SATHEE: 3D Geometry 3 Question 25

3D Geometry 3 Question 25

####25. Let $A$ be a point on the line $r = (1 - 3 μ) \hat{i} + (μ - 1) \hat{j} + (2 + 5 μ) \hat{k}$ and $B (3, 2, 6)$ be a point in the space. Then, the value of $μ$ for which the vector $A B$ is parallel to the plane $x - 4 y + 3 z = 1$ is

(a) $\frac{1}{4}$

(b) $- \frac{1}{4}$

(d) $\frac{1}{2}$

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Answer:

(a)

Solution:

Given equation of line is

$r = (1 - 3 μ) \hat{i} + (μ - 1) \hat{j} + (2 + 5 μ) \hat{k}$

Clearly, any point on the above line is of the form $(1 - 3 μ, μ - 1, 2 + 5 μ)$

Let $A$ be $(- 3 μ + 1, μ - 1, 5 μ + 2)$ for some $μ \in R$ . Then, $A B = (3 - (- 3 μ + 1)) \hat{i} + (2 - (μ - 1)) \hat{j}$

$\begin{aligned} + (6 - (5 μ + 2)) \hat{k} [∵ A B = O B - O A] \\ = (3 μ + 2) \hat{i} + (3 - μ) \hat{j} + (4 - 5 μ) \hat{k} \end{aligned}$

Normal vector $(n)$ of the plane $x - 4 y + 3 z = 1$ is

$n = \hat{i} - 4 \hat{j} + 3 \hat{k}$

$∵ AB$ is parallel to the plane.

$∴ n$ is perpendicular to the $A B$ .

$\Rightarrow A B \cdot n = 0$

$\Rightarrow [(3 μ + 2) \hat{i} + (3 - μ) \hat{j} + (4 - 5 μ) \hat{k}] \cdot [\hat{i} - 4 \hat{j} + 3 \hat{k}] = 0$

[From Eqs. (i) and (ii)]

$\begin{aligned} \Rightarrow (3 μ + 2) - 4 (3 - μ) + 3 (4 - 5 μ) = 0 \\ \Rightarrow - 8 μ + 2 = 0 \\ \Rightarrow μ = \frac{1}{4} \end{aligned}$

3D Geometry 3 Question 25

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3D Geometry 3 Question 25

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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