Find the intersection of the line x=y−12=z−23 and the shere x2+(y−3)2+(z−2)2=4 a) (1, 1, 0) b) (1, 2, 3) c) (0, 2, 3) d) (16, 236, 176) e) (47, 157, 267)

Question

Find the intersection of the line x=y−12=z−23 and the shere x2+(y−3)2+(z−2)2=4  a) (1, 1, 0)  b) (1, 2, 3)  c) (0, 2, 3)  d) (16, 236, 176)  e) (47, 157, 267)

boronganfh · Accepted Answer

Step 1  In analytic geometry, a line and a sphere can intersect in three ways:  No intersection at all  Intersection in exactly one point  Intersection in two points  Consider a line given by,  x−x1x2−x1=y−y1y2−y1=z−z1z2−z1  and a shere with center (x0, y0, z0) and radius r given by,  (x−x0)2+(y−y0)2+(z−z0)2=r2  We parametrize the given line as,  x−x1x2−x1=y−y1y2−y1=z−z1z2−z1=s(say)  ⇒x=x1+s(x2−x1), y=y1+s(y2−y1), z=z1+s(z2−z1)  Now, substituting these values in the equation of the sphere, we get a quadratic equation. Based on the solution of this quadratic equation, we have,  If real number solution doesnt

Pansdorfp6 · Accepted Answer

Given  x=y−12=z−23  Spher  x2+(y−3)2+(z−2)2=4  Ary Pont on the line is given by  x=y−12=z−23=t(≤t) t∈R  So,  x=t, y=2t+1, z=3t+2  Hence P(t, 2t+1, 3t+2) is Parametric Point on the line  Now we find value of t such thet point P is on the Shere, so it satisfy the sphere eqn  Hence,  (t)2+(2t+1−3)2+(3t+2−2)2=4  t2+4t2+4−8t+9t2=4  14t2−8t=0⇒t(14t−8)=0  t=0S is t=814=47  Hence the Intersection Points Are  For t=0; P(0, 1, 2)  For t=47; P(47, 157, 267)  So, option E is correct

nick1337 · Accepted Answer

Step 1 For the given line, let: x=y−12=z−23=t Hence we get: x = t -(1) y−12=t y-1=2t y=1+2t-(2) z−23=t  z-2=3t z=2+3t z=2+3t−(3) Substituting the values of x,y and z from equation (1), (2) and (3) respectively into the equation of sphere: t2+(1+2t−3)2+(2t+3t−2)2=4 t2+(2t−2)2+(3t)2=4 t2+4t2+4−8t+9t2=4 14t2−8t+4−4=0  14t2−8t=0 t(14t-8)=0 t=0 or 14t-8=0 t=0 or t=814  t=0 or t=47 1) For t=0 From equation (1),(2) and (3): x=0 y=1+2×0=1 z=2+3×0=2 The point of intersection is (0,1,2) 2) For t=47 From equation (1),(2) and (3): x=47 y=1+2×47 =1+87 =157 z=2+3×47 =2+127 =267 The point of intersection is (47, 157, 267) Answer: Option E

Find the intersection of the line x=\frac{y-1}{2}=\frac{z-2}{3} and the shere

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