The image of the line x-1-3-y-3-1-z-4-5 in the plane 2 x-y-z-3-0 is the linea x-3-3-y-5-1-z-2-5b x-3-3-y-5-1-z-2-5c x-3-3-y-5-1-z-2-5d x-3-3-y-5-1-z-2-5

Point (1,3,4) Solution of normal to plane is x 1/2=y 3/ 1=z 4/j=k Any point on the normal is ⇒ (2k+1, k+3, k+4) (It lies on plane) ⇒ 2(k+1) ( 6 k/2 )+8+k/2+3 ...

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Byju's Answer

Standard VI

Mathematics

Collinear Points

The image of ...

Question

The image of the line $\frac{x - 1}{3} = \frac{y - 3}{1} = \frac{z - 4}{- 5}$ in the plane $2 x - y + z + 3 = 0$ is the line
a) $\frac{x + 3}{3} = \frac{y - 5}{1} = \frac{z - 2}{- 5}$
b) $\frac{x + 3}{- 3} = \frac{y - 5}{- 1} = \frac{z + 2}{5}$
c) $\frac{x - 3}{3} + \frac{y + 5}{1} = \frac{z - 2}{- 5}$
d) $\frac{x - 3}{- 3} = \frac{y + 5}{- 1} = \frac{z - 2}{5}$

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Solution

Point (1,3,4)
Solution of normal to plane is
$\frac{x - 1}{2} = \frac{y - 3}{- 1} = \frac{z - 4}{j} = k$
Any point on the normal is
$\Rightarrow (2 k + 1, - k + 3, k + 4)$ (It lies on plane)
$\Rightarrow 2 (k + 1) - (\frac{6 - k}{2}) + \frac{8 + k}{2} + 3 = 0$
$\Rightarrow k = - 2$
$\Rightarrow$ Point through which image passing (-3, 5, 2)
So, required equation of line
$\frac{x + 3}{3} = \frac{y - 5}{1} = \frac{z - 5}{- 5}$

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Similar questions

Q. The equation of the plane through the line x + y + z + 3 = 0 = 2x − y + 3z + 1 and parallel to the line

\frac{x}{1} = \frac{y}{2} = \frac{z}{3}

is
(a) x − 5y + 3z = 7
(b) x − 5y + 3z = −7
(c) x + 5y + 3z = 7
(d) x + 5y + 3z = −7

Q. 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

Q. The image of line

\frac{x - 1}{3} = \frac{y - 3}{1} = \frac{z - 4}{- 5}

in the plane

2 x - y + z + 3 = 0

Perpendicular are drawn from points on the line $\frac{x + 2}{2} = \frac{y + 1}{- 1} = \frac{z}{3}$ to the plane x + y + z = 3. The feet of perpendiculars lie on the line

Q. Equation of the plane containing the line

x + 2 y + 3 z - 5 = 0 = 3 x + 2 y + z - 5

which is parallel to the line

x - 1 = 2 - y = z - 3

, is-

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The image of the line x−13=y−31=z−4−5 in the plane 2x−y+z+3=0 is the line a) x+33=y−51=z−2−5 b) x+3−3=y−5−1=z+25 c) x−33+y+51=z−2−5 d) x−3−3=y+5−1=z−25

Point (1,3,4) Solution of normal to plane is x−12=y−3−1=z−4j=k Any point on the normal is ⇒(2k+1,−k+3,k+4) (It lies on plane) ⇒2(k+1)−(6−k2)+8+k2+3=0 ⇒k=−2 ⇒ Point through which image passing (-3, 5, 2) So, required equation of line x+33=y−51=z−5−5