SATHEE: 3D Geometry 3 Question 10

3D Geometry 3 Question 10

####10. If the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{4}$ meets the plane, $x + 2 y + 3 z = 15$ at a point $P$ , then the distance of $P$ from the origin is

(2019 Main, 9 April I)

(a) $7 / 2$

(b) $9 / 2$

(d) $2 \sqrt{5}$

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Answer:

(b)

Solution:

Equation of given plane is

$x + 2 y + 3 z = 15$ and line is, $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{4} = r$ (let)

So, the coordinates of any point on line (ii) is

$P (1 + 2 r, - 1 + 3 r, 2 + 4 r)$ .

$∵$ Point $P$ is intersecting point of plane (i) and line (ii)

$∴ (1 + 2 r) + 2 (- 1 + 3 r) + 3 (2 + 4 r) = 15$

$\Rightarrow 1 + 2 r - 2 + 6 r + 6 + 12 r = 15 \Rightarrow 20 r = 10$

$\Rightarrow r = \frac{1}{2}$

$∴$ Coordinates of $P = 1 + 1, - 1 + \frac{3}{2}, 2 + 2 = 2, \frac{1}{2}, 4$

Now, distance of the point $P$ from the origin

$= \sqrt{4 + \frac{1}{4} + 16} = \sqrt{20 + \frac{1}{4}} = \sqrt{\frac{81}{4}} = \frac{9}{2}$ units

3D Geometry 3 Question 10

Answer:

Solution:

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3D Geometry 3 Question 10

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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