First slide

Transformation of axes

Question

Consider triangle AOB ln, the x-y plane ,where A $\equiv$ (1,0,0),,B $\equiv$ (0,2,0), and O $\equiv$ (0,0,0).The new position of O, when triangle is rotated about side AB by 90^o can be

Moderate

A

$(\frac{4}{5}, \frac{3}{5}, \frac{2}{\sqrt{5}})$
B

$(\frac{- 3}{5}, \frac{\sqrt{2}}{5}, \frac{2}{\sqrt{5}})$
C

$(\frac{4}{5}, \frac{2}{5}, \frac{2}{\sqrt{5}})$
D

$(\frac{4}{5}, \frac{2}{5}, \frac{1}{\sqrt{5}})$

Solution

Equation of line AB is $\frac{x - 1}{1} = \frac{y}{- 2} = \frac{z}{0} = λ$
Now
$\begin{array}{l} AB ⊥ OC \Rightarrow 1 (λ + 1) + (- 2 λ) (- 2) = 0 \\ 5 λ = - 1 or λ = - \frac{1}{5} \\ C is (\frac{4}{5}, \frac{2}{5}, 0) . Now \\ x_{1}^{2} + {(y_{1} - 2)}^{2} + z_{1}^{2} = 4 - - - - (i) \\ {(x_{1} - 1)}^{2} + y_{1}^{2} + z_{1}^{2} = 1 - - - - - (ii) \end{array}$
$Now O C ⊥ C D$
$\Rightarrow (x_{1} - \frac{4}{5}) \frac{4}{5} + (y_{1} - \frac{2}{5}) \frac{2}{5} + (z_{1} - 0) 0 = 0 - - - - (iii)$
From (i) and (ii), we get
From (iii), putting $x_{1} = 2 y_{1} \Rightarrow 2 y_{1} = \frac{4}{5} or y_{1} = \frac{2}{5}$
$\Rightarrow x_{1} = \frac{4}{5}$ Putting this value of x₁, and y₁ in (i),
we get
$z_{1} = \pm \frac{2}{\sqrt{5}}$

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