Question

Answer this Question of iit jee . Chapter → Three dimensional Geometry ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given equation of line is

$\rm :\longmapsto\:x = \dfrac{y - 1}{2} = \dfrac{z - 3}{\lambda }$

can be rewritten as

$\rm :\longmapsto \: \dfrac{x}{1} = \dfrac{y - 1}{2} = \dfrac{z - 3}{\lambda }$

So, Direction ratios of line is given by

$\rm :\longmapsto\:d.r's \: = (1, \: 2, \: \lambda ) - - - (1)$

So,

$\rm :\longmapsto\: \vec{b} \: = \hat{i} + 2\hat{j} + \lambda \hat{k}$

So,

$\rm :\longmapsto\: |\vec{b} | = \sqrt{1 + 4 + {\lambda }^{2} } = \sqrt{5 + {\lambda }^{2} }$

Again,

Equation of plane is

$\rm :\longmapsto\:x + 2y + 3z = 4$

So, normal to the plane or direction ratios of plane is

$\rm :\longmapsto\:d.r's \: = (1, \: 2, \: 3 ) - - - (2)$

So,

$\rm :\longmapsto\: \vec{n} \: = \hat{i} + 2\hat{j} + 3 \hat{k}$

So,

$\rm :\longmapsto\: | \vec{n}| = \sqrt{1 + 4 + 9} = \sqrt{14}$

Let us assume that angle between line and plane is p.

Now, it is given that

$\rm :\longmapsto\:p = {cos}^{ - 1} \sqrt{\dfrac{5}{14} }$

So, by using triangle method, we get

$\rm :\longmapsto\:p = {sin}^{ - 1} \dfrac{3}{ \sqrt{14} }$

$\rm :\longmapsto\:sinp =\dfrac{3}{ \sqrt{14} }$

Now, we know that

Angle p, between line and plane is given by

$\rm :\longmapsto\:sinp = \dfrac{\vec{b} \: . \: \vec{n}}{ |\vec{b} | \: | \vec{n}| }$

On substituting all the values, evaluated above we get

$\rm :\longmapsto\:\dfrac{3}{ \sqrt{14} } = \dfrac{(\hat{i} + 2\hat{j} + \lambda \hat{k}) \: . \: (\hat{i} + 2\hat{j} + 3\hat{k})}{ \sqrt{14} \times \sqrt{5 + {\lambda }^{2} } }$

$\rm :\longmapsto\:3 = \dfrac{1 + 4 + 3\lambda }{ \sqrt{5 + {\lambda }^{2} } }$

$\rm :\longmapsto\:3 = \dfrac{5+ 3\lambda }{ \sqrt{5 + {\lambda }^{2} } }$

$\rm :\longmapsto\:3 \sqrt{5 + {\lambda }^{2} } = 5 + 3\lambda$

On squaring both sides, we get

$\rm :\longmapsto\:9(5 + {\lambda }^{2} ) = 25 + 9 {\lambda }^{2} + 30\lambda$

$\rm :\longmapsto\:45+ 9{\lambda }^{2} = 25 + 9 {\lambda }^{2} + 30\lambda$

$\rm :\longmapsto\:45 = 25 + 30\lambda$

$\rm :\longmapsto\:45 - 25 = 30\lambda$

$\rm :\longmapsto\:20= 30\lambda$

$\rm :\longmapsto\:2= 3\lambda$

$\bf\implies \: \lambda = \dfrac{2}{3}$

HENCE

• Option (b) is correct

Additional Information :-

Let us consider two planes,

$\rm :\longmapsto\:\vec{r}. \vec{n_1} = c$

and

$\rm :\longmapsto\:\vec{r}. \vec{n_2} = d$

Then,

1. Two planes are parallel iff

$\rm :\longmapsto\:\vec{n_1} = k \: \vec{n_2}$

2. Two planes are perpendicular off

$\rm :\longmapsto\:\vec{n_1} \: .\: \vec{n_2} = 0$

3. Angle p, between two planes is given by

$\rm :\longmapsto\:cosp = \dfrac{\vec{n_1} \: . \: \vec{n_2}}{ |\vec{n_1} | \: |\vec{n_2}| }$

Answer 2

let the angle between the line and plane be θ.

then,

$θ = { \cos }^{ - 1} ( \sqrt{ \frac{5}{14} } )$

$→ \cosθ = \sqrt{ \frac{5}{14} }$

$→ \sqrt{1 - { \sin }^{2}θ } = \sqrt{ \frac{5}{14} }$

squaring both sides, we get,

$→1 - { \sin }^{2} θ = \frac{5}{14}$

$→ { \sin}^{2} θ = \frac{9}{14}$

$→ \sinθ = \frac{3}{√14}$ ----eqn(1). (∵θ is acute angle)

Now,

$The \: direction \: ratios \: of \: given \: line \: \frac{x}{1} = \frac{y - 1}{2} = \frac{z - 3}{λ} \: are \: 1,2,λ.$

and the direction ratio of the given plane x+2y+3z=4 is 1, 2, 3.

→a₁=1, b₁=2, c₁=λ

→a₂=1, b₂=2, c₂=3

$→ \sin(θ) = \frac{1 + 4 + 3λ}{ \sqrt{ {1}^{2} + {2}^{2} + {λ}^{2} } \sqrt{ {1}^{2} + {2}^{2} + {3}^{2} } }$

from eqn(1)

$→ \frac{3}{ \sqrt{14} } = \frac{5 + 3λ}{ \sqrt{5 + {λ}^{2} } \sqrt{14} }$

→ $3 \sqrt{ {λ}^{2} + 5} = 5 + 3λ$

squaring both sides, we get,

→ $9( {λ}^{2} + 5) = ({5 + 3λ})^{2}$

→ $9 {λ}^{2} + 45 = 25 + 9 {λ}^{2} + 30λ$

→ $30λ = 45 - 25$

→ $λ = \frac{20}{30}$

∴ $λ = \frac{2}{3}$