hwsoln2

# hwsoln2 - CHAPTER 10 HOMEWORK SOLUTIONS Section 10.4 2 Find...

This preview shows pages 1–3. Sign up to view the full content.

CHAPTER 10 HOMEWORK SOLUTIONS Section 10.4 2. Find the length and direction of A × B and B × A . Where A = 2 i + 3 j and B = - i + j A × B = i j k 2 3 0 - 1 0 0 = 0 i +0 j +(2 - ( - 3)) k = 5 k length=5 and direction= k . B × A = - ( A × B ) length=5 and direction= - k . 8. Find the length and direction of A × B and B × A . Where A = 3 2 i - 1 2 j and B = i + j + 2 k A × B = i j k 3 2 - 1 2 1 1 1 0 = ( - 1 2 * 2 - 1 * 1) i - ( 3 2 * 2 - 1 * 1) j + ( 3 2 * 1 - ( - 1 2 * 1)) k = - 2 i - 2 j + 2 k length= 4 + 4 + 4 = 2 3 and direction= - 1 3 i - 1 3 j + 1 3 k . B × A = - ( A × B ) length=2 3 and direction= 1 3 i + 1 3 j - 1 3 k . 12. Sketch the vectors A = 2 i - j , B = i + 2 j , and A × B with all three vectors starting at the origin. A × B = i j k 2 - 1 0 1 2 0 = 5 k 0.0 0.5 1.0 1.5 2.0 x - 1 0 1 2 y 0 2 4 z 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 CHAPTER 10 HOMEWORK SOLUTIONS 16. a) Find the area of the triangle determined by the points P (1 , 1 , 1), Q (2 , 0 , - 1), and R (3 , - 1 , 1). Two vectors that connect the points: -→ PQ = (1 , 0 , 2) and -→ PR = (2 , - 2 , 0) The cross product -→ PQ × -→ PR has magnitude equal to the parallelogram created by -→ PQ and -→ PR which is twice the area of the triangle given by P , Q , and R . -→ PQ × -→ PR = i j k 1 0 2 2 - 2 0 = 4 i + 4 j - 2 k Area of triangle = 1 2 | -→ PQ × -→ PR | = 1 2 16 + 16 + 4 = 3 b)Find a unit vector perpendicular to the plane PQR . The unit vector in the directions of -→ PQ × -→ PR and - -→ PQ × -→ PR, namely ± ( 4 3 i + 4 3 j - 2 3 k ) 28. a)True: A · B = a 1 b 1 + a 2 b 2 + a 3 b 3 = b 1 a 1 + b 2 a 2 + b 3 a 3 = B · A b)True: A × B = i j k a 1 a 2 a 3 b 1 b 2 0 = - i j k b 1 b 2 b 3 a 1 a 2 a 3 = - B × A c)True: ( - A ) × B = i j k - a 1 - a 2 - a 3 b 1 b 2 b 3 = - i j k a 1 a 2 a 3 b 1 b 2 b 3 = - ( A × B ) d)True: ( c A ) · B = ( ca 1 ) b 1 + ( ca 2 ) b 2 + ( ca 3 ) b 3 = a 1 ( cb 1 ) + a 2 ( cb 2 ) + a 3 ( cb 3 ) = A · ( c B ) = ca 1 b 1 + ca 2 b 2 + ca 3 b 3 = c ( A · B ) e)True: c ( A × B ) = c i j k a 1 a 2 a 3 b 1 b 2 b 3 = i j k ca 1 ca 2 ca 3 b 1 b 2 b 3 = ( c A ) × B And: c ( A × B ) = c i j k a 1 a 2 a 3 b 1 b 2 b 3 = i j k a 1 a 2 a 3 cb 1 cb 2 cb 3 = A × ( c B )
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern