SATHEE: 3D Geometry 3 Question 23

3D Geometry 3 Question 23

####23. On which of the following lines lies the point of intersection of the line, $\frac{x - 4}{2} = \frac{y - 5}{2} = \frac{z - 3}{1}$ and the plane, $x + y + z = 2$ ?

(2019 Main, 10 Jan II)

(a) $\frac{x - 4}{1} = \frac{y - 5}{1} = \frac{z - 5}{- 1}$

(b) $\frac{x + 3}{3} = \frac{4 - y}{3} = \frac{z + 1}{- 2}$

(d) $\frac{x - 1}{1} = \frac{y - 3}{2} = \frac{z + 4}{- 5}$

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Answer:

(d)

Solution:

Given equation of line is

$\begin{aligned} \frac{x - 4}{2} = \frac{y - 5}{2} = \frac{z - 3}{1} = r (let) \\ \Rightarrow x = 2 r + 4; y = 2 r + 5 and z = r + 3 \end{aligned}$

$∴$ General point on the line (i) is

$P (2 r + 4, 2 r + 5, r + 3)$

So, the point of intersection of line (i) and plane $x + y + z = 2$ will be of the form $P (2 r + 4, 2 r + 5, r + 3)$ for some $r \in R$ .

$\Rightarrow (2 r + 4) + (2 r + 5) + (r + 3) = 2$

$[∵$ the point will lie on the plane]

$\Rightarrow 5 r = - 10 \Rightarrow r = - 2$

So, the point of intersection is $P (0, 1, 1)$

[putting $r = - 2$ in $(2 r + 4, 2 r + 5, r + 3)$ ]

Now, on checking the options, we get

$\frac{x - 1}{1} = \frac{y - 3}{2} = \frac{z + 4}{- 5}$ contain the point $(0, 1, 1)$

3D Geometry 3 Question 23

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Solution:

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3D Geometry 3 Question 23

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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