(i) A force of magnitude $5$ units acting parallel to $2 \to i - 2 \to j + \to k$ displaces the point of application from $(1, 2, 3)$ to $(5, 3, 7)$ . Find the work done by the force.
(ii) If $A (- 1, 4, - 3)$ is one end of a diameter $A B$ of the sphere $x^{2} + y^{2} + z^{2} - 3 x - 2 y + 2 z - 15 = 0$ , then find the coordinates of $B$

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Solution

(i) Let $\to r$ be the unit vector acting along the force $\to F$
Given, $\to r = 2 \to i + 2 \to j + \to k$
$| \to r | = \sqrt{2^{2} + (- 2)^{2} + 1^{2}} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$
$^r = \frac{\to r}{| \to r |} \frac{2 \to i - 2 \to j + \to k}{3}$
Force $\to F = 5^r = \frac{5}{3} (2 \to i - 2 \to j + 3 \to k)$
Also, Let $\to O A = \to i + 2 \to j + 3 \to k$ and $\to O B = 5 \to i + 3 \to j + 7 \to k$
Let $\to A B = \to O B - \to O A = (5 \to i + 3 \to j + 7 \to k) - (\to i + 2 \to j + 3 \to k)$
$\to d = 4 \to i + \to j + 4 \to k$
$∴$ work done $= \to F \cdot \to d = \frac{5}{3} (2 \to i + 2 \to j + \to k)$
$(4 \to i + \to j + 4 \to k)$
$= \frac{5}{3} [2 (4) + - 2 + 1 (4)]$
$= \frac{5}{3} [8 - 2 + 4] = \frac{50}{3}$ units.

(ii) The equation of the given sphere is
$x^{2} + y^{2} + z^{2} - 3 x - 2 y + 2 z - 15 = 0$
$2 u =$ Co-efficient of $x = - 3 \Rightarrow u = \frac{- 3}{2}$
$2 v =$ Co-efficient of $y = - 2 \Rightarrow v = \frac{- 2}{2} = - 1$
$2 w =$ Co-efficient of $z = 2 \Rightarrow w = \frac{2}{2} = 1$
$d = - 15$
$∴$ Centre is $(- u, - v, - w) = (\frac{3}{2}, 1, - 1)$
Given one end of the diameter $A$ is $(- 1, 4, 3)$ and the centre (mid point of the diameter $A B$ ) is $(\frac{- 3}{2}, 1, - 1)$
$∴$ Let the other end be $B (x, y, z)$
$(\frac{x - 1}{2}, \frac{y - 4}{2}, \frac{- 3}{2}) = (\frac{- 3}{2}, 1, - 1)$ (using mid point formula)
Equating the co-ordinates of $x, y$ and $z$ , we get
$\frac{x - 1}{2} = \frac{3}{2} \Rightarrow x = 3 + 1 \Rightarrow x = 4$
$\frac{y + 4}{2} = 1 \Rightarrow y + 4 = 2 \Rightarrow y = 2 - 4 = - 2$
$\frac{- 3}{2} = 1 \Rightarrow z - 3 = 2 \Rightarrow z = 2 + 3 = 5$
Other end of the diameter $A B$ is $(4, - 2, 1)$

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