cosines of the angles $\alpha,\beta,\gamma$ it makes with the positive $x,y,z$ axes, and they satisfy $l^2+m^2+n^2=1$.

   Given: $\alpha=90^\circ,\ \beta=135^\circ,\ \gamma=45^\circ$.

   1. $l=\cos 90^\circ = 0$ (perpendicular to the $x$-axis).
   2. $m=\cos 135^\circ=\cos(180^\circ-45^\circ)=-\cos 45^\circ=-\dfrac{1}{\sqrt2}$.
   3. $n=\cos 45^\circ=\dfrac{1}{\sqrt2}$.
   4. Verify: $0^2+\left(-\tfrac{1}{\sqrt2}\right)^2+\left(\tfrac{1}{\sqrt2}\right)^2=0+\tfrac12+\tfrac12=1$ ✓. …

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4. CBSE 2026Set ANNUAL1 markMCQ
   Q.A line passing through $(2, -1, 3)$ has direction ratio (d.r.) $(3, -1, 2)$, then its equation is

   (a) $\dfrac{x+2}{3} = \dfrac{y-1}{-1} = \dfrac{z-3}{2}$

   (b) $\dfrac{x+2}{3} = \dfrac{y+1}{-1} = \dfrac{z-3}{2}$

   (c) $\dfrac{x-2}{3} = \dfrac{y+1}{-1} = \dfrac{z-3}{2}$

   (d) None of these

   ›Reveal solutionSolution

   A line through point $(x_1,y_1,z_1)$ with direction ratios $(a,b,c)$ has equation $\dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}$.

   Here $(x_1,y_1,z_1)=(2,-1,3)$ and $(a,b,c)=(3,-1,2)$.

   …

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5. CBSE 2026Set ANNUAL1 mark
   Q.Find the vector equation of the line passing through the point (2, 3, 4) and parallel to the vector 2î + 5ĵ − 3k̂.

   ›Reveal solutionSolution

   The vector equation of a line through a point with position vector $\vec a$, parallel to $\vec b$, is $\vec r = \vec a + \lambda \vec b$.

   Position vector of the given point: $\vec a = 2\hat i+3\hat j+4\hat k$.

   Direction vector: $\vec b = 2\hat i+5\hat j-3\hat k$.

   …

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6. CBSE 2025Set 65/2/11 markMCQ
   Q.The line $x = 1 + 5\mu,\ y = -5 + \mu,\ z = -6 - 3\mu$ passes through which of the following point? (A) $(1, -5, 6)$ (B) $(1, 5, 6)$ (C) $(1, -5, -6)$ (D) $(-1, -5, 6)$

   ›Reveal solutionSolution

   To check if a point lies on a line given by parametric equations, substitute the point's coordinates into the equations and verify if a single, consistent value of the parameter $\mu$ is obtained for all three coordinates. The point $(1, -5, -6)$ yields $\mu=0$ for all equations, so it lies on the line.

   Concept and Intuition

   A line in three-dimensional space can be described using parametric equations. These equations express the $x, y,$ and $z$ coordinates of any point on the line in terms of a single parameter, often denoted by $\mu$ (or $t, \lambda$, etc.).

   The given equations are:

   $x = 1 + 5\mu$

   $y = -5 + \mu$

   $z = -6 - 3\mu$

   This means that as $\mu$ varies over all real numbers, the point $(x, y, z)$ traces out the entire line. Each specific value of $\mu$ corresponds to a unique point on the line.

   For a given point $(x_0, y_0, z_0)$ to lie on this line, there must exist one specific value of the parameter $\mu$ such that when this $\mu$ is substituted into all three equations, it simultaneously produces $x_0, y_0,$ and $z_0$. If we substitute the coordinates of a candidate point into the equations and solve for $\mu$ from each equation, we must get the same value of $\mu$ from all three equations. If the $\mu$ values are different, the point does not lie on the line.

   Step-by-Step Solution

   1. Understand the condition for a point to be on the line:

      A point $(x_0, y_0, z_0)$ lies on the line $x = 1 + 5\mu,\ y = -5 + \mu,\ z = -6 - 3\mu$ if and only if there exists a single real value of $\mu$ that satisfies all three equations simultaneously when $x=x_0, y=y_0, z=z_0$.

   2. Test Option (A): $(1, -5, 6)$

      Substitute $x=1, y=-5, z=6$ into the parametric equations:

      • For $x$: $1 = 1 + 5\mu \implies 5\mu = 0 \implies \mu = 0$
      • For $y$: $-5 = -5 + \mu \implies \mu = 0$
      • For $z$: $6 = -6 - 3\mu \implies 12 = -3\mu \implies \mu = -4$ Since the values of $\mu$ obtained are $0, 0,$ and $-4$, they are not consistent. Therefore, the point $(1, -5, 6)$ does not lie on the line.

   3. Test Option (B): $(1, 5, 6)$

      Substitute $x=1, y=5, z=6$ into the parametric equations:

      • For $x$: $1 = 1 + 5\mu \implies 5\mu = 0 \implies \mu = 0$
      • For $y$: $5 = -5 + \mu \implies \mu = 10$ The values of $\mu$ obtained are $0$ and $10$, which are not consistent. There is no need to check the $z$-coordinate. Therefore, the point $(1, 5, 6)$ does not lie on the line. …

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7. CBSE 2025Set X11 markMCQ
   Q.The equation of $y$-axis in space is

   (a) $x = 0,\ y = 0$

   (b) $x = 0,\ z = 0$

   (c) $y = 0,\ z = 0$

   (d) $y = 0$

   ›Reveal solutionSolution

   Coordinate axis as intersection of two planes — correct option (b).

   A point lies on the $y$-axis exactly when its $x$ and $z$ coordinates are both zero, with $y$ arbitrary. Henc …

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8. CBSE 2025Set ANNUAL1 markMCQ
   Q.The cartesian equation of the line passing through point $(1,2,3)$ and parallel to the line $\dfrac{x+3}{3}=\dfrac{y-4}{5}=\dfrac{z+8}{6}$ will be -

   (a) $\dfrac{x-1}{3}=\dfrac{y-2}{5}=\dfrac{z-3}{6}$

   (b) $\dfrac{x+1}{3}=\dfrac{y+2}{5}=\dfrac{z+3}{6}$

   (c) $\dfrac{x+3}{1}=\dfrac{y-4}{2}=\dfrac{z+8}{3}$

   (d) $\dfrac{x+2}{3}=\dfrac{y-6}{5}=\dfrac{z+5}{6}$

   ›Reveal solutionSolution

   Parallel lines share the same direction ratios; only the point through which the line passes changes.

   The given line $\dfrac{x+3}{3}=\dfrac{y-4}{5}=\dfrac{z+8}{6}$ has direction ratios $(3,5,6)$.

   A line parallel to it and passing through $(1,2,3)$ has the same direction ratios: …

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9. CBSE 2025Set ANNUAL1 mark
   Q.Find the vector equation of the line passing through the points $(3,2,0)$ and $(1,2,5)$.

   ›Reveal solutionSolution

   The vector equation of a line through two points $A$ and $B$ is $\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})$.

   Let $A(3,2,0)$ and $B(1,2,5)$, so $\vec{a}=3\hat{i}+2\hat{j}+0\hat{k}$ and $\vec{b}=1\hat{i}+2\hat{j}+5\hat{k}$.

   $\vec{b}-\vec{a}=(1-3)\hat{i}+(2-2)\hat{j}+(5-0)\hat{k}=-2\hat{i}+0\hat{j}+5\hat{k}$

   So the vector equation of the line is: …

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10. CBSE 2025Set ANNUAL1 markMCQ
    Q.The vector equation of the $x$-axis is

    (a) $\vec{r} = \hat{i}$

    (b) $\vec{r} = \hat{j} + \hat{k}$

    (c) $\vec{r} = \lambda\hat{i}$

    (d) none of these

    ›Reveal solutionSolution

    The x-axis consists of all points of the form (λ, 0, 0), which as a position vector is simply λî.

    A point on the x-axis has coordinates $(\lambda, 0, 0)$ for some real λ. As a position vector this is:

    $\vec r = \lambda\hat i + 0\hat j + 0\hat k = \lambda \hat i$

    …

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11. CBSE 2025Set ANNUAL1 mark
    Q.Find the vector equation of the line through the points A(3, 4, −7) and B(1, −1, 6).

    ›Reveal solutionSolution

    The vector equation of a line through two given points $A$ and $B$ is $\vec r = \vec a + \lambda(\vec b - \vec a)$, where $\vec a, \vec b$ are the position vectors of $A, B$.

    Given: $A(3, 4, -7)$, $B(1, -1, 6)$

    Step 1 — position vectors:

    $\vec a = 3\hat i + 4\hat j - 7\hat k, \qquad \vec b = \hat i - \hat j + 6\hat k$

    Step 2 — direction vector $\vec b - \vec a$:

    $\vec b - \vec a = (1-3)\hat i + (-1-4)\hat j + (6-(-7))\hat k = -2\hat i - 5\hat j + 13\hat k$

    …

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12. CBSE 2024Set ANNUAL1 markMCQ
    Q.Equation of a line parallel to x-axis and passing through the origin is -

    (a) $\dfrac{x}{0} = \dfrac{y}{0} = \dfrac{z}{0}$

    (b) $\dfrac{x}{0} = \dfrac{y}{1} = \dfrac{z}{1}$

    (c) $\dfrac{x}{1} = \dfrac{y}{0} = \dfrac{z}{0}$

    (d) $\dfrac{x}{1} = \dfrac{y}{1} = \dfrac{z}{1}$

    ›Reveal solutionSolution

    The x-axis direction is $(1,0,0)$; a line through the origin with this direction has equation $x/1=y/0=z/0$.

    A line parallel to the x-axis has direction ratios proportional to $(1,0,0)$. The symmetric (cartesian) form of a line passing through a point $(x_1,y_1,z_1)$ with direction ratios $(a,b,c)$ is:

    $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$ …

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