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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) × (i + k) = h1, −1, 0i × 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) × (i − j + k)) · 3k = (h1, 1, 1i × h1, −1, 1i) · h0, 0, 3i = h2, 0, −2i · 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 × 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