SATHEE: 3D Geometry 3 Question 19

3D Geometry 3 Question 19

####19. If the point $(2, α, β)$ lies on the plane which passes through the points $(3, 4, 2)$ and

$(7, 0, 6)$ and is perpendicular to the plane $2 x - 5 y = 15$ , then $2 α - 3 β$ is equal to

(2019 Main, 11 Jan II)

(a) 17

(b) 7

(d) 12

Show Answer

Answer:

(b)

Solution:

According to given information, we have the following figure.

From figure, it is clear that

$(A B \times B C) = p$ and $n = 2 \hat{i} - 5 \hat{j} + 0 \hat{k}$

$∴ p = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & - 4 & 4 \\ - 5 & α & β - 6 \end{array} |$

$[∵ A B = (7 - 3) \hat{i} + (0 - 4) \hat{j} + (6 - 2) \hat{k}$

$= 4 \hat{i} - 4 \hat{j} + 4 \hat{k}$ and $B C = (2 - 7) \hat{i} + (α - 0) \hat{j} + (β - 6) \hat{k}$

$= - 5 \hat{i} + α \hat{j} + (β - 6) \hat{k}]$

$= \hat{i} (- 4 β + 24 - 4 α) - \hat{j} (4 β - 24 + 20) + \hat{k} (4 α - 20)$

$\Rightarrow p = (24 - 4 α - 4 β) \hat{i} + \hat{j} (4 - 4 β) + \hat{k} (4 α - 20)$

Now, as the planes are perpendicular, therefore $p \cdot n = 0$

$\Rightarrow ((24 - 4 α - 4 β) \hat{i} + (4 - 4 β) \hat{j} + (4 α - 20) \hat{k})$

$\cdot (2 \hat{i} - 5 \hat{j} + 0 \hat{k})) = 0$

$\Rightarrow 2 (24 - 4 α - 4 β) - 5 (4 - 4 β) + 0 = 0$

$\Rightarrow 8 (6 - α - β) - 4 (5 - 5 β) = 0$

$\Rightarrow 12 - 2 α - 2 β - 5 + 5 β = 0 \Rightarrow 2 α - 3 β = 7$

3D Geometry 3 Question 19

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

3D Geometry 3 Question 19

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu