SATHEE: 3D Geometry 3 Question 17

3D Geometry 3 Question 17

17. The perpendicular distance from the origin to the plane containing the two lines, $\frac{x + 2}{3} = \frac{y - 2}{5} = \frac{z + 5}{7}$ and $\frac{x - 1}{1} = \frac{y - 4}{4} = \frac{z + 4}{7}$ , is

(2019 Main, 12 Jan I)

(a) $11 \sqrt{6}$

(b) $\frac{11}{\sqrt{6}}$

(d) $6 \sqrt{11}$

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Answer:

(b)

Solution:

Let the equation of plane, containing the two lines

$\frac{x + 2}{3} = \frac{y - 2}{5} = \frac{z + 5}{7} and \frac{x - 1}{1} = \frac{y - 4}{4} = \frac{z + 4}{7} is$

$a (x + 2) + b (y - 2) + c (z + 5) = 0$

$∵$ Plane (i) contain lines, so

$3 a + 5 b + 7 c = 0$

and $a + 4 b + 7 c = 0$

From Eqs. (ii) and (iii), we get

$\begin{aligned} \frac{a}{35 - 28} = \frac{b}{7 - 21} & = \frac{c}{12 - 5} \\ \Rightarrow \frac{a}{7} & = \frac{b}{- 14} = \frac{c}{7} \Rightarrow \frac{a}{1} = \frac{b}{- 2} = \frac{c}{1} \end{aligned}$

So, equation of plane will be

$\begin{aligned} 1 (x + 2) - 2 (y - 2) + 1 (z + 5) & = 0 \\ \Rightarrow & x - 2 y + z + 11 = 0 \end{aligned}$

Now, perpendicular distance from origin to plane is

$= \frac{11}{\sqrt{1 + 4 + 1}} = \frac{11}{\sqrt{6}}$

$[∵$ perpendicular distance from origin to the plane

$a x + b y + c z + d = 0, is \frac{| d |}{\sqrt{a^{2} + b^{2} + c^{2}}}]$

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3D Geometry 3 Question 17

17. The perpendicular distance from the origin to the plane containing the two lines, $\frac{x + 2}{3} = \frac{y - 2}{5} = \frac{z + 5}{7}$ and $\frac{x - 1}{1} = \frac{y - 4}{4} = \frac{z + 4}{7}$ , is

Answer:

Solution:

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3D Geometry 3 Question 17

17. The perpendicular distance from the origin to the plane containing the two lines, x+23=y−25=z+57 and x−11=y−44=z+47, is

Answer:

Solution:

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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17. The perpendicular distance from the origin to the plane containing the two lines, $\frac{x + 2}{3} = \frac{y - 2}{5} = \frac{z + 5}{7}$ and $\frac{x - 1}{1} = \frac{y - 4}{4} = \frac{z + 4}{7}$ , is