SATHEE: 3D Geometry 3 Question 28

3D Geometry 3 Question 28

####28. The equation of the line passing through $(- 4, 3, 1)$ , parallel to the plane $x + 2 y - z - 5 = 0$ and intersecting the line

(2019 Main, 9 Jan I) $\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z - 2}{- 1}$ is

(a) $\frac{x + 4}{3} = \frac{y - 3}{- 1} = \frac{z - 1}{1}$

(b) $\frac{x + 4}{- 1} = \frac{y - 3}{1} = \frac{z - 1}{1}$

(d) $\frac{x - 4}{2} = \frac{y + 3}{1} = \frac{z + 1}{4}$

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Answer:

(a)

Solution:

Any point on the line $\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z - 2}{- 1}$ is of the form $(- 3 λ - 1, 2 λ + 3, - λ + 2)$

$\begin{array}{r} [take \frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z - 2}{- 1} = λ \Rightarrow x = - 3 λ - 1, \\ y = 2 λ + 3 and z = - λ + 2] \end{array}$

So, the coordinates of point of intersection of two lines will be $(- 3 λ - 1, 2 λ + 3, - λ + 2)$ for some $λ \in R$ .

Let the point $A \equiv (- 3 λ - 1, 2 λ + 3, - λ + 2)$ and $B \equiv (- 4, 3, 1)$

$\begin{aligned} Then, A B = O B - O A \\ = (- 4 \hat{i} + 3 \hat{j} + \hat{k}) - (- 3 λ - 1) \hat{i} + (2 λ + 3) \hat{j} + (- λ + 2) \hat{k} \\ = (3 λ - 3) \hat{i} - 2 λ \hat{j} + (λ - 1) \hat{k} \end{aligned}$

Now, as the line is parallel to the given plane, therefore $A B$ will be parallel to the given plane and so $A B$ will be perpendicular to the normal of plane.

$\Rightarrow A B \cdot λ = 0$ , where $n = \hat{i} + 2 \hat{j} - \hat{k}$ is normal to the plane.

$\begin{aligned} \Rightarrow ((3 λ - 3) \hat{i} - 2 λ \hat{j} + (λ - 1) \hat{k}) \cdot (\hat{i} + 2 \hat{j} - \hat{k}) = 0 \\ \Rightarrow 3 (λ - 1) - 4 λ + (- 1) (λ - 1) = 0 \\ [∵ If a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k} and b = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}, \\ then a \cdot b = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}] \\ \Rightarrow 3 λ - 3 - 4 λ - λ + 1 = 0 \\ \Rightarrow - 2 λ = 2 \Rightarrow λ = - 1 \end{aligned}$

Now, the required equation is the equation of line joining $A (2, 1, 3)$ and $B (- 4, 3, 1)$ , which is

$\frac{x - (- 4)}{2 - (- 4)} = \frac{y - 3}{1 - 3} = \frac{z - 1}{3 - 1}$

$[∵$ Equation of line joining $(x_{1}, y_{1}, z_{1})$ and $(x_{2}, y_{2}, z_{2})$ is

$\begin{aligned} \frac{x - x_{1}}{x_{2} - x_{1}} = \frac{y - y_{1}}{y_{2} - y_{1}} = \frac{z - z_{1}}{z_{2} - z_{1}} \\ \Rightarrow \frac{x + 4}{6} & = \frac{y - 3}{- 2} = \frac{z - 1}{2} \\ or & \frac{x + 4}{3} = \frac{y - 3}{- 1} = \frac{z - 1}{1} [multiplying by 2] \end{aligned}$

3D Geometry 3 Question 28

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3D Geometry 3 Question 28

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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