Mathematics-I (3110014)

Question-1a

BE | Semester-1 Winter-2019 | 17-01-2020

Q1) (a)

$Here, f (x, y, z) = x^{2} + y^{2} + z^{2} - 3$ $and$ $p = (x_{1}, y_{1}, z_{1}) = (1, 1, 1)$ .

$Now,$

$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} (x^{2} + y^{2} + z^{2} - 3) = 2 x$

$⟹ \frac{\partial f}{\partial x} = {(2 x)}_{(1, 1, 1)} [Replace, p = (1, 1, 1)]$

$⟹ {(\frac{\partial f}{\partial x})}_{p} = 2 (1) = 2$
$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} [x^{2} + y^{2} + z^{2} - 3] = 2 y$

$⟹ \frac{\partial f}{\partial y} = {(2 y)}_{(1, 1, 1)} [Replace, p = (1, 1, 1)]$

$⟹ {(\frac{\partial f}{\partial y})}_{p} = 2 (1) = 2$
$\frac{\partial f}{\partial z} = \frac{\partial}{\partial z} [x^{2} + y^{2} + z^{2} - 3] = 2 z$

$⟹ \frac{\partial f}{\partial z} = {(2 z)}_{(1, 1, 1)} [Replace, p = (1, 1, 1)]$

$⟹ {(\frac{\partial f}{\partial z})}_{p} = 2 (1) = 2$

Equation of TANGENT PLANE is,

(x - x_{1}) {(\frac{\partial f}{\partial x})}_{p} + (y - y_{1}) {(\frac{\partial f}{\partial x})}_{p} + (z - z_{1}) {(\frac{\partial f}{\partial x})}_{p} = 0

⟹ (x - 1) (2) + (y - 1) (2) + (z - 1) (2) = 0

⟹ 2 x - 2 + 2 y - 2 + 2 z - 2 = 0

⟹ 2 x + 2 y + 2 z - 6 = 0

⟹ 2 x + 2 y + 2 z = 6

⟹ x + y + z = 3 [Cancelling by "2" from eqch term]

Equation of NORMAL LINE is,

\frac{x - x_{1}}{{(\frac{\partial f}{\partial x})}_{p}} = \frac{y - y_{1}}{{(\frac{\partial f}{\partial y})}_{p}} = \frac{z - z_{1}}{{(\frac{\partial f}{\partial z})}_{p}}

⟹ \frac{x - 1}{2} = \frac{y - 1}{2} = \frac{z - 1}{2} [Cancelling " 2 " from denomenator]

⟹ x - 1 = y - 1 = z - 1 [Adding " 1 " in all terms]

⟹ x = y = z