QJEEM20Ma3b69: JEE Main 2020 Mathematics - Class 12 JEEM20Ma3b69 Solution: Let A = {X = (x, y, z) T : PX = 0 and x 2 + y 2 + z 2 = 1}…

Let $$X=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$$

PX = O

$$\left[\begin{array}{ccc}1& 2& 1\\ -2& 3& -4\\ 1& 9& -1\end{array}\right]\left[\begin{array}{c}x\\ y\\ z\end{array}\right]=\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]$$

x + 2y + z = 0........(1)

-2x + 3y – 4z = 0....(2)

x + 9y - z = 0..........(3)

from (1) & (3)
$\Rightarrow$ 2x+11y =0

from (1) & (2)
$\Rightarrow$ 2x + 11y = 0

from (2) & (3)
–6x –33y = 0
$\Rightarrow$ 2x +11y = 0

putting value of x in (1), we get
–7y + 2z = 0

Now ${\left( {{{11y} \over 2}} \right)^2} + {y^2} + {\left( {{{7y} \over 2}} \right)^2} = 1$

y²(121 + 1 + 49) = 4

y²(171) = 4

$y = \pm {2 \over {\sqrt {171} }}$

$\Rightarrow x = \pm {7 \over {\sqrt {171} }}$

$\Rightarrow z = \pm {{11} \over {\sqrt {171} }}$

$\therefore$ So, there are 2 solution set of (x, y, z)