SATHEE: 3D Geometry 3 Question 44

3D Geometry 3 Question 44

####44. Equation of the plane containing the straight line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the staight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is

(2010)

(a) $x + 2 y - 2 z = 0$

(b) $3 x + 2 y - 2 z = 0$

(d) $5 x + 2 y - 4 z = 0$

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Answer:

Correct Answer: 44. (c)

Solution:

The DR’s of normal to the plane containing $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{z} = \frac{z}{3}$ .

$n_{1} = | \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & 4 & 2 \\ 4 & 2 & 3 \end{array} | = (8 \hat{i} - \hat{j} - 10 \hat{k})$

Also, equation of plane containing $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and DR’s of normal to be $n_{1} = a \hat{i} + b \hat{j} + c \hat{k}$

$\Rightarrow a x + b y + c z = 0$

where, $\mathbf{n}{1}: \mathbf{n}{2}=0$

$\Rightarrow 8 a - b - 10 c = 0$

and $n_{2} ⊥ (2 \hat{i} + 3 \hat{j} + 4 \hat{k})$

$\Rightarrow 2 a + 3 b + 4 c = 0$

From Eqs (ii) and (iii),

$\begin{aligned} \frac{a}{- 4 + 30} & = \frac{b}{- 20 - 32} = \frac{c}{24 + 2} \\ \Rightarrow \frac{a}{26} & = \frac{b}{- 52} = \frac{c}{26} \end{aligned}$

$\Rightarrow \frac{a}{1} = \frac{b}{- 2} = \frac{c}{1}$

From Eqs. (i) and (iv), required equation of plane is $x - 2 y + z = 0$

3D Geometry 3 Question 44

Answer:

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3D Geometry 3 Question 44

Answer:

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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