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Ncert Solutions Maths class 12th Maths Class 12th Three Dimensional Geometry

23. In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

$(a$ $)$ $z = 2$

$(b) x + y + z = 1$

$(c) 2 x + 3 y - z = 5$

$(d) 5 y + 8 = 0$

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Vishal Baghel | Contributor-Level 10

4 months ago

(a) The equation of the plane is $z = 2 o r 0 x + 0 y + z = 2 . . . . . . . . . . (1)$

The direction ratios of normal are $0, 0, a n d 1 .$

$∴ + 02 + 12 = 1$

Dividing both sides of equation (1) by 1, we obtain

$0 . x + 0 . y + 1 . z = 2$

This is of the form $l x + m y + n z = d$ , where l, m, n are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the origin is 2 units.

(b) $x + y + z = 1 . . . . . . . . . . (1)$

The direction ratios of normal are 1, 1, and 1.

$∴ + (1) ² + (1) ² =$

Dividing both sides of equation (1) by , we obtain

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{} . . . . . . . . . (2)$

This equation is of the form $l x + m y + n z = d$ , where l, m, n

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(a) The equation of the plane is $z = 2 o r 0 x + 0 y + z = 2 . . . . . . . . . . (1)$

The direction ratios of normal are $0, 0, a n d 1 .$

$∴ + 02 + 12 = 1$

Dividing both sides of equation (1) by 1, we obtain

$0 . x + 0 . y + 1 . z = 2$

This is of the form $l x + m y + n z = d$ , where l, m, n are the direction cosines of normal to the plane and d is the distance of the perpendicular drawn from the origin.

Therefore, the direction cosines are 0, 0, and 1 and the distance of the plane from the origin is 2 units.

(b) $x + y + z = 1 . . . . . . . . . . (1)$

The direction ratios of normal are 1, 1, and 1.

$∴ + (1) ² + (1) ² =$

Dividing both sides of equation (1) by , we obtain

$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{} . . . . . . . . . (2)$

This equation is of the form $l x + m y + n z = d$ , where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal are $\frac{1}{}$ , $\frac{1}{}$ and $\frac{1}{}$ the distance of normal from the origin is $\frac{1}{}$ units.

(c) $2 x + 3 y - z = 5 . . . . . . . . . . (1)$

The direction ratios of normal are 2, 3, and −1.

$∴ + (3) ² + (- 1) ² =$

Dividing both sides of equation (1) by , we obtain

$\frac{2}{x} + \frac{3}{y} - \frac{1}{z} = \frac{5}{}$

This equation is of the form $l x + m y + n z = d$ where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal to the plane $\frac{2}{x} + \frac{3}{y} - \frac{1}{z}$ and the distance of normal from the origin is $\frac{5}{}$ units.

(d)

$\begin{array}{l} 5 y + 8 = 0 & \Rightarrow 0 x - 5 y + 0 z = 8 . . . . . . . . . . (1) \end{array}$

The direction ratios of normal are 0, −5, and 0.

$∴ + (- 5) ² + 0 = 5$

Dividing both sides of equation (1) by 5, we obtain

$- y = \frac{8}{5}$

This equation is of the form $l x + m y + n z = d$ where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.

Therefore, the direction cosines of the normal to the plane are 0, −1, and 0 and the distance of normal from the origin is $\frac{8}{5}$ units.

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23. In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

$(a$ $)$ $z = 2$

$(b) x + y + z = 1$

$(c) 2 x + 3 y - z = 5$

$(d) 5 y + 8 = 0$

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23. In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin. (a) z=2 (b) x+y+z=1 (c) 2x+3y−z=5 (d) 5y+8=0

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23. In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

$(a$ $)$ $z = 2$

$(b) x + y + z = 1$

$(c) 2 x + 3 y - z = 5$

$(d) 5 y + 8 = 0$