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QUIZ 1 : MATH 251, Section 516
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Write up your result, detail your calculations if necessary and BOX your final answer
√
1. [25pts] Find the equation of the sphere with center (6, -1, 2) and radius 2 2.
2. [25pts] Find a unit vector which is orthogonal to both i + j and j + k.
3. [25pts] Find the cross product of h3, 1, −1i and h−6, −2, 2i.
4. [25pts] Find parametric equations and symmetric equations for the line passing through A (3, 1, −1)
and B (3, 2, −6).
Answers :
1. The equation of the sphere with center (0, 1, −1) and radius
√
2 is :
√
(x − 6)2 + (y − (−1))2 + (z − 2)2 = (2 2)2 ,
so,
(x − 6)2 + (y + 1)2 + (z − 2)2 = 8.
2. There are 2 ways to obtain the result :
The shortest one : We saw that the cross product of 2 vectors is an orthogonal vector to both of them.
We have ~i + ~j = h1, 1, 0i and ~j + ~k = h0, 1, 1i.
* 1 + * 0 + 1 1 1 0 1 0
= h1, −1, 1i.
1 × 1
=
+ (−1) +
0 1
0 1 1 1 1
0
√
1
1 1
√ , −√ , √
.
The length of this vector is 3. So, the wanted unit vector is
3
3 3
Another Way : Let hx, y, zi be a vector which is orthogonal to both previous vectors, it should verify
the following system :
hx, y, zi • h1, 1, 0i = 0
x+y =0
x = −y
⇔
⇔
hx, y, zi • h0, 1, 1i = 0
y+z =0
z = −y
so, hx, y, zi = h−1, 1, −1i.y. The vector h−1, 1, −1i is orthogonal to both vectors. To obtain a unit
vector, we divide by its length, the wanted vector is
h−1, 1, −1i
−1 1 −1
p
= √ ,√ ,√
.
3 3 3
(−1)2 + 12 + (−1)2
3. We observe that h−6, −2, 2i = −2.h3, 1, −1i, so these vectors are collinear and its cross product is zero.
~ = h3 − 3, 2 − 1, −6 − (−1)i = h0, 1, −5i. Let t ∈ R and (x, y, z)
4. The line is directed by the vector AB
a point of the line, parametric equations of the line are :
* x + * 3
+ * 0 +
* 3
+
y
z
=
2
(−6)
+
1
−5
.t =
2+t
−6 − 5t
Parametric equations are : hx, y, zi = h3, 2 + t, −6 − 5ti.
Symmetric equations are obtained from the previous equations :
t=y−1=
−1 − z
,
5
x = 3.
so, symmetric equations are
5y + z = 4,
x = 3.