SATHEE: 3D Geometry 3 Question 18

3D Geometry 3 Question 18

####18. Two lines $\frac{x - 3}{1} = \frac{y + 1}{3} = \frac{z - 6}{- 1}$ and $\frac{x + 5}{7} = \frac{y - 2}{- 6} = \frac{z - 3}{4}$ intersect at the point $R$ . The reflection of $R$ in the $x y$ -plane has coordinates

(2019 Main, 11 Jan II)

(a) $(2, - 4, - 7)$

(b) $(2, - 4, 7)$

(d) $(2, 4, 7)$

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Answer:

(a)

Solution:

Let $\frac{x - 3}{1} = \frac{y + 1}{3} = \frac{z - 6}{- 1} = a$ (say). Then, any point on this line is of the form $P (a + 3, 3 a - 1, - a + 6)$

Similarly, any point on the line. $\frac{x + 5}{7} = \frac{y - 2}{- 6} = \frac{z - 3}{4} =$ $b$ (say), is of the form $Q (7 b - 5, - 6 b + 2, 4 b + 3)$

Now, if the lines are intersect, then $P = Q$ for some $a$ and $b$ .

$\begin{aligned} \Rightarrow a + 3 & = 7 b - 5 \\ 3 a - 1 = - 6 b + 2 \\ and - a + 6 & = 4 b + 3 \end{aligned}$

$\Rightarrow a - 7 b = - 8, a + 2 b = 1 and a + 4 b = 3$

On solving $a - 7 b = - 8$ and $a + 2 b = 1$ , we get $b = 1$ and $a = - 1$ ,

which also satisfy $a + 4 b = 3$

$∴ P = Q \equiv (2, - 4, 7) for a = - 1 and b = 1$

Thus, coordinates of point $R$ are $(2, - 4, 7)$

and reflection of $R$ in $x y$ -plane is $(2, - 4, - 7)$

3D Geometry 3 Question 18

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3D Geometry 3 Question 18

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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