The equation of the line passing through the point $(1, 1, 1)$ and parallel to the $z$-axis is: (a) $\frac{x}{1} = \frac{y}{1} = \frac{z}{1}$ (b) $\frac{x-1}{1} = \frac{y-1}{1} = \frac{z-1}{1}$ (c) $\frac{x}{0} = \frac{y}{0} = \frac{z}{z-1}$ (d) $\frac{x-1}{0} = \frac{y-1}{0} = \frac{z-1}{1}$

> [!TLDR] > A line parallel to the z-axis has a direction vector of the form $\langle 0, 0, k \rangle$. Using the given point $(1,1,1)$ and this direction, the equation of the line is $\boxed{\frac{x-1}{0} = \frac{y-1}{0} = \frac{z-1}{1}}$. The equation of a line in three-dimensional space requires two pieces of information: a point that the line passes through, and a vector that gives the direction of the line. **Concept and Intuition** 1. **General Form of a Line:** If a line passes through a point $P_1(x_1, y_1, z_1)$ and is parallel to a direction vector $\vec{b} = \langle a, b, c \rangle$, its vector equation is given by $\vec{r} = \vec{p_1} + \lambda \vec{b}$, where $\vec{r} = \langle x, y, z \rangle$ is a general point on the line, $\vec{p_1} = \langle x_1, y_1, z_1 \rangle$ is the position vector of the given point, and $\lambda$ is a scalar parameter. This expands to: $\langle x, y, z \rangle = \langle x_1, y_1, z_1 \rangle + \lambda \langle a, b, c \rangle$ $\langle x, y, z \rangle = \langle x_1 + \lambda a, y_1 + \lambda b, z_1 + \lambda c \rangle$ From this, we get the parametric equations: $x = x_1 + \lambda a$ $y = y_1 + \lambda b$ $z = z_1 + \lambda c$ If $a, b, c$ are all non-zero, we can solve for $\lambda$ from each equation and equate them to get the symmetric (Cartesian) form: > [!FORMULA] > $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$ 2. **Understanding "Parallel to the z-axis":** A line parallel to the z-axis means its direction vector must be aligned with the z-axis. The standard unit vector along the z-axis is $\hat{k} = \langle 0, 0, 1 \rangle$. Therefore, any vector parallel to the z-axis will have the form $\langle 0, 0, k \rangle$ for some non-zero scalar $k$. We can choose the simplest one, $\langle 0, 0, 1 \rangle$, as our direction vector. This implies that the x and y components of the direction vector are zero. 3. **Handling Zero Denominators in Cartesian Form:*

The equation of the line passing through the point (1, 1, 1) and parallel to the z-axis is: (a) (x)/(1) = (y)/

Question

(1, 1, 1)

(a)

\frac{x}{1} = \frac{y}{1} = \frac{z}{1}

(b)

\frac{x-1}{1} = \frac{y-1}{1} = \frac{z-1}{1}

(c)

\frac{x}{0} = \frac{y}{0} = \frac{z}{z-1}

(d)

\frac{x-1}{0} = \frac{y-1}{0} = \frac{z-1}{1}

CBSECBSE Class XII Board 2023MCQ· 1mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

A line parallel to the z-axis has a direction vector of the form $\langle 0, 0, k \rangle$ . Using the given point $(1,1,1)$ and this direction, the equation of the line is $\boxed{\frac{x-1}{0} = \frac{y-1}{0} = \frac{z-1}{1}}$ .

The equation of a line in three-dimensional space requires two pieces of information: a point that the line passes through, and a vector that gives the direction of the line.

Concept and Intuition

General Form of a Line:

If a line passes through a point $P_1(x_1, y_1, z_1)$ and is parallel to a direction vector $\vec{b} = \langle a, b, c \rangle$ , its vector equation is given by $\vec{r} = \vec{p_1} + \lambda \vec{b}$ , where $\vec{r} = \langle x, y, z \rangle$ is a general point on the line, $\vec{p_1} = \langle x_1, y_1, z_1 \rangle$ is the position vector of the given point, and $\lambda$ is a scalar parameter.

This expands to:

$\langle x, y, z \rangle = \langle x_1, y_1, z_1 \rangle + \lambda \langle a, b, c \rangle$

$\langle x, y, z \rangle = \langle x_1 + \lambda a, y_1 + \lambda b, z_1 + \lambda c \rangle$

From this, we get the parametric equations:

$x = x_1 + \lambda a$

$y = y_1 + \lambda b$

$z = z_1 + \lambda c$

If $a, b, c$ are all non-zero, we can solve for $\lambda$ from each equation and equate them to get the symmetric (Cartesian) form:
$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$
Understanding "Parallel to the z-axis":

A line parallel to the z-axis means its direction vector must be aligned with the z-axis. The standard unit vector along the z-axis is $\hat{k} = \langle 0, 0, 1 \rangle$ . Therefore, any vector parallel to the z-axis will have the form $\langle 0, 0, k \rangle$ for some non-zero scalar $k$ . We can choose the simplest one, $\langle 0, 0, 1 \rangle$ , as our direction vector. This implies that the x and y components of the direction vector are zero.
Handling Zero Denominators in Cartesian Form:

When one or more components of the direction vector are zero, the symmetric form needs careful interpretation. For example, if $a=0$ , then from $x = x_1 + \lambda a$ , we get $x = x_1$ . This means the x-coordinate of every point on the line is fixed at $x_1$ . In the symmetric form, we write $\frac{x - x_1}{0}$ to represent the condition $x - x_1 = 0$ . This is a notational convention and does not imply actual division by zero. It simply means that the numerator must be zero, fixing that coordinate.

Now, let's apply these concepts to the given problem.

Step-by-step Solution

Identify the given point:

The line passes through the point $(1, 1, 1)$ .

So, $(x_1, y_1, z_1) = (1, 1, 1)$ .
Determine the direction vector:

The line is parallel to the z-axis.

Important

A direction vector for a line parallel to the z-axis has its x and y components equal to zero, and its z component non-zero.

We can choose the simplest such vector: $\vec{b} = \langle 0, 0, 1 \rangle$ .

Thus, the direction ratios are $(a, b, c) = (0, 0, 1)$ .
Write the parametric equations of the line:

Using $x = x_1 + \lambda a$ , $y = y_1 + \lambda b$ , $z = z_1 + \lambda c$ :

$x = 1 + \lambda(0) \implies x = 1$ …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 25 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$(2, -1, 3)$
(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 $(a,b,c)=(3,-1,2)$ .

…

Start your 14-day free trial to see the full solution →
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: $\vec b = 2\hat i+5\hat j-3\hat k$ .

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set 65/2/11 markMCQ
$3\hat{i}+\hat{j}+2\hat{k}$
›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:

A point through which the line passes.

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, $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ .
$\vec{a}$
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.

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, $\vec{b} = 3\hat{i} + \hat{j} + 2\hat{k}$ .

Formulate the vector equation of the line.

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

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

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:

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

$x\hat{i} + y\hat{j} + z\hat{k} = (4 + 3t)\hat{i} + (-3 + t)\hat{j} + (7 + 2t)\hat{k}$ …
Start your 14-day free trial to see the full solution →
CBSE 2025Set 65/2/11 markMCQ
$x = 1 + 5\mu,\ y = -5 + \mu,\ z = -6 - 3\mu$
›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

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 $x=x_0, y=y_0, z=z_0$ .

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.

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. …
Start your 14-day free trial to see the full solution →
CBSE 2025Set X11 markMCQ
$y$
(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 …
Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$(1,2,3)$
(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: …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$(3,2,0)$
›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: …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$x$
(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:

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

…

Start your 14-day free trial to see the full solution →
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 $\vec b - \vec a$ :

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

…

Start your 14-day free trial to see the full solution →
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:

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

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 markMCQ
$x$
(a) $y=0, z=0$
(b) $x=0, y=0$
(c) $x=0, z=0$
(d) $x=0, y=0, z=0$
›Reveal solutionSolution

Points on the x-axis have the form (x,0,0), so the axis is described by y=0 and z=0 together.

Any point on the x-axis is of the form $(x,0,0)$ - the y and z coordinates are always zero while x varies freely. So t …

Start your 14-day free trial to see the full solution →
CBSE 2024Set ANNUAL1 mark
$(-2, 4, -5)$
›Reveal solutionSolution

A parallel line shares the same direction ratios; use the point–direction form.

The given line is

$\dfrac{x+3}{3}=\dfrac{y-4}{5}=\dfrac{z+8}{6},$

so its direction ratios are $\langle 3, 5, 6\rangle$ .

A line parallel to it has the same direction ratios $\langle 3,5,6\rangle$ .

The Cartesian equation of a line through $(x_1,y_1,z_1)=(-2,4,-5)$ with direction ratios $\langle a,b,c\rangle$ is …

Start your 14-day free trial to see the full solution →

Q.The equation of the line passing through the point $(1, 1, 1)$ and parallel to the $z$ -axis is:

Concept understanding — Vector Equation Of Line

Vector Equation of a Line

How to read it

Line through two points

Example

Concept and Intuition

Step-by-Step Solution

Q.The equation of the line passing through the point (1,1,1)(1, 1, 1)(1,1,1) and parallel to the zzz-axis is:

🎓Unlock everything free for 14 days

Q.The equation of the line passing through the point $(1, 1, 1)$ and parallel to the $z$ -axis is:

Unlock everything free for 14 days