3D Geometry — Page 25 | JEE Mathematics

Q241

Let the line L pass through

(1,1,1)

and intersect the lines

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

and

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

. Then, which of the following points lies on the line

L

?

A

(7,15,13)

B

(4,22,7)

C

(10,-29,-50)

D

(5,4,3)

Correct Answer

Option A

Solution

\begin{aligned} & L: \frac{x-1}{a}=\frac{y-1}{b}=\frac{z-1}{c} \\ & L_1: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=\lambda \text{ (say) } \end{aligned}

Any point on $L_1$ be $A(2 \lambda+1,3 \lambda+1,4 \lambda+1)$

L_2: \frac{x-3}{1}=\frac{y-4}{2}=\frac{z}{1}=\mu \text{ (say) }

Any point on $L_2$ be $B(\mu+3,2 \mu-4, \mu)$ $D$ of $L$ be : $<2 \lambda, 3 \lambda-2,4 \lambda>$ or $<\mu+2,2 \mu+3$ , $\mu-1>$ Now $\dfrac{2 \lambda}{\mu+2}=\dfrac{3 \lambda-7}{2 \mu+3}=\dfrac{4 \lambda}{\mu-1}$

\begin{aligned} &\begin{aligned} & \Rightarrow \lambda=\frac{-6}{5} \quad \mu=-5 \\ & \therefore< a, b, c >\equiv\langle-3,-7,-6 >\text{ or }<3,7,6> \\ & \therefore \quad L: \frac{x-1}{3}=\frac{y-1}{7}=\frac{z-1}{6} \end{aligned}\\ &(7,15,13) \text{ lies on the line. } \end{aligned}

Q242

If the shortest distance between the lines

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

and

\dfrac{x}{1}=\dfrac{y}{\alpha}=\dfrac{z-5}{1}

is

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

, then the sum of all possible values of

\alpha

is

A

\dfrac{3}{2}

B

3

C

-3

D

-\dfrac{3}{2}

Correct Answer

Option C

Solution

\begin{aligned} & d=\left|\frac{(\vec{a}-\vec{b}) \cdot \vec{p}_1 \times \vec{p}_2}{\left|\vec{p}_1 \times \vec{p}_2\right|}\right|=\frac{5}{\sqrt{6}} \\ & \vec{a}-\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k} \\ & \overrightarrow{p_1} \times \overrightarrow{p_2}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & \alpha & 1 \end{array}\right| \\ & \quad=\hat{i}(3-4 \alpha)-\hat{j}(-2)+\hat{k}(2 \alpha-3) \\ & (\vec{a}-\vec{b}) \cdot\left|\overrightarrow{p_1} \times \overrightarrow{p_2}\right|=3-4 \alpha+4-4 \alpha+6 \\ & =13-8 \alpha \end{aligned}

\begin{aligned} & \left|\frac{13-8 \alpha}{\sqrt{(3-4 \alpha)^2+4+(2 \alpha-3)^2}}\right|=\frac{5}{\sqrt{6}} \\ & \left|\frac{13-8 \alpha}{\sqrt{20 \alpha^2-36 \alpha+22}}\right|=\frac{5}{\sqrt{6}} \\ & =6(13-8 \alpha)^2=25\left(20 \alpha^2-36 \alpha+22\right) \\ & =116 \alpha^2+348 \alpha-464=0 \\ & \text{ Sum of roots }=-3 \end{aligned}

Q243

If the image of the point

\mathrm{P}(1,0,3)

in the line joining the points

\mathrm{A}(4,7,1)

and

\mathrm{B}(3,5,3)

is

Q(\alpha, \beta, \gamma)

, then

\alpha+\beta+\gamma

is equal to :

A

\dfrac{46}{3}

B 18

C 13

D

\dfrac{47}{3}

Correct Answer

Option A

Solution

Let $X$ be mid point of $P$ and $Q$ , which would be also feet of perpendicular.

Let $X$ divides $A$ and $B$ in $\lambda: 1, \lambda \neq-1$

\begin{aligned} &X=\left(\frac{3 \lambda+4}{\lambda+1}, \frac{5 \lambda+7}{\lambda+1}, \frac{3 \lambda+1}{\lambda+1}\right)\\ &\text{ Now } P X \perp A B \Rightarrow \overrightarrow{P X} \cdot \overrightarrow{A B}=0\\ &\begin{aligned} &\left(\frac{3 \lambda+4}{\lambda+1}-1\right) \cdot(4-3)+\left(\frac{5 \lambda+7}{\lambda+1}-0\right)(7-5) \\ &+\left(\frac{3 \lambda+1}{\lambda+1}-3\right) \cdot(1-3)=0 \end{aligned} \end{aligned}

\begin{aligned} &\begin{aligned} & \frac{2 \lambda+3}{\lambda+1}+\frac{10 \lambda+14}{\lambda+1}+\frac{4}{\lambda+1}=0 \\ & \Rightarrow \frac{12 \lambda+21}{\lambda+1}=0 \Rightarrow \lambda=\frac{-7}{4} \\ & X=\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right) \end{aligned}\\ &X \text{ is mid point of } P Q\\ &\begin{aligned} & Q \equiv\left(2 \cdot \frac{5}{3}-1,2 \cdot \frac{7}{3}-0,2 \cdot \frac{17}{3}-3\right) \equiv(\alpha, \beta, \gamma) \\ & \Rightarrow \quad \alpha+\beta+\gamma=\frac{2(5+7+17)}{3}-4=\frac{58}{3}-4=\frac{46}{3} \end{aligned} \end{aligned}

Q244

The line

\mathrm{L}_1

is parallel to the vector

\overrightarrow{\mathrm{a}}=-3 \hat{i}+2 \hat{j}+4 \hat{k}

and passes through the point

(7,6,2)

and the line

\mathrm{L}_2

is parallel to the vector

\overrightarrow{\mathrm{b}}=2 \hat{i}+\hat{j}+3 \hat{k}

and passes through the point

(5,3,4)

. The shortest distance between the lines

L_1

and

L_2

is :

A

\dfrac{23}{\sqrt{38}}

B

\dfrac{21}{\sqrt{38}}

C

\dfrac{23}{\sqrt{57}}

D

\dfrac{21}{\sqrt{57}}

Correct Answer

Option A

Solution

The line $L_1$ is parallel to the vector $\overrightarrow{\mathrm{a}} = -3 \hat{i} + 2 \hat{j} + 4 \hat{k}$ and passes through the point $(7, 6, 2)$ .

The line $L_2$ is parallel to the vector $\overrightarrow{\mathrm{b}} = 2 \hat{i} + \hat{j} + 3 \hat{k}$ and passes through the point $(5, 3, 4)$ .

Equations of the Lines: Equation of $L_1$ : $\mathbf{r} = 7 \hat{i} + 6 \hat{j} + 2 \hat{k} + \lambda(-3 \hat{i} + 2 \hat{j} + 4 \hat{k})$ Equation of $L_2$ : $\mathbf{r} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k} + \mu(2 \hat{i} + \hat{j} + 3 \hat{k})$ Shortest Distance Between $L_1$ and $L_2$ : The formula for the shortest distance between two skew lines is given by: $\text{Distance} = \left|\dfrac{\left(\vec{a}_2 - \vec{a}_1\right) \cdot \left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|$ Calculations: Vector between the points: $\vec{a}_2 - \vec{a}_1 = (5 \hat{i} + 3 \hat{j} + 4 \hat{k}) - (7 \hat{i} + 6 \hat{j} + 2 \hat{k}) = -2 \hat{i} - 3 \hat{j} + 2 \hat{k}$ Cross product of direction vectors: $\vec{b}_1 = -3 \hat{i} + 2 \hat{j} + 4 \hat{k}, \quad \vec{b}_2 = 2 \hat{i} + \hat{j} + 3 \hat{k}$ $\vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 2 & 4 \\ 2 & 1 & 3 \end{vmatrix} = 2 \hat{i} + 17 \hat{j} - 7 \hat{k}$ Magnitude of the cross product: $\left|\vec{b}_1 \times \vec{b}_2\right| = \sqrt{2^2 + 17^2 + (-7)^2} = \sqrt{342}$ Dot product: $\left(\vec{a}_2 - \vec{a}_1\right) \cdot \left(\vec{b}_1 \times \vec{b}_2\right) = (-2 \hat{i} - 3 \hat{j} + 2 \hat{k}) \cdot (2 \hat{i} + 17 \hat{j} - 7 \hat{k})$ $= (-2)(2) + (-3)(17) + (2)(-7) = -4 - 51 - 14 = -69$ Shortest Distance: $\text{Distance} = \left|\dfrac{-69}{\sqrt{342}}\right| = \dfrac{23}{\sqrt{38}}$

Q245

Let

A

and

B

be two distinct points on the line

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

. Both

A

and

B

are at a distance

2 \sqrt{17}

from the foot of perpendicular drawn from the point

(1,2,3)

on the line

L

. If

O

is the origin, then

\overrightarrow{O A} \cdot \overrightarrow{O B}

is equal to

A 49

B 21

C 47

D 62

Correct Answer

Option C

Solution

L: \frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}

Point $A(3 \lambda+6,2 \lambda+7,7-2 \lambda)$

B(3 \mu+6,2 \mu+7,7-2 \mu)

Let $P(3 k+6,2 k+7,7-2 k)$ be foot of perpendicular from $P^{\prime}(1,2,3)$

\begin{aligned} & \therefore \quad P P^{\prime}\langle 3,2,-2\rangle=0 \\ & 3(3 k+5)+(2 k+5) 2+(4-2 k)(-2)=0 \\ & 9 k+15+4 k+10-8+4 k=0 \\ & 17 k+17=0 \\ & \Rightarrow k=-1 \qquad \therefore \quad P(3,5,9)\\ & |\overrightarrow{A P}|=2 \sqrt{17} \end{aligned}

\begin{aligned} & \Rightarrow(3 \lambda+3)^2+(2 \lambda+2)^2+(-2-2 \lambda)^2=17 \times 4 \\ & =17(\lambda+1)^2=17 \times 4 \\ & \Rightarrow \lambda+1= \pm 2 \Rightarrow \lambda=1 \text{ or } \lambda=-3 \\ & \therefore A(9,9,5), B(-3,1,13) \\ & \overrightarrow{O A} \cdot \overrightarrow{O B}=(9 \hat{i}+9 \hat{j}+5 \hat{k}) \cdot(-3 \hat{i}+\hat{j}+13 \hat{k}) \\ & =-27+9+65=47 \end{aligned}

Q246

Each of the angles

\beta

and

\gamma

that a given line makes with the positive

y

- and

z

-axes, respectively, is half of the angle that this line makes with the positive

x

-axes. Then the sum of all possible values of the angle

\beta

is

A

\dfrac{\pi}{2}

B

\pi

C

\dfrac{3 \pi}{4}

D

\dfrac{3 \pi}{2}

Correct Answer

Option C

Solution

Given: Each of the angles $\beta$ and $\gamma$ is half of the angle that the line makes with the positive $x$ -axis, i.e., $\beta = \gamma = \dfrac{\alpha}{2}$ .

The equation for the direction cosines of angles a line makes with the coordinate axes is given by: $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$ Since $\beta = \gamma$ , we substitute to get: $\cos^2 \alpha + 2 \cos^2 \beta = 1$ Substitute $\cos \beta = \cos \gamma$ : $\cos \alpha = 2 \cos^2 \beta - 1$ Replacing back into the equation: $(2 \cos^2 \beta - 1)^2 + 2 \cos^2 \beta = 1$ Simplify: $(2 \cos^2 \beta - 1)(2 \cos^2 \beta + 1) = 0$ Solving for $\cos^2 \beta$ , we have: $2 \cos^2 \beta - 1 = 0 \quad \text{or} \quad 2 \cos^2 \beta + 1 = 0$ The latter gives no real solutions, thus: $\cos^2 \beta = \dfrac{1}{2}$ Therefore, $\beta = \dfrac{\pi}{4}$ or $\beta = \dfrac{\pi}{2}$ .

Thus, the sum of all possible values of $\beta$ is: $\dfrac{\pi}{4} + \dfrac{\pi}{2} = \dfrac{3\pi}{4}$

Q247

The distance of the point

(7,10,11)

from the line

\dfrac{x-4}{1}=\dfrac{y-4}{0}=\dfrac{z-2}{3}

along the line

\dfrac{x-9}{2}=\dfrac{y-13}{3}=\dfrac{z-17}{6}

is

A 16

B 12

C 18

D 14

Correct Answer

Option D

Solution

Equation of line passing through $P(7,10,11)$ along the line $\dfrac{x-9}{2}=\dfrac{y-13}{3}=\dfrac{z-17}{6}$ is

\frac{x-7}{2}=\frac{y-10}{3}=\frac{z-11}{6}=\lambda

Let the point on the line is

Q(2 \lambda+7,3 \lambda+10,6 \lambda+11)

$Q$ lies on line $\dfrac{x-4}{1}=\dfrac{y-4}{0}=\dfrac{z-2}{3}$

\begin{aligned} & 3 \lambda+10=4 \Rightarrow \lambda=-2 \\ & \therefore \quad Q(3,4,-1) \\ & P Q=\sqrt{16+36+144}=14 \end{aligned}

Q248

The shortest distance from the plane

12x+4y+3z=327

to the sphere

{x^2} + {y^2} + {z^2} + 4x - 2y - 6z = 155

is

A

39

B

26

C

11{4 \over {13}}

D

13

Correct Answer

Option D

Solution

Shortest distance $=$ perpendicular distance between the plane and sphere $=$ distance of plane from center of sphere $-$ radius

= \left| {{{ - 2 \times 12 + 4 \times 1 + 3 \times 3 - 327} \over {\sqrt {144 + 9 + 16} }}} \right| - \sqrt {4 + 1 + 9 + 155}

= 26 - 13 = 13

Q249

If the line,

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

lies in the planes,

lx+my-z=9,

then

{l^2} + {m^2}

is equal to :

A

5

B

2

C

26

D

18

Correct Answer

Option B

Solution

Line lies in the plane

\Rightarrow \left( {3, - 2, - 4} \right)

lie in the plane

\Rightarrow 3\ell - 2m + 4 = 9

or

3\ell - 2m = 5....\left( 1 \right)

Also,

\ell ,

m, - 1

are dr's of line perpendicular to plane and

2, - 1,3

are dr's of line lying in the plane

\Rightarrow 2\ell - m - 3 = 0\,\,\,

or

\,\,\,2\ell - m = 3....\left( 2 \right)

Solving

(1)

and

(2)

we get

\ell = 1

and

m=-1

\Rightarrow {\ell ^2} + {m^2} = 2.

Q250

The projections of a vector on the three coordinate axis are

6,-3,2

respectively. The direction cosines of the vector are :

A

{6 \over 5},{{ - 3} \over 5},{2 \over 5}

B

{6 \over 7 },{{ - 3} \over 7},{2 \over 7}

C

{- 6 \over 7 },{{ - 3} \over 7},{2 \over 7}

D

6, -3, 2

Correct Answer

Option B

Solution

Let

P\left( {{x_1},{y_1},{z_1}} \right)

and

Q\left( {{x_2},{y_2},{z_2}} \right)

be the initial and final points of the vector whose projections on the three coordinates axes are

{6, - 3,2}

then

{x_2} - {x_1}, = 6;\,\,{y_2} - {y_1} = - 3;\,\,{z_2} - {z_1} = 2

So that directions ratios of

\overrightarrow {PQ}

are

{6, - 3,2}

$\therefore$ Direction cosines of

\overrightarrow {PQ}

are

{6 \over {\sqrt {{6^2} + {{\left( { - 3} \right)}^2} + {2^2}} }},{{ - 3} \over {\sqrt {{6^2} + {{\left( { - 3} \right)}^2} + {2^2}} }},

\,\,\,\,\,\,\,\,

{2 \over {\sqrt {{6^2} + {{\left( { - 3} \right)}^2} + {2^2}} }} = {6 \over 7},{{ - 3} \over 7},{2 \over 7}