Find a vector of magnitude 18 which is perpendicular to both the vectors 4i^-j^-3^ and -2i^+j^-2k^.
To find a vector that is perpendicular to both the given vectors, we can take their cross product.
Let A = 4i - j - 3k and B = -2i + j - 2k.
A x B = (Ai, Aj, Ak) x (Bi, Bj, Bk) = (Aj * Bk - Ak * Bj, Ak * Bi - Ai * Bk, Ai * Bj - Aj * Bi) = ((-1 * -2) - (-3 * 1), (-3 * -2) - (4 * -2), (4 * 1) - (-1 * -2)) = (2 - (-3), 6 - (-8), 4 - 2)
A x B = (5, 14, 2)
Now, we need to find a vector with magnitude 18 in the direction of the cross product.
First, let's find the magnitude of the cross product.
|A x B| = sqrt(5² + 14² + 2²) = sqrt(25 + 196 + 4) = sqrt(225)
|A x B| = 15
Now, let's find the unit vector in the direction of the cross product.
u = (5/15, 14/15, 2/15)
Finally, let's scale the unit vector to have a magnitude of 18.
v = 18 * u = (18 * 5/15, 18 * 14/15, 18 * 2/15) = (6, 16.8, 2.4)
So the vector with magnitude 18 that is perpendicular to both the given vectors is 6i + 16.8j + 2.4k.
To find a vector that is perpendicular to two given vectors, you can take their cross product. The cross product of two vectors will give you a vector that is perpendicular to both of them.
Given the two vectors:
v1 = 4i^ - j^ - 3k^
v2 = -2i^ + j^ - 2k^
Step 1: Take the cross product of v1 and v2.
To do this, use the following formula:
v1 × v2 = (a1b2 - a2b1)i^ + (a2b0 - a0b2)j^ + (a0b1 - a1b0)k^
where v1 = a1i^ + a2j^ + a3k^ and v2 = b1i^ + b2j^ + b3k^
Let's plug in the values:
v1 × v2 = [(4)(1) - (-1)(-2)]i^ + [(-3)(1) - (4)(-2)]j^ + [(4)(-2) - (-3)(-2)]k^
= (10)i^ - (-5)j^ + (2)k^
= 10i^ + 5j^ + 2k^
Step 2: Normalize the resulting vector.
To find a vector with a magnitude of 1, we need to normalize the vector.
The magnitude of the vector 10i^ + 5j^ + 2k^ can be found using the formula:
|v| = sqrt((v1)^2 + (v2)^2 + (v3)^2)
|10i^ + 5j^ + 2k^| = sqrt((10)^2 + (5)^2 + (2)^2)
= sqrt(100 + 25 + 4)
= sqrt(129)
Step 3: Divide the vector by its magnitude to normalize it.
To normalize the vector, divide each component of the vector by its magnitude:
normalized_vector = vector / |vector|
normalized_vector = (10i^ + 5j^ + 2k^) / sqrt(129)
So, the vector of magnitude 1 that is perpendicular to both v1 and v2 is:
(10i^ + 5j^ + 2k^) / sqrt(129)
To find a vector of magnitude 18, simply multiply the normalized vector by 18:
(10i^ + 5j^ + 2k^) / sqrt(129) * 18
Thus, the vector of magnitude 18 that is perpendicular to both v1 and v2 is:
18(10i^ + 5j^ + 2k^) / sqrt(129)
To find a vector that is perpendicular to both given vectors, we can calculate their cross product. The cross product of two vectors will give us a vector that is perpendicular to both of them.
Given vectors:
Vector A = 4i^-j^-3^
Vector B = -2i^+j^-2k^
Step 1: Calculate the cross product of the two given vectors.
To calculate the cross product, we can use the determinant of the following matrix:
| i^ j^ k^ |
| 4 -1 -3 |
| -2 1 -2 |
i-component: (-1 * -2) - (1 * -3) = -2 + 3 = 1
j-component: (4 * -2) - (1 * -2) = -8 + 2 = -6
k-component: (4 * 1) - (-1 * -2) = 4 + 2 = 6
So, the cross product of Vector A and Vector B is: Vector C = 1i^-6j^+6k^.
Step 2: Normalize the cross product vector to get a vector of magnitude 1.
To find a vector of magnitude 18, we need to scale the cross product vector by the desired magnitude. Divide each component of the vector by its magnitude and then multiply by the desired magnitude.
Magnitude of Vector C = sqrt(1^2 + (-6)^2 + 6^2) = sqrt(1 + 36 + 36) = sqrt(73)
Scaling factor = 18 / sqrt(73)
Scaled vector = (1/sqrt(73)) * (1i^-6j^+6k^)
So, the vector of magnitude 18 that is perpendicular to Vector A and Vector B is:
(18/sqrt(73)) * (1i^-6j^+6k^)