The distance of the line $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}$ from the point $(1,4,0)$ along the line $\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}$ is :

To find the distance from the line $ \frac{x-2}{2} = \frac{y-6}{3} = \frac{z-3}{4} $ to the point $(1,4,0)$ along the line $ \frac{x}{1} = \frac{y-2}{2} = \frac{z+3}{3}, $ we first consider a parallel line passing through the point $(1,4,0)$. The equation of this parallel line is: $ \frac{x-1}{1} = \frac{y-4}{2} = \frac{z-0}{3}. $ Next, we find the point of intersection between the given line and the parallel line, which can be expressed in parameter form as: $ (\lambda + 1, 2\lambda + 4, 3\lamb

3D Geometry — Page 23 | JEE Mathematics

Q221

If the square of the shortest distance between the lines

\dfrac{x-2}{1}=\dfrac{y-1}{2}=\dfrac{z+3}{-3}

and

\dfrac{x+1}{2}=\dfrac{y+3}{4}=\dfrac{z+5}{-5}

is

\dfrac{m}{n}

, where

m

,

n

are coprime numbers, then

m+n

is equal to :

A 14

B 6

C 21

D 9

Correct Answer

Option D

Solution

\begin{aligned} & \overrightarrow{\mathrm{a}}=(2,1,-3) \\ & \overrightarrow{\mathrm{b}}=(-1,-3,-5) \\ & \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 2 & -3 \\ 2 & 4 & -5 \end{array}\right| \\ & =2 \hat{\mathrm{i}}-\hat{\mathrm{j}} \\ & \overrightarrow{\mathrm{~b}}-\overrightarrow{\mathrm{a}}=-3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}-2 \hat{\mathrm{k}} \\ & \mathrm{~S}_{\mathrm{d}}=\frac{|(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}}) \cdot(\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}})|}{|\overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}|} \\ & =\frac{2}{\sqrt{5}} \\ & \left(\mathrm{~S}_{\mathrm{d}}\right)^2=\frac{4}{5} \\ & \mathrm{~m}=4, \mathrm{n}=5 \Rightarrow \mathrm{~m}+\mathrm{n}=9 \end{aligned}

Q222

Let P be the foot of the perpendicular from the point

(1,2,2)

on the line

\mathrm{L}: \dfrac{x-1}{1}=\dfrac{y+1}{-1}=\dfrac{z-2}{2}

. Let the line

\vec{r}=(-\hat{i}+\hat{j}-2 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}), \lambda \in \mathbf{R}

, intersect the line L at Q . Then

2(\mathrm{PQ})^2

is equal to :

A 25

B 27

C 19

D 29

Correct Answer

Option B

Solution

\begin{aligned} & \mathrm{L}: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}+1}{-1}=\frac{\mathrm{z}-2}{2}=\mu \\ & \mathrm{P}(\mu+1,-\mu-1,2 \mu+2) \\ & \overrightarrow{\mathrm{AP}} \cdot \overrightarrow{\mathrm{~d}}=0 \Rightarrow(\mu,-\mu-3,2 \mu) \cdot(1,-1,2)=0 \\ & \Rightarrow \mu+\mu+3+4 \mu=0 \Rightarrow \mu=-\frac{1}{2} \\ & \therefore \mathrm{P}\left(\frac{-1}{2}+1,+\frac{1}{2}-1,2\left(\frac{-1}{2}\right)+2\right) \\ & \mathrm{P}\left(\frac{1}{2}, \frac{-1}{2}, 1\right) \end{aligned}

\begin{array}{c|c|c} \mu+1=-1+\lambda & -\mu-1=1-\lambda & 2 \mu+2=-2+\lambda \\ \mu=\lambda-2 & \mu=\lambda-2 & \downarrow \end{array}

\qquad\qquad\qquad\qquad\qquad\qquad\qquad\begin{aligned} & 2(\lambda-2)+2=-2+\lambda \\ & 2 \lambda-4+2=-2+\lambda \end{aligned}

\begin{aligned} & \therefore \quad \mu=-2, \lambda=0 \\ & \therefore \quad \mathrm{Q} \equiv(-1,1-2) \\ & \mathrm{P}\left(\frac{1}{2}, \frac{-1}{2}, 1\right) \text{ and } \mathrm{Q}(-1,1,-2) \\ & \mathrm{PQ}=\sqrt{\left(\frac{1}{2}+1\right)^2+\left(\frac{-1}{2}-1\right)^2+(1+2)^2} \\ & =\sqrt{\frac{9}{4}+\frac{9}{4}+9}=\sqrt{\frac{54}{4}} \\ & \therefore \quad 2(\mathrm{PQ})^2=2\left(\frac{54}{4}\right)=27 \end{aligned}

Q223

Let

\mathrm{L}_1: \dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}

and

\mathrm{L}_2: \dfrac{x-2}{3}=\dfrac{y-4}{4}=\dfrac{z-5}{5}

be two lines. Then which of the following points lies on the line of the shortest distance between

\mathrm{L}_1

and

\mathrm{L}_2

?

A

\left(\dfrac{14}{3},-3, \dfrac{22}{3}\right)

B

\left(2,3, \dfrac{1}{3}\right)

C

\left(\dfrac{8}{3},-1, \dfrac{1}{3}\right)

D

\left(-\dfrac{5}{3},-7,1\right)

Correct Answer

Option A

Solution

$\begin{aligned} & L_1: \dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4} \\\\ & L_2: \dfrac{x-2}{3}=\dfrac{y-4}{4}=\dfrac{z-5}{5}\end{aligned}$

\begin{aligned} & P(2 \lambda+1,3 \lambda+2,4 \lambda+3) \\ & Q(3 \mu+2,4 \mu+4,5 \mu+5) \end{aligned}

Dr's of $P Q<2 \lambda-3 \mu-1,3 \lambda-4 \mu-2$ ,

\begin{aligned} & 4 \lambda-5 \mu-2> \\ & P Q=\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{array}\right|=-\hat{i}+2 \hat{j}-\hat{k} \end{aligned}

\begin{aligned} & \Rightarrow \quad \frac{2 \lambda-3 \mu-1}{-1}=\frac{3 \lambda-4 \mu-2}{2}=\frac{4 \lambda-5 \mu-2}{-1} \\ & \Rightarrow \quad \lambda=\frac{1}{3} \mu=\frac{-1}{6} \\ & \Rightarrow \quad P\left(\frac{5}{3}, 3, \frac{13}{3}\right) \quad Q\left(\frac{3}{2}, \frac{10}{3}, \frac{25}{6}\right) \end{aligned}

Dr's $P Q\langle 1,-2,1\rangle$ $\therefore$ Line

\frac{y-\frac{5}{3}}{1}=\frac{y-3}{-2}=\frac{y-\frac{13}{3}}{1}

Q224

Let a line pass through two distinct points

P(-2,-1,3)

and

Q

, and be parallel to the vector

3 \hat{i}+2 \hat{j}+2 \hat{k}

. If the distance of the point Q from the point

\mathrm{R}(1,3,3)

is 5 , then the square of the area of

\triangle P Q R

is equal to :

A 148

B 144

C 136

D 140

Correct Answer

Option C

Solution

P = (-2, -1, 3)

Since the line through $P$ is parallel to the vector

\vec{d} = (3, 2, 2),

any point $Q$ on this line can be expressed as:

Q = P + t\,\vec{d} = (-2 + 3t,\, -1 + 2t,\, 3 + 2t),

where $t$ is a real number.

Given that $P$ and $Q$ must be distinct, we require $t \neq 0$ .

The point $R$ is given by:

R = (1, 3, 3).

The distance from $Q$ to $R$ is $5$ , so we have:

\left[ (-2 + 3t - 1)^2 + (-1 + 2t - 3)^2 + (3 + 2t - 3)^2 \right] = 5^2.

Simplify each coordinate difference: For the $x$ -coordinate:

-2 + 3t - 1 = 3t - 3 = 3(t - 1).

For the $y$ -coordinate:

-1 + 2t - 3 = 2t - 4 = 2(t - 2).

For the $z$ -coordinate:

3 + 2t - 3 = 2t.

So the equation becomes:

[3(t - 1)]^2 + [2(t - 2)]^2 + (2t)^2 = 25.

Expanding the squares:

9(t - 1)^2 + 4(t - 2)^2 + 4t^2 = 25.

Expand each term:

9(t^2 - 2t + 1) + 4(t^2 - 4t + 4) + 4t^2 = 25,

which simplifies to:

9t^2 - 18t + 9 + 4t^2 - 16t + 16 + 4t^2 = 25.

Combine like terms:

(9t^2 + 4t^2 + 4t^2) - (18t + 16t) + (9 + 16) = 25,

17t^2 - 34t + 25 = 25.

Subtract 25 from both sides:

17t^2 - 34t = 0.

Factor out $17t$ :

17t(t - 2) = 0.

Since $t \neq 0$ , we must have:

t = 2.

Substitute $t = 2$ back into the equation for $Q$ :

Q = (-2 + 3(2),\, -1 + 2(2),\, 3 + 2(2)) = (4, 3, 7).

Next, to find the square of the area of triangle $PQR$ , first compute the vectors:

\vec{PQ} = Q - P = (4 - (-2),\, 3 - (-1),\, 7 - 3) = (6, 4, 4),

\vec{PR} = R - P = (1 - (-2),\, 3 - (-1),\, 3 - 3) = (3, 4, 0).

The area of the triangle is given by:

\text{Area} = \frac{1}{2} \, \| \vec{PQ} \times \vec{PR} \|.

Calculate the cross product:

\vec{PQ} \times \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 4 & 4 \\ 3 & 4 & 0 \\ \end{vmatrix}.

Expanding the determinant: $\hat{i}$ -component:

4 \times 0 - 4 \times 4 = -16,

$\hat{j}$ -component:

-(6 \times 0 - 4 \times 3) = 12,

$\hat{k}$ -component:

6 \times 4 - 4 \times 3 = 24 - 12 = 12.

Thus,

\vec{PQ} \times \vec{PR} = (-16,\, 12,\, 12).

Find the magnitude squared of the cross product:

\| \vec{PQ} \times \vec{PR} \|^2 = (-16)^2 + 12^2 + 12^2 = 256 + 144 + 144 = 544.

The square of the area of triangle $PQR$ is:

(\text{Area})^2 = \left(\frac{1}{2}\right)^2 \| \vec{PQ} \times \vec{PR} \|^2 = \frac{1}{4} \times 544 = 136.

Thus, the square of the area of $\triangle PQR$ is

136.

Q225

The perpendicular distance, of the line

\dfrac{x-1}{2}=\dfrac{y+2}{-1}=\dfrac{z+3}{2}

from the point

\mathrm{P}(2,-10,1)

, is :

A

6

B

4 \sqrt{3}

C

3 \sqrt{5}

D

5 \sqrt{2}

Correct Answer

Option C

Solution

To find the perpendicular distance from the point

P(2,-10,1)

to the line given by

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

follow these steps: Parametrize the Line: From the given symmetric equations, set the common parameter as

t

:

x = 1 + 2t

y = -2 - t

z = -3 + 2t

This shows that: A point on the line is

A(1,-2,-3)

(when

t=0

). The direction vector is

\vec{d} = \langle 2, -1, 2 \rangle.

Determine the Vector from Point A to P: Calculate

\vec{AP} = P - A = \langle 2 - 1, \; -10 - (-2), \; 1 - (-3) \rangle = \langle 1, \; -8, \; 4 \rangle.

Compute the Cross Product: The formula for the perpendicular distance from a point to a line in 3D is:

d = \frac{\|\vec{AP} \times \vec{d}\|}{\|\vec{d}\|}.

First, find the cross product

\vec{AP} \times \vec{d}

:

\vec{AP} \times \vec{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -8 & 4 \\ 2 & -1 & 2 \end{vmatrix} = \Big( (-8)(2) - (4)(-1), \; (4)(2) - (1)(2), \; (1)(-1) - (-8)(2) \Big).

Evaluate each component: First component:

(-8 \times 2) - (4 \times -1) = -16 + 4 = -12.

Second component:

(4 \times 2) - (1 \times 2) = 8 - 2 = 6.

Third component:

(1 \times -1) - (-8 \times 2) = -1 + 16 = 15.

So,

\vec{AP} \times \vec{d} = \langle -12, 6, 15 \rangle.

Calculate the Magnitudes: For the cross product:

\|\vec{AP} \times \vec{d}\| = \sqrt{(-12)^2 + 6^2 + 15^2} = \sqrt{144 + 36 + 225} = \sqrt{405} = 9\sqrt{5}.

For the direction vector:

\|\vec{d}\| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3.

Compute the Distance: Substitute the magnitudes into the distance formula:

d = \frac{9\sqrt{5}}{3} = 3\sqrt{5}.

Thus, the perpendicular distance from the point

P(2,-10,1)

to the line is

3\sqrt{5}

. Comparing with the options given, the correct answer is: Option C:

3\sqrt{5}

.

Q226

The square of the distance of the point

\left( \dfrac{15}{7}, \dfrac{32}{7}, 7 \right)

from the line

\dfrac{x + 1}{3} = \dfrac{y + 3}{5} = \dfrac{z + 5}{7}

in the direction of the vector

\hat{i} + 4\hat{j} + 7\hat{k}

is:

A 66

B 54

C 41

D 44

Correct Answer

Option A

Solution

\begin{aligned} & L=\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7} \\ & P Q=\frac{x-\frac{15}{7}}{1}=\frac{y-\frac{32}{7}}{4}=\frac{z-7}{7}=\lambda \end{aligned}

\begin{aligned} &\Rightarrow \mathrm{Q}\left(\lambda+\frac{15}{7}, 4 \lambda+\frac{32}{7}, 7 \lambda+7\right)\\ &\text{ Since } \mathrm{Q} \text{ lies on line } \mathrm{L}\\ &\begin{aligned} & \text{ So, } \frac{\lambda+\frac{15}{7}+1}{3}=\frac{7 \lambda+7+5}{7} \\ & \Rightarrow 7 \lambda+22=21 \lambda+36 \\ & \Rightarrow \lambda=-1 \\ & \therefore \text{ Point } Q\left(\frac{8}{7}, \frac{4}{7}, 0\right) \\ & \mathrm{PQ}=\sqrt{\left(\frac{15}{7}-\frac{8}{7}\right)^2+\left(\frac{32}{7}-\frac{4}{7}\right)^2+(7-0)} \\ & \mathrm{PQ}=\sqrt{66} \\ & \Rightarrow(\mathrm{PQ})^2=66 \end{aligned} \end{aligned}

Q227

The distance of the line

\dfrac{x-2}{2}=\dfrac{y-6}{3}=\dfrac{z-3}{4}

from the point

(1,4,0)

along the line

\dfrac{x}{1}=\dfrac{y-2}{2}=\dfrac{z+3}{3}

is :

A

\sqrt{17}

B

\sqrt{13}

C

\sqrt{15}

D

\sqrt{14}

Correct Answer

Option D

Solution

To find the distance from the line $\dfrac{x-2}{2} = \dfrac{y-6}{3} = \dfrac{z-3}{4}$ to the point $(1,4,0)$ along the line $\dfrac{x}{1} = \dfrac{y-2}{2} = \dfrac{z+3}{3},$ we first consider a parallel line passing through the point $(1,4,0)$ .

The equation of this parallel line is: $\dfrac{x-1}{1} = \dfrac{y-4}{2} = \dfrac{z-0}{3}.$ Next, we find the point of intersection between the given line and the parallel line, which can be expressed in parameter form as: $(\lambda + 1, 2\lambda + 4, 3\lambda) = (2t + 2, 3t + 6, 4t + 3).$ Solving for $\lambda$ in terms of $t$ , from the first coordinate: $\lambda = 2t + 1.$ Using the second coordinates, we equate: $2\lambda + 4 = 3t + 6 \quad \Rightarrow \quad 2(2t + 1) + 4 = 3t + 6.$ Simplifying gives: $4t + 2 + 4 = 3t + 6 \quad \Rightarrow \quad 4t + 6 = 3t + 6 \quad \Rightarrow \quad t = 0.$ Substituting $t = 0$ into either line equation yields the point of intersection (POI) as: $(2, 6, 3).$ The distance from the point $(1, 4, 0)$ to the POI $(2, 6, 3)$ is computed as: $\sqrt{(2-1)^2 + (6-4)^2 + (3-0)^2} = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}.$

Q228

Let in a

\triangle A B C

, the length of the side

A C

be 6 , the vertex

B

be

(1,2,3)

and the vertices

A, C

lie on the line

\dfrac{x-6}{3}=\dfrac{y-7}{2}=\dfrac{z-7}{-2}

. Then the area (in sq. units) of

\triangle A B C

is:

A 42

B 17

C 56

D 21

Correct Answer

Option D

Solution

\begin{aligned} & \text{ Let } \mathrm{M}(3 \lambda+6,2 \lambda+7,-2 \lambda+7) \\ & \overrightarrow{\mathrm{BM}}=(3 \lambda+5) \hat{\mathrm{i}}+(2 \lambda+5) \hat{\mathrm{j}}+(-2 \lambda+4) \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{AC}} \cdot \overrightarrow{\mathrm{BM}}=0=3(3 \lambda+5)+2(2 \lambda+5)-2(-2 \lambda+4) \\ & \overrightarrow{\mathrm{BM}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+6 \hat{\mathrm{k}} \\ & \mid \overrightarrow{\mathrm{BM}}=7 \\ & \text{ Area }=\frac{1}{2} \times 6 \times 7=21 \end{aligned}

Q229

Let the line passing through the points

(-1,2,1)

and parallel to the line

\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z}{4}

intersect the line

\dfrac{x+2}{3}=\dfrac{y-3}{2}=\dfrac{z-4}{1}

at the point

P

. Then the distance of

P

from the point

Q(4,-5,1)

is

A

5 \sqrt{6}

B

5

C

5 \sqrt{5}

D

10

Correct Answer

Option C

Solution

Equation of line through point $(-1,2,1)$ is $\rightarrow$

\begin{aligned} &\Rightarrow \frac{x+1}{2}=\frac{y-2}{3}=\frac{z-1}{4}-(2)=\lambda\\ &\text{ So, }\left[\begin{array}{l} x=2 \lambda-1 \\ y=3 \lambda+2 \\ z=4 \lambda+1 \end{array}\right. \end{aligned}

By $(1) \rightarrow \dfrac{x+2}{3}=\dfrac{y-3}{2}=\dfrac{z-4}{1}=\mu($ Let $)$ So, $\left[\begin{array}{l}x=3 \mu-2 \\ y=2 \mu+3 \\ z=\mu+4\end{array}\right.$ For intersection point 'P'

\begin{aligned} & \mathrm{x}=2 \lambda-1=3 \mu-2 \\ & \mathrm{y}=3 \lambda+2=2 \mu+3 \quad\left[\begin{array}{l} \lambda=1 \\ \mu=1 \end{array}\right] \\ & \mathrm{z}=4 \lambda+1=\mu+4 \\ & \text{ So, point } \mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})=(1,5,5) \\ & \& \mathrm{Q}(4,-5,1) \\ & \therefore \mathrm{PQ}=\sqrt{9+100+16} \\ & =\sqrt{125}=5 \sqrt{5} \end{aligned}

Q230

Let

\mathrm{A}(x, y, z)

be a point in

x y

-plane, which is equidistant from three points

(0,3,2),(2,0,3)

and

(0,0,1)

. Let

\mathrm{B}=(1,4,-1)

and

\mathrm{C}=(2,0,-2)

. Then among the statements (S1) :

\triangle \mathrm{ABC}

is an isosceles right angled triangle, and (S2) : the area of

\triangle \mathrm{ABC}

is

\dfrac{9 \sqrt{2}}{2}

,

A both are false

B only (S2) is true

C only (S1) is true

D both are true

Correct Answer

Option C

Solution

\begin{aligned} & A(x, y, z) \text{ Let } P(0,3,2), Q(2,0,3), R(0,0,1) \\ & A P=A Q=A R \\ & x^2+(y-3)^2+(z-2)^2=(x-2)^2+y^2+(z-3)^2=x^2+ \\ & y^2+(z-1)^2 \end{aligned}

In xy plane $\mathrm{z}=0$ So, $x^2-4 x+4+y^2+9=x^2+y^2+1$

\begin{aligned} & \Rightarrow y=2 \\ & x=3 \\ & 9+y^2-6 y+9+4=x^2+y^2+1 \end{aligned}

So, $A(3,2,0)$ also $B(1,4,-1) \& C(2,0,-2)$ Now $A B=\sqrt{4+4+1}=3$

\begin{aligned} &\begin{aligned} & \mathrm{AC}=\sqrt{1+4+4}=3 \\ & \mathrm{BC}=\sqrt{1+16+1}=\sqrt{18} \\ & \mathrm{AB}=\mathrm{AC} \\ & \text{ isosceles } \Delta \& \mathrm{AB}^2+\mathrm{AC}^2=\mathrm{BC}^2 \\ & \text{ right angle } \Delta \\ & \text{ Area of } \triangle \mathrm{ABC}=\frac{1}{2} \times \text{ base.height } \\ & \frac{1}{2} \times 3 \times 3=\frac{9}{2} \end{aligned}\\ &\text{ So only } S_1 \text{ is true } \end{aligned}