Equation of line tangent to sphere from a point
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Sphere B with center (3,2,1) and radius 3.
Show that P = (1,0,2) is a point on the sphere.
Find the cartesian equation of the plane tangent to the sphere through P.
Show that line
l
l
given by
x=(0,3,0)+λ(−3,2,4)
x=(0,3,0)+λ(−3,2,4)
has no point of intersection with B.
Find the cartesian equations of each plane through
l
l
and tangent to the sphere.
What I have now (abbreviated):
1) equation for B is
|x−(3,2,1)|≤3
|x−(3,2,1)|≤3
, for x = (1,0,2) we get
|(−2,−2,1)|≤3⇒
4+4+1
−
−
−
−
−
−
−
√
≤3
|(−2,−2,1)|≤3⇒4+4+1≤3
. So P is on the surface of the sphere.
2) vector from P to the center of B is
P
→
−(3,2,1)=(−2,−2,1)
P→−(3,2,1)=(−2,−2,1)
, which is the normal vector of the plane we want. So the plane is
−2
x
1
−2
x
2
+
x
3
=a
−2x1−2x2+x3=a
. We find
a
a
by filling in the values for P: -2*1 - 2*0 + 2 = 0. (answer section of the booklet gives
−2
x
1
−2
x
2
+
x
3
=1
−2x1−2x2+x3=1
but I am convinced that is wrong... hope you people agree)
3) say the center of B is P, the point on
l
l
that intersects with the line perpedicular to
l
l
through P is Q. Q is on
l
l
so the vector to Q is
⎛
⎝
⎜
−3λ
3+2λ
4λ
⎞
⎠
⎟
(−3λ3+2λ4λ)
for a certain value of
λ
λ
. Vector
PQ
−
→
−
PQ→
is Q - P is
⎛
⎝
⎜
−3λ−3
2λ+1
4λ−1
⎞
⎠
⎟
(−3λ−32λ+14λ−1)
. Perpedicular to
l
l
so the dot product of
PQ
−
→
−
PQ→
and (-3, 2, 4) is zero. Working that out gives
λ=
−7
29
λ=−729
. So Q is
⎛
⎝
⎜
⎜
21
/
29
73
/
29
−28
/
29
⎞
⎠
⎟
⎟
(21/2973/29−28/29)
. The length of
PQ
−
→
−
PQ→
is
(
66
/
29
)
2
+
(
15
/
29
)
2
+
(
−57
/
29
)
2
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
√
(66/29)2+(15/29)2+(−57/29)2
which is larger than 3.
4) What I could think of: if I can find the lines through Q, tangent to B, perpendicular to
l
l
, the equation for each tangent plane can be calculated from the vector representation of
l
l
by adding an extra direction vector, the direction vector from each tangent line. I call the direction vector
m
m
here. The vector equation for the tangent lines is (with each a different
m
m
)
x=
Q
→
+λm
x=Q→+λm
These tangent lines (I believe there are two) go through a point on sphere B. That point thus adheres to | x - (3,2,1) | = 3. That intersection point is on the tangent line, so
∣
∣
∣
∣
∣
⎛
⎝
⎜
⎜
21
/
29
73
/
29
−28
/
29
⎞
⎠
⎟
⎟
+λ
⎛
⎝
⎜
m
1
m
2
m
3
⎞
⎠
⎟
−
⎛
⎝
⎜
3
2
1
⎞
⎠
⎟
∣
∣
∣
∣
∣
=3
⇒
∣
∣
∣
∣
∣
⎛
⎝
⎜
⎜
−66
/
29
15
/
29
−57
/
29
⎞
⎠
⎟
⎟
+λ
⎛
⎝
⎜
m
1
m
2
m
3
⎞
⎠
⎟
∣
∣
∣
∣
∣
=3
|(21/2973/29−28/29)+λ(m1m2m3)−(321)|=3⇒|(−66/2915/29−57/29)+λ(m1m2m3)|=3
⇒
(
−66
/
29
+λ
m
1
)
2
+
(
15
/
29
+λ
m
2
)
2
+
(
−57
/
29
+λ
m
3
)
2
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
⎷
=3
⇒(−66/29+λm1)2+(15/29+λm2)2+(−57/29+λm3)2=3
The lines are perpendicular to
l
l
, so the dot-product of
m
m
and the direction vector of
l
l
is zero. That gives
−3
m
1
+2
m
2
+4
m
3
=0⇒
m
1
=
2
m
2
+4
m
3
3
−3m1+2m2+4m3=0⇒m1=2m2+4m33
.
I can then substitute
m
1
m1
in the equation of the last paragraph, but that leaves me with three unknowns:
λ
λ
,
m
2
m2
and
m
3
m3
. I thought I could assume
λ
λ
to be 1 in the intersection point on B, as any change in
λ
λ
will only make me find some scalar multiple of
m
m
. (and
λ
λ
is zero in Q already)
Still that ends up with an equation far more complicated than in any previous exercise, and with that fraction with the very impractical denominator 29 I suspect I am both following the wrong approach, and I can have made some calculation error.
The booklet gives me the answer (
4
x
1
+3
x
3
=0
4x1+3x3=0
and
2
x
1
−
x
2
+2
x
3
=−3
2x1−x2+2x3=−3
) but I cannot find a clue in that. Only that probably there should not be such a complicated fraction in my calculations :)
Show that P = (1,0,2) is a point on the sphere.
Find the cartesian equation of the plane tangent to the sphere through P.
Show that line
l
l
given by
x=(0,3,0)+λ(−3,2,4)
x=(0,3,0)+λ(−3,2,4)
has no point of intersection with B.
Find the cartesian equations of each plane through
l
l
and tangent to the sphere.
What I have now (abbreviated):
1) equation for B is
|x−(3,2,1)|≤3
|x−(3,2,1)|≤3
, for x = (1,0,2) we get
|(−2,−2,1)|≤3⇒
4+4+1
−
−
−
−
−
−
−
√
≤3
|(−2,−2,1)|≤3⇒4+4+1≤3
. So P is on the surface of the sphere.
2) vector from P to the center of B is
P
→
−(3,2,1)=(−2,−2,1)
P→−(3,2,1)=(−2,−2,1)
, which is the normal vector of the plane we want. So the plane is
−2
x
1
−2
x
2
+
x
3
=a
−2x1−2x2+x3=a
. We find
a
a
by filling in the values for P: -2*1 - 2*0 + 2 = 0. (answer section of the booklet gives
−2
x
1
−2
x
2
+
x
3
=1
−2x1−2x2+x3=1
but I am convinced that is wrong... hope you people agree)
3) say the center of B is P, the point on
l
l
that intersects with the line perpedicular to
l
l
through P is Q. Q is on
l
l
so the vector to Q is
⎛
⎝
⎜
−3λ
3+2λ
4λ
⎞
⎠
⎟
(−3λ3+2λ4λ)
for a certain value of
λ
λ
. Vector
PQ
−
→
−
PQ→
is Q - P is
⎛
⎝
⎜
−3λ−3
2λ+1
4λ−1
⎞
⎠
⎟
(−3λ−32λ+14λ−1)
. Perpedicular to
l
l
so the dot product of
PQ
−
→
−
PQ→
and (-3, 2, 4) is zero. Working that out gives
λ=
−7
29
λ=−729
. So Q is
⎛
⎝
⎜
⎜
21
/
29
73
/
29
−28
/
29
⎞
⎠
⎟
⎟
(21/2973/29−28/29)
. The length of
PQ
−
→
−
PQ→
is
(
66
/
29
)
2
+
(
15
/
29
)
2
+
(
−57
/
29
)
2
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
√
(66/29)2+(15/29)2+(−57/29)2
which is larger than 3.
4) What I could think of: if I can find the lines through Q, tangent to B, perpendicular to
l
l
, the equation for each tangent plane can be calculated from the vector representation of
l
l
by adding an extra direction vector, the direction vector from each tangent line. I call the direction vector
m
m
here. The vector equation for the tangent lines is (with each a different
m
m
)
x=
Q
→
+λm
x=Q→+λm
These tangent lines (I believe there are two) go through a point on sphere B. That point thus adheres to | x - (3,2,1) | = 3. That intersection point is on the tangent line, so
∣
∣
∣
∣
∣
⎛
⎝
⎜
⎜
21
/
29
73
/
29
−28
/
29
⎞
⎠
⎟
⎟
+λ
⎛
⎝
⎜
m
1
m
2
m
3
⎞
⎠
⎟
−
⎛
⎝
⎜
3
2
1
⎞
⎠
⎟
∣
∣
∣
∣
∣
=3
⇒
∣
∣
∣
∣
∣
⎛
⎝
⎜
⎜
−66
/
29
15
/
29
−57
/
29
⎞
⎠
⎟
⎟
+λ
⎛
⎝
⎜
m
1
m
2
m
3
⎞
⎠
⎟
∣
∣
∣
∣
∣
=3
|(21/2973/29−28/29)+λ(m1m2m3)−(321)|=3⇒|(−66/2915/29−57/29)+λ(m1m2m3)|=3
⇒
(
−66
/
29
+λ
m
1
)
2
+
(
15
/
29
+λ
m
2
)
2
+
(
−57
/
29
+λ
m
3
)
2
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
−
⎷
=3
⇒(−66/29+λm1)2+(15/29+λm2)2+(−57/29+λm3)2=3
The lines are perpendicular to
l
l
, so the dot-product of
m
m
and the direction vector of
l
l
is zero. That gives
−3
m
1
+2
m
2
+4
m
3
=0⇒
m
1
=
2
m
2
+4
m
3
3
−3m1+2m2+4m3=0⇒m1=2m2+4m33
.
I can then substitute
m
1
m1
in the equation of the last paragraph, but that leaves me with three unknowns:
λ
λ
,
m
2
m2
and
m
3
m3
. I thought I could assume
λ
λ
to be 1 in the intersection point on B, as any change in
λ
λ
will only make me find some scalar multiple of
m
m
. (and
λ
λ
is zero in Q already)
Still that ends up with an equation far more complicated than in any previous exercise, and with that fraction with the very impractical denominator 29 I suspect I am both following the wrong approach, and I can have made some calculation error.
The booklet gives me the answer (
4
x
1
+3
x
3
=0
4x1+3x3=0
and
2
x
1
−
x
2
+2
x
3
=−3
2x1−x2+2x3=−3
) but I cannot find a clue in that. Only that probably there should not be such a complicated fraction in my calculations :)
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