Section 8‐4
Perpendicular Vectors
Perpendicular Vectors
y In physics, torque is the measure of the effectiveness of a force in turning an object around a pivot point. We can use perpendicular vectors to find the torque of a force.
Inner Product of vectors
y a and b are two vectors a1,a2 and b1,b2
The inner product of a and b is defined as a ∙ b = a1b1 +a2b2
a ∙ b is read “a dot b” and is often called the dot product. Two vectors are perpendicular if and only if their inner product is zero. Just as in a plane, two vectors in three dimensional space are perpendicular if and only if their inner product a1b1+a2b2+a3b3 = 0
Example # 1
y Find each inner product if x = 2,‐5 y = 4, 1 and z = 10, 4 . Are any pairs of vectors perpendicular?
y Solution: x ∙y= 2(4)+‐5(1) =3
x ∙z=2(10)+‐5(4)=0 perpendicular
y ∙z=4(10)+1(4)=44
Example # 2
y Find the inner product of v and w if v = 2, ‐3, ‐4 and w = 8, 3, 2 . Are they perpendicular?
y Solution: y v ∙ w= 2(8)+(‐3)(3)+(‐4)(2)= ‐1 Not perpendicular
Cross Product
y Another important product involving vectors in space is the cross product. Unlike the inner product, the cross product of two vectors is a vector. The cross product of a and b is written a x b. The vector does not lie in the plane of the given vectors, but is perpendicular to the plane containing the two vectors. In other words, the cross product of two vectors is perpendicular to both vectors!
Cross Product
y If a = a1,a2,a3 and b = b1,b2,b3 then the cross product of a x b = a2 a3 a1 a3 a1 a2
b2 b3 i ‐ b1 b3 j + b1 b2 k
An easy way to remember the coefficients of i, j, k is to write it as a determinant i j k
a1 a2 a3
b1 b2 b3
Example # 3
y Find the cross product of a and b if a = ‐4,1,0 and
b = 5, 4,‐2 . Verify that the resulting vector is perpendicular to a and b.
Solution
a x b = 1 0 ‐4 0 ‐4 1 = ‐2i ‐ 8j ‐21k or ‐2,‐8, ‐21
4 ‐2 i ‐
5 ‐2 j + 5 4 k
Since the inner products are zero, the cross product a x b is perpendicular to both a and b.
Inner products ‐2, ‐8, ‐21 ∙
‐4, 1, 0 and ‐2, ‐8, ‐21 ∙ 5, 4,‐2 ‐2(‐4)+‐8(1)+‐21(0) = 0 and ‐2(5)+‐8(4)+‐21(‐2)=0 Relationship in Physics
y In Physics, the Torque T about a point A created by a force F at a point B is given by T = AB x F. The magnitude of T represents the torque in foot‐pounds.
The Monster Truck
A monster truck driver is applying a force of 28
pounds along the positive z axis to the gearshift
of her truck. If the center of the gearshift
connection is at the point x, the force is applied
at the point 0.70,0.31 . Find the torque.
We need to find |T|, the torque at the force 0.70, 0.31 where each value is the distance from the origin in feet and F represents the force in pounds.
AB= .70, 0, .31 ‐ 0,0,o = .70, 0, .31 Any upward force is measured along the z axis so F = 28k or 0, 0, 28 . Now T = AB x F = 0 .31 .70 .31 .70 0 = 0i ‐ 19.6 j + 0k or 0, ‐19.6, 0
0 28 i ‐
0 28 j + 0 0 k
The magnitude of T = = 19.6 ft‐lb
384.16
HW # 23
ySection 8-4
yPg 509 #11-25 odd and
yWS J: Vectors in Space