If the mirror image of the point $P(3,4,9)$ in the line $\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-2}{1}$ is $(\alpha, \beta, \gamma)$, then 14 $(\alpha+\beta+\gamma)$ is :

$\begin{aligned} & \overrightarrow{\mathrm{PN}}. \overrightarrow{\mathrm{b}}=0\\\\ & 3(3 \lambda-2)+2(2 \lambda-5)+(\lambda-7)=0 \\\\ & 14 \lambda=23 \Rightarrow \lambda=\frac{23}{14}\end{aligned}$ $\begin{aligned} & \mathrm{N}\left(\frac{83}{14}, \frac{32}{14}, \frac{51}{14}\right) \\\\ & \therefore \frac{\alpha+3}{2}=\frac{83}{14} \Rightarrow \alpha=\frac{62}{7}\end{aligned}$ $\begin{aligned} & \frac{\beta+4}{2}=\frac{32}{14} \Rightarrow \beta=\frac{4}{7} \\\\ & \fr

3D Geometry — Page 20 | JEE Mathematics

Q191

Consider a

\triangle A B C

where

A(1,3,2), B(-2,8,0)

and

C(3,6,7)

. If the angle bisector of

\angle B A C

meets the line

B C

at

D

, then the length of the projection of the vector

\overrightarrow{A D}

on the vector

\overrightarrow{A C}

is :

A

\dfrac{37}{2 \sqrt{38}}

B

\sqrt{19}

C

\dfrac{39}{2 \sqrt{38}}

D

\dfrac{\sqrt{38}}{2}

Correct Answer

Option A

Solution

$D$ divides $B C$ in ratio $1: 1$

D:\left(\frac{1}{2}, 7, \frac{7}{2}\right)

$\begin{aligned} & \overrightarrow{A D}=\left(\dfrac{1}{2}-1\right) \hat{i}+(7-3) \hat{j}+\left(\dfrac{7}{2}-2\right) \hat{k} \\\\ & =-\dfrac{1}{2} \hat{i}+4 \hat{j}+\dfrac{3}{2} \hat{k} \\\\ & \overrightarrow{A C}=2 \hat{i}+3 \hat{j}+5 \hat{k}\end{aligned}$ Projection of $\overrightarrow{A D}$ on $\overrightarrow{A C}$

=\frac{-1+12+\frac{15}{2}}{\sqrt{4+9+25}}=\frac{37}{2 \sqrt{38}}

Q192

Let

\mathrm{P}

and

\mathrm{Q}

be the points on the line

\dfrac{x+3}{8}=\dfrac{y-4}{2}=\dfrac{z+1}{2}

which are at a distance of 6 units from the point

\mathrm{R}(1,2,3)

. If the centroid of the triangle PQR is

(\alpha, \beta, \gamma)

, then

\alpha^2+\beta^2+\gamma^2

is :

A 18

B 24

C 26

D 36

Correct Answer

Option A

Solution

Any point on line $\dfrac{x+3}{8}=\dfrac{y-4}{2}=\dfrac{z+1}{2}$ can be taken as $(8 \lambda-3,2 \lambda+4,2 \lambda-1)$ If at a distance of 6 units from $R(1,2,3)$

\begin{aligned} & \Rightarrow(8 \lambda-3-1)^2+(2 \lambda+4-2)^2+(2 \lambda-1-3)^2=36 \\\\ & \left.\Rightarrow \lambda^2-\lambda=0 \text{ \{on simplification }\right\} \\\\ & \Rightarrow \lambda=0, \lambda=1 \end{aligned}

Here $P \& Q$ are $(-3,4,-1)$ and $(5,6,1)$ Centroid of $\triangle P Q R$

\begin{aligned} & (\alpha, \beta, \gamma) \equiv\left(\frac{5-3+1}{3}, \frac{6+4+2}{3}, \frac{1-1+3}{3}\right) \\\\ & \Rightarrow \alpha=1, \beta=4, \gamma=1 \\\\ & \Rightarrow \alpha^2+\beta^2+\gamma^2=18 \end{aligned}

Q193

If the mirror image of the point

P(3,4,9)

in the line

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

is

(\alpha, \beta, \gamma)

, then 14

(\alpha+\beta+\gamma)

is :

A 102

B 138

C 132

D 108

Correct Answer

Option D

Solution

$\begin{aligned} & \overrightarrow{\mathrm{PN}}. \overrightarrow{\mathrm{b}}=0\\\\ & 3(3 \lambda-2)+2(2 \lambda-5)+(\lambda-7)=0 \\\\ & 14 \lambda=23 \Rightarrow \lambda=\dfrac{23}{14}\end{aligned}$ $\begin{aligned} & \mathrm{N}\left(\dfrac{83}{14}, \dfrac{32}{14}, \dfrac{51}{14}\right) \\\\ & \therefore \dfrac{\alpha+3}{2}=\dfrac{83}{14} \Rightarrow \alpha=\dfrac{62}{7}\end{aligned}$ $\begin{aligned} & \dfrac{\beta+4}{2}=\dfrac{32}{14} \Rightarrow \beta=\dfrac{4}{7} \\\\ & \dfrac{\gamma+9}{2}=\dfrac{51}{14} \Rightarrow \gamma=\dfrac{-12}{7}\end{aligned}$ Now, $14(\alpha+\beta+\gamma)=14\left(\dfrac{62+4-12}{7}\right)=108$

Q194

Let

L_1: \vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\lambda(\hat{i}-\hat{j}+2 \hat{k}), \lambda \in \mathbb{R}

,

L_2: \vec{r}=(\hat{j}-\hat{k})+\mu(3 \hat{i}+\hat{j}+p \hat{k}), \mu \in \mathbb{R} \text{, and } L_3: \vec{r}=\delta(\ell \hat{i}+m \hat{j}+n \hat{k}), \delta \in \mathbb{R}

be three lines such that

L_1

is perpendicular to

L_2

and

L_3

is perpendicular to both

L_1

and

L_2

. Then, the point which lies on

L_3

is

A

(1,7,-4)

B

(1,-7,4)

C

(-1,7,4)

D

(-, 1-7,4)

Correct Answer

Option C

Solution

\mathrm{L}_1 \perp \mathrm{L}_2 \quad \mathrm{~L}_3 \perp \mathrm{L}_1, \mathrm{~L}_2

\begin{aligned} & 3-1+2 \mathrm{P}=0 \\ & \mathrm{P}=-1 \\ & \left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & -1 & 2 \\ 3 & 1 & -1 \end{array}\right|=-\hat{\mathrm{i}}+7 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\ & \therefore(-\delta, 7 \delta, 4 \delta) \text{ will lie on } \mathrm{L}_3 \end{aligned}

For

\delta=1

the point will be

(-1,7,4)

Q195

Let

(\alpha, \beta, \gamma)

be the foot of perpendicular from the point

(1,2,3)

on the line

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

. Then

19(\alpha+\beta+\gamma)

is equal to :

A 99

B 102

C 101

D 100

Correct Answer

Option C

Solution

Let foot

P(5 k-3,2 k+1,3 k-4)

DR's $\rightarrow$ AP :

5 \mathrm{k}-4,2 \mathrm{k}-1,3 \mathrm{k}-7

DR's $\rightarrow$ Line:

5,2,3

Condition of perpendicular lines

(25 k-20)+(4 k-2)+(9 k-21)=0

Then

\mathrm{k}=\frac{43}{38}

Then

19(\alpha+\beta+\gamma)=\mathbf{1 0 1}

Q196

Consider the line

\mathrm{L}

passing through the points

(1,2,3)

and

(2,3,5)

. The distance of the point

\left(\dfrac{11}{3}, \dfrac{11}{3}, \dfrac{19}{3}\right)

from the line

\mathrm{L}

along the line

\dfrac{3 x-11}{2}=\dfrac{3 y-11}{1}=\dfrac{3 z-19}{2}

is equal to

A 6

B 3

C 5

D 4

Correct Answer

Option B

Solution

L: \frac{x-1}{1}=\frac{y-2}{1}=\frac{z-3}{2}=\mu

Measured along

L_2: \frac{x-\frac{11}{3}}{\frac{2}{3}}=\frac{y-\frac{11}{3}}{\frac{1}{3}}=\frac{z-\frac{19}{3}}{\frac{2}{3}}=\lambda

Any point on

L_1:(\mu+1, \mu+2,2 \mu+3)

Any point on

L_2\left(\frac{2}{3} \lambda+\frac{11}{3}, \frac{\lambda}{3}+\frac{11}{3}, \frac{2}{3} \lambda+\frac{19}{3}\right)

Now

\begin{aligned} & \mu+1=\frac{2}{3} \lambda+\frac{11}{3} \\ & \frac{\mu+2=\frac{\lambda}{3}+\frac{11}{3}}{\lambda=-3} \\ & \mu=\frac{2}{3} \\ \end{aligned}

\begin{aligned} & \text{ Point on } L=\left(\frac{5}{3}, \frac{8}{3}, \frac{13}{3}\right) \\ & d=\sqrt{\left(\frac{11}{3}-\frac{5}{3}\right)^2+\left(\frac{8}{3}-\frac{11}{3}\right)^2+\left(\frac{19}{3}-\frac{13}{3}\right)^2} \\ & d=\sqrt{4+1+4} \\ & d=3 \end{aligned}

Q197

If the shortest distance between the lines

\dfrac{x-\lambda}{-2}=\dfrac{y-2}{1}=\dfrac{z-1}{1}

and

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

is 1 , then the sum of all possible values of

\lambda

is :

A 0

B

2 \sqrt{3}

C

3 \sqrt{3}

D

-2 \sqrt{3}

Correct Answer

Option B

Solution

Given the two lines:

L_1: \frac{x-\lambda}{-2} = \frac{y-2}{1} = \frac{z-1}{1}

L_2: \frac{x-\sqrt{3}}{1} = \frac{y-1}{-2} = \frac{z-2}{1}

We observe that these lines are not parallel as their directional vectors are not proportional.

The directional vector for $L_1$ is $(-2,1,1)$ and for $L_2$ is $(1,-2,1)$ .

The shortest distance between two skew (non-intersecting and non-parallel) lines in the three-dimensional space is along the line that is perpendicular to both lines.

This implies we can find a vector that is perpendicular to both directional vectors by taking their cross product.

The directional vector for $L_1$ is $d_1 = \langle -2, 1, 1 \rangle$ , and for $L_2$ is $d_2 = \langle 1, -2, 1 \rangle$ .

The cross product of $d_1$ and $d_2$ , which will be perpendicular to both lines, is given by:

d = d_1 \times d_2 = \left| \begin{matrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -2 & 1 & 1 \\ 1 & -2 & 1 \\ \end{matrix} \right|

Expanding the determinate gives:

d = \mathbf{i}((1)(1) - (1)(-2)) - \mathbf{j}((-2)(1) - (1)(1)) + \mathbf{k}((-2)(-2) - (1)(1))

d = \mathbf{i}(1 + 2) - \mathbf{j}(-2 - 1) + \mathbf{k}(4 - 1)

d = 3\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}

d = \langle 3, 3, 3 \rangle

The shortest distance $D$ between the two lines can then be given by the formula:

D = \frac{\left| (\mathbf{a_2} - \mathbf{a_1}) \cdot d \right|}{\|d\|}

Where $\mathbf{a_1}$ and $\mathbf{a_2}$ are position vectors to any points on line $L_1$ and line $L_2$ , respectively, and ' $\cdot$ ' denotes the dot product.

From the lines' equations, we can choose a point on each line (when the parameter is zero).

Thus, for $L_1$ , let's choose the point $A(\lambda, 2, 1)$ , and for $L_2$ , let's choose the point $B(\sqrt{3}, 1, 2)$ .

These points correspond to the vectors $\mathbf{a_1} = \langle \lambda, 2, 1 \rangle$ and $\mathbf{a_2} = \langle \sqrt{3}, 1, 2 \rangle$ , respectively.

The vector $\mathbf{a_2} - \mathbf{a_1}$ is:

\mathbf{a_2} - \mathbf{a_1} = \langle \sqrt{3}, 1, 2 \rangle - \langle \lambda, 2, 1 \rangle

\mathbf{a_2} - \mathbf{a_1} = \langle \sqrt{3} - \lambda, 1 - 2, 2 - 1 \rangle

\mathbf{a_2} - \mathbf{a_1} = \langle \sqrt{3} - \lambda, -1, 1 \rangle

We can now substitute this, along with $d$ , into the distance formula:

D = \frac{\left| (\langle \sqrt{3} - \lambda, -1, 1 \rangle) \cdot \langle 3, 3, 3 \rangle \right|}{\|\langle 3, 3, 3 \rangle\|}

D = \frac{\left| 3(\sqrt{3} - \lambda) + 3(-1) + 3(1) \right|}{\sqrt{3^2 + 3^2 + 3^2}}

D = \frac{\left| 3\sqrt{3} - 3\lambda - 3 + 3 \right|}{\sqrt{27}}

D = \frac{\left| 3\sqrt{3} - 3\lambda \right|}{3\sqrt{3}}

D = \frac{\left| \sqrt{3} - \lambda \right|}{\sqrt{3}}

Given that the shortest distance $D$ between the lines is 1, we can equate the above result to 1, and solve for $\lambda$ :

\frac{\left| \sqrt{3} - \lambda \right|}{\sqrt{3}} = 1

\left| \sqrt{3} - \lambda \right| = \sqrt{3}

This absolute value equation gives us two possible cases: Case 1: $\sqrt{3} - \lambda = \sqrt{3}$ , which gives $\lambda = 0$ .

Case 2: $\sqrt{3} - \lambda = -\sqrt{3}$ , which gives $\lambda = 2\sqrt{3}$ .

Therefore, the sum of all possible values of $\lambda$ is:

\lambda_{\sum} = \lambda_1 + \lambda_2 = 0 + 2\sqrt{3} = 2\sqrt{3}

Hence, option B ( $2 \sqrt{3}$ ) is the correct answer.

Q198

The distance, of the point

(7,-2,11)

from the line

\dfrac{x-6}{1}=\dfrac{y-4}{0}=\dfrac{z-8}{3}

along the line

\dfrac{x-5}{2}=\dfrac{y-1}{-3}=\dfrac{z-5}{6}

, is :

A 12

B 18

C 21

D 14

Correct Answer

Option D

Solution

\mathrm{B}=(2 \lambda+7,-3 \lambda-2,6 \lambda+11)

Point B lies on

\frac{x-6}{1}=\frac{y-4}{0}=\frac{z-8}{3}

\frac{2 \lambda+7-6}{1}=\frac{-3 \lambda-2-4}{0}=\frac{6 \lambda+11-8}{3}

-3 \lambda-6=0

\lambda=-2

\mathrm{B} \Rightarrow(3,4,-1)

\mathrm{AB}=\sqrt{(7-3)^2+(4+2)^2+(11+1)^2}

=\sqrt{16+36+144}

=\sqrt{196}=14

Q199

If the shortest distance between the lines

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

and

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

is

\dfrac{6}{\sqrt{5}}

, then the sum of all possible values of

\lambda

is :

A 10

B 5

C 7

D 8

Correct Answer

Option D

Solution

\begin{aligned} & \frac{x-4}{1}=\frac{y+1}{2}=\frac{z}{-3} \\ & \frac{x-\lambda}{2}=\frac{y+1}{4}=\frac{z-2}{-5} \end{aligned}

the shortest distance between the lines

=\left|\frac{(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}) \cdot\left(\overrightarrow{\mathrm{d}_1} \times \overrightarrow{\mathrm{d}_2}\right)}{\left|\overrightarrow{\mathrm{d}_1} \times \overrightarrow{\mathrm{d}_2}\right|}\right|

=\left|\frac{\left|\begin{array}{ccc}\lambda-4 & 0 & 2 \\ 1 & 2 & -3 \\ 2 & 4 & -5\end{array}\right|}{\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 4 & -5\end{array}\right|}\right|

=\left|\frac{(\lambda-4)(-10+12)-0+2(4-4)}{|2 \hat{i}-1 \hat{j}+0 \hat{k}|}\right|

\begin{aligned} & \frac{6}{\sqrt{5}}=\left|\frac{2(\lambda-4)}{\sqrt{5}}\right| \\ & 3=|\lambda-4| \\ & \lambda-4= \pm 3 \\ & \lambda=7,1 \end{aligned}

Sum of all possible values of $\lambda$ is

=8

Q200

Let the image of the point

(1,0,7)

in the line

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

be the point

(\alpha, \beta, \gamma)

. Then which one of the following points lies on the line passing through

(\alpha, \beta, \gamma)

and making angles

\dfrac{2 \pi}{3}

and

\dfrac{3 \pi}{4}

with

y

-axis and

z

-axis respectively and an acute angle with

x

-axis ?

A

(1,-2,1+\sqrt{2})

B

(3,-4,3+2 \sqrt{2})

C

(3,4,3-2 \sqrt{2})

D

(1,2,1-\sqrt{2})

Correct Answer

Option C

Solution

\mathrm{L}_1=\frac{\mathrm{x}}{1}=\frac{\mathrm{y}-1}{2}=\frac{\mathrm{z}-2}{3}=\lambda

\begin{aligned} & \mathrm{M}(\lambda, 1+2 \lambda, 2+3 \lambda) \\ & \overrightarrow{\mathrm{PM}}=(\lambda-1) \hat{\mathrm{i}}+(1+2 \lambda) \hat{\mathrm{j}}+(3 \lambda-5) \hat{\mathrm{k}} \end{aligned}

\overrightarrow{\mathrm{PM}}

is perpendicular to line

\mathrm{L}_1

\begin{aligned} & \overrightarrow{\mathrm{PM}} \overrightarrow{\mathrm{b}}=0 \quad(\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \\ & \Rightarrow \lambda-1+4 \lambda+2+9 \lambda-15=0 \\ & 14 \lambda=14 \Rightarrow \lambda=1 \\ & \therefore \mathrm{M}=(1,3,5) \\ & \overrightarrow{\mathrm{Q}}=2 \overrightarrow{\mathrm{M}}-\overrightarrow{\mathrm{P}}[\mathrm{M} \text{ is midpoint of } \overrightarrow{\mathrm{P}} \& \overrightarrow{\mathrm{Q}}] \end{aligned}

\begin{aligned} & \overrightarrow{\mathrm{Q}}=2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+10 \hat{\mathrm{k}}-\hat{\mathrm{i}}-7 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{Q}}=\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\ & \therefore(\alpha, \beta, \gamma)=(1,6,3) \end{aligned}

Required line having direction cosine

(l, \mathrm{~m}, \mathrm{n})

\begin{aligned} & l^2+m^2+n^2=1 \\ & \Rightarrow l^2+\left(-\frac{1}{2}\right)^2+\left(-\frac{1}{\sqrt{2}}\right)^2=1 \end{aligned}

l^2=\frac{1}{4}

\therefore l=\frac{1}{2}

[Line make acute angle with

\mathrm{x}

-axis] Equation of line passing through

(1,6,3)

will be

\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+6 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\mu\left(\frac{1}{2} \hat{\mathrm{i}}-\frac{1}{2} \hat{\mathrm{j}}-\frac{1}{\sqrt{2}} \hat{\mathrm{k}}\right)

Option (3) satisfying for

\mu=4