Question

Answer this Question of iit jee . Chapter → Three dimensional Geometry​

Answer 1

Answer:

I can't understand

Step-by-step explanation:

byyyyyy

Answer 2

Answer:

Let the radius of the circle be 'r' and radius of sphere be 'R'.

The plane x+2y-z=4 cuts the sphere x²+y²+z²-x+z-2 = 0 in a circle which is at a distance 'd'.

→ R²= d²+r²

→ r² = R²-d². -------(1)

Now,

Eqn of sphere : x²+y²+z²-x+z-2 = 0

converting it in standard form we will get :-

${(x - \frac{1}{2} )}^{2} + {(y - 0)}^{2} + {(z + \frac{1}{2} })^{2} = \frac{5}{2}$

$where, \: ( \frac{1}{2} ,0, - \frac{1}{2} ) \: is \: the \: centre \: of \: the \: sphere.$

$and \: radius, \: R = \sqrt{ \frac{5}{2} }$

Also,,

we have to find 'd' which is the shortest perpendicular distance from the plane x+2y-z=4 to the centre (1/2,0,-1/2)

$→d = | \frac{ \frac{1}{2} + 2 \times 0 -( \frac{ - 1}{2} ) - 4}{ \sqrt{ {1}^{2} + {2}^{2} + { (- 1)}^{2} } } |$

$→d = | \frac{1 - 4}{ \sqrt{6} } |$

$→d = \frac{3}{ \sqrt{6} }$

from eqn(1),,

→r² = R²-d²

$→ {r}^{2} = ({ \sqrt{ \frac{5}{2} } } \: \: )^{2} - ( { \frac{3}{ \sqrt{6} } })^{2}$

$→ {r}^{2} = \frac{5}{2} - \frac{9}{6}$

$→ {r}^{2} = \frac{5}{2} - \frac{3}{2}$

$→ {r}^{2} = \frac{5 - 3}{2}$

$→ {r}^{2} = \frac{2}{2}$

$→ {r}^{2} = 1$

∴r = 1

Hence, the radius of the circle is 1.