Find the area of the parallelogram determined by the vectors v and w where v=2i+3k and w=2j-3k. Area?
If a=i+1j+k and b=i+5j+k, find a unit vector with positive first coordinate orthogonal to both a and b. _i+_j+_k?
the area is |vxw|
recall that axb is perpendicular to both a and b
To find the area of the parallelogram determined by the vectors v and w, you can use the cross product of the two vectors. Here's how to do it step by step:
1. Start by finding the cross product of v and w. The cross product of two vectors is given by the formula:
v x w = |i j k|
|2 0 -3|
|0 2 3|
Calculate the determinant of the 3x3 matrix:
v x w = (2 * 3) - (0 * 3)i - [(2 * -3) - (0 * 3)]j + [(2 * 2) - (0 * 0)]k
= 6i + 6k
2. The magnitude of the cross product represents the area of the parallelogram, so calculate the magnitude:
|v x w| = sqrt((6)^2 + (0)^2 + (6)^2)
= sqrt(36 + 36)
= sqrt(72)
= 6√2
Therefore, the area of the parallelogram determined by the vectors v and w is 6√2.
For the second question:
To find a unit vector orthogonal to both a and b, you can use the cross product of the two vectors again. Here's how to do it step by step:
1. Calculate the cross product of vectors a and b. Using the same formula as before:
a x b = |i j k|
|1 1 1|
|1 5 1|
Calculate the determinant of the 3x3 matrix:
a x b = [(1 * 1) - (1 * 1)]i - [(1 * 1) - (1 * 1)]j + [(1 * 5) - (1 * 1)]k
= 0i + 0j + 4k
= 4k
The cross product result is 4k.
2. Since we want a unit vector, we need to normalize it. The magnitude of 4k is 4, so divide each component by 4:
(0/4)i + (0/4)j + (4/4)k
= 0i + 0j + k
Therefore, the unit vector with a positive first coordinate orthogonal to both a and b is _i + _j + _k.