Show that vector A=_6i^+9j^_12k^ and vector B=2i^_3j^+4k^ are parellel to each other
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Sure, let's put on our clown shoes and show it in a fun and humorous way!
To determine if two vectors are parallel, we can compare their components and see if there is a constant ratio between them. If there is, they're like two clowns wearing the same costume, parallel to each other!
For vector A (_6i^ + 9j^ - 12k^) and vector B (2i^ - 3j^ + 4k^), let's compare their respective components:
The x-component of vector A is -6, and the x-component of vector B is 2. Now, let's divide them: -6/2 = -3.
You know what, this is like having two clowns with different shoe sizes trying to match each other's shoes. It's just not going to work!
Now let's compare the y-components: 9/(-3) = -3. Hey, that's the same ratio as before! Our clowns might have mismatched shoes, but they're definitely parallel to each other in terms of their y-components.
Lastly, let's compare the z-components: (-12)/4 = -3. Would you look at that, another -3! These clowns just can't help being parallel to each other in the z-direction too!
So, while vector A and vector B may not have the same shoe size (I mean, component) in every direction, they are indeed parallel to each other, at least in the y and z directions! Quite the pair of parallel clowns, don't you think?
To show that two vectors A and B are parallel, we need to check if their direction vectors are proportional.
Let's find the direction vector of vector A. The direction vector is obtained by dividing each component of the vector by its magnitude.
Magnitude of vector A (|A|) is calculated using the formula:
|A| = √(A_x² + A_y² + A_z²),
where A_x, A_y, and A_z are the components of vector A in the x, y, and z directions, respectively.
Given vector A = 6i^ + 9j^ - 12k^, we can calculate the magnitude as follows:
|A| = √(6² + 9² + (-12)²)
= √(36 + 81 + 144)
= √(261)
≈ 16.155
Now, let's calculate the direction vector of A:
A_direction = (A_x/|A|)i^ + (A_y/|A|)j^ + (A_k/|A|)k^
= (6/16.155)i^ + (9/16.155)j^ + (-12/16.155)k^
≈ 0.371i^ + 0.557j^ - 0.743k^
Similarly, let's find the direction vector of vector B:
Magnitude of vector B (|B|) can be calculated in the same way:
|B| = √(2² + (-3)² + 4²)
= √(4 + 9 + 16)
= √(29)
≈ 5.385
Then, the direction vector of B is:
B_direction = (B_x/|B|)i^ + (B_y/|B|)j^ + (B_k/|B|)k^
= (2/5.385)i^ + (-3/5.385)j^ + (4/5.385)k^
≈ 0.371i^ - 0.557j^ + 0.742k^
Comparing the direction vectors of A and B, we can see that the components along each direction (i, j, and k) are proportional. Therefore, vector A and vector B are parallel to each other.