(a) Find the value of $b$ so that the lines $\frac{x-1}{2} = \frac{y-b}{3} = \frac{z-3}{4}$ and $\frac{x-4}{5} = \frac{y-1}{2} = z$ are intersecting lines. Also, find the point of intersection of these given lines. OR (b) Find the equations of all the sides of the parallelogram $ABCD$ whose vertices are $A(4, 7, 8)$, $B(2, 3, 4)$, $C(-1, -2, 1)$ and $D(1, 2, 5)$. Also, find the coordinates of the foot of the perpendicular from $A$ to $CD$.

(a) Find the value of b so that the lines (x-1)/(2) = (y-b)/(3) = (z-3)/(4) and (x-4)/(5) = (y-1)/(2) = z are

Question

b

(OR)

(b) Find the equations of all the sides of the parallelogram

ABCD

whose vertices are

A(4, 7, 8)

B(2, 3, 4)

C(-1, -2, 1)

and

D(1, 2, 5)

. Also, find the coordinates of the foot of the perpendicular from

A

CD

CBSECBSE Class XII Board 2023Subjective· 5mImportance★★★★★

🔒 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 →

Part (a)Concept understanding — Skew Lines

Skew Lines

In a plane, two straight lines have only two possibilities: they meet, or they are parallel. In three dimensions a third possibility appears — lines that neither meet nor run parallel. These are skew lines.

What Makes Lines Skew

Two lines in space are skew if they are not parallel and do not intersect. The deeper reason is that skew lines do not lie in the same plane — they are non-coplanar. Parallel lines and intersecting lines always share a plane; skew lines never do.

Note

A classic picture: one edge along the top of a room and a different edge along the floor, running in a different direction. Extend them forever and they still never touch, yet they are clearly not parallel.

The Three Cases in Space

Lines	Directions	Do they meet?	Coplanar?
Intersecting	different	yes, at one point	yes
Parallel	same (proportional)	no	yes
Skew	different	no	no

How to Test for Skew Lines

Take two lines $\vec{r} = \vec{a_1} + \lambda\vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu\vec{b_2}$ .

Not parallel: $\vec{b_1}$ and $\vec{b_2}$ are not proportional (so $\vec{b_1}\times\vec{b_2}\neq\vec{0}$ ).
Do not intersect: no values of $\lambda,\mu$ make the points coincide.

Both conditions are captured by one scalar triple product. The lines are skew exactly when

(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2}) \neq 0.

If this value is zero, the lines are coplanar (they intersect or are parallel); if it is non-zero, they are skew.

Shortest Distance Between Skew Lines

Because skew lines miss each other, there is a well-defined shortest distance between them, measured along their common perpendicular:

$d = \frac{\left|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})\right|}{\left|\vec{b_1}\times\vec{b_2}\right|}.$ …

Part (b)Concept understanding — Vector Equation Of Line

Vector Equation of a Line

A line is fixed by two pieces of information: one point it passes through and the direction it runs in. The vector equation packages both.

Let $\vec{a}$ be the position vector of a known point $A$ on the line, and let $\vec{b}$ be any vector parallel to the line (its direction). For any point $P$ on the line with position vector $\vec{r}$ , the displacement $\overrightarrow{AP}$ points along the line, so it is a scalar multiple of $\vec{b}$ : $\overrightarrow{AP}=\lambda\vec{b}$ . Since $\vec{r}=\vec{a}+\overrightarrow{AP}$ ,

\vec{r} = \vec{a} + \lambda\,\vec{b},\qquad \lambda\in\mathbb{R}

How to read it

As the parameter $\lambda$ runs through all real numbers, $\vec{r}$ traces every point of the line. At $\lambda=0$ you sit at $A$ ; positive $\lambda$ moves one way along $\vec{b}$ , negative $\lambda$ the other. Think of $\vec{a}$ as "where you start" and $\lambda\vec{b}$ as "how far and which way you walk."

Line through two points

If the line passes through points with position vectors $\vec{a}$ and $\vec{b}$ , its direction is $\vec{b}-\vec{a}$ , so

\vec{r} = \vec{a} + \lambda\,(\vec{b}-\vec{a})

Example

The line through $A(1,2,-1)$ parallel to $\vec{b}=2\hat{i}-\hat{j}+3\hat{k}$ is

$\vec{r} = (\hat{i}+2\hat{j}-\hat{k}) + \lambda(2\hat{i}-\hat{j}+3\hat{k}).$ …

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

The lines intersect when $b=2$ , meeting at $(-1,-1,-1)$ .
The sides of parallelogram $ABCD$ are the four lines listed below, and the foot of the perpendicular from $A$ to $CD$ is $\left(\tfrac{28}{9},\tfrac{56}{9},\tfrac{83}{9}\right)$ .

Part (a)

Two lines in space intersect only if some parameter values make all three coordinates agree. Writing each line parametrically:

L_1:\ \frac{x-1}{2}=\frac{y-b}{3}=\frac{z-3}{4}=t\ \Rightarrow\ (1+2t,\ b+3t,\ 3+4t),

L_2:\ \frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{1}=s\ \Rightarrow\ (4+5s,\ 1+2s,\ s).

Match $z$ : $3+4t=s$ .

Match $x$ : $1+2t=4+5s$ . Substitute $s=3+4t$ :

1+2t=4+5(3+4t)=19+20t\ \Rightarrow\ -18t=18\ \Rightarrow\ t=-1,\quad s=3+4(-1)=-1.

Match $y$ to find $b$ : $b+3t=1+2s\Rightarrow b+3(-1)=1+2(-1)\Rightarrow b-3=-1\Rightarrow b=2.$ …

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 26 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.(a) Find the value of bbb so that the lines x−12=y−b3=z−34\frac{x-1}{2} = \frac{y-b}{3} = \frac{z-3}{4}2x−1​=3y−b​=4z−3​ and x−45=y−12=z\frac{x-4}{5} = \frac{y-1}{2} = z5x−4​=2y−1​=z are intersecting lines. Also, find the point of intersection of these given lines.

Part (a)

🎓Unlock everything free for 14 days

Q.(a) Find the value of $b$ so that the lines $\frac{x-1}{2} = \frac{y-b}{3} = \frac{z-3}{4}$ and $\frac{x-4}{5} = \frac{y-1}{2} = z$ are intersecting lines. Also, find the point of intersection of these given lines.

Unlock everything free for 14 days