SATHEE: 3D Geometry 3 Question 27

3D Geometry 3 Question 27

####27. The plane through the intersection of the planes $x + y + z = 1$ and $2 x + 3 y - z + 4 = 0$ and parallel to $Y$ -axis also passes through the point (2019 Main, 9 Jan I)

(a) $(3, 3, - 1)$

(b) $(- 3, 1, 1)$

(d) $(- 3, 0, - 1)$

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Answer:

(c)

Solution:

The plane through the intersection of the planes $x + y + z - 1 = 0$ and $2 x + 3 y - z + 4 = 0$ is given by $(x + y + z - 1) + λ (2 x + 3 y - z + 4) = 0$ , where $λ \in R$

$\Rightarrow (1 + 2 λ) x + (1 + 3 λ) y + (1 - λ) z + (4 λ - 1) = 0$ , where

$λ \in R$

Since, this plane is parallel to $Y$ -axis, therefore its normal is perpendicular to $Y$ -axis.

$\begin{array}{lc} \Rightarrow & (1 + 2 λ) \hat{i} + (1 + 3 λ) \hat{j} + (1 - λ) \hat{k} \cdot \hat{j} = 0 \\ \Rightarrow & 1 + 3 λ = 0 \Rightarrow λ = - \frac{1}{3} \end{array}$

Now, required equation of plane is

$\begin{aligned} 1 - \frac{2}{3} x + 1 - \frac{3}{3} y + 1 + \frac{1}{3} z + - \frac{4}{3} - 1 & = 0 \\ [substituting λ & = \frac{- 1}{3} in Eq. (i)] \end{aligned}$

$\Rightarrow x + 4 z - 7 = 0$

Here, only $(3, 2, 1)$ satisfy the above equation.

3D Geometry 3 Question 27

Answer:

Solution:

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3D Geometry 3 Question 27

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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