# MATH 1260 - Quiz 2 Solution

```MATH 1260 - Quiz 2
Solution
(1) Find a unit vector perpendicular to i-j and to i+k and with a positive k component.
The vector
(i − j) &times; (i + k) = h1, −1, 0i &times; h1, 0, 1i = h−1, −1, 1i
is perpendicular to both i-j and i+k. To make it a unit vector, we divide by its length
obtain
1
1 1
h− √ , − √ , √ i.
3
3 3
√
3 to
(2) Find the volume of the parallelepiped spanned by the vectors i+j+k, i-j+k, and 3k.
The volume is the absolute value of the triple product
((i + j + k) &times; (i − j + k)) &middot; 3k = (h1, 1, 1i &times; h1, −1, 1i) &middot; h0, 0, 3i = h2, 0, −2i &middot; h0, 0, 3i = −6,
and therefore the volume is 6.
(3) Find the equation of the plane through the point (2, 1, 1) and containing the line x = t − 1,
y = 2t + 1, z = −t − 1.
Let us find two vectors in the plane. Since the plane contains the line x = t − 1, y = 2t + 1,
z = −t − 1, it also contains the vector of the direction of the line, which is h1, 2, −1i. By taking a
point on the line, say (−1, 1, −1), and taking its difference with (2, 1, 1), we obtain another vector
on the plane: h3, 0, 2i.
The cross product of these two vectors gives us the normal to the plane:
h1, 2, −1i &times; h3, 0, 2i = h4, −5, −6i.
So the equation of the plane is
4x − 5y − 6z = d.
To find d, we plug in the point (2, 1, 1): (4)(2) − (5)(1) − (6)(1) = −3.
The equation of the plane is
4x − 5y − 6z = −3.
1
```