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. (a)

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, $n_{1} : n_{2} = 0$

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

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 $x - 2 y + z = 0$

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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

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3D Geometry 3 Question 44

44. Equation of the plane containing the straight line x2=y3=z4 and perpendicular to the plane containing the staight lines x3=y4=z2 and x4=y2=z3 is

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इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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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