Find the vector projection of the vector 6i+3j+2k onto thevector 3i-2j+4k.
The projection of u onto v is u•v * v/|v|
So in this case, that would be
(6*3 + 3(-2) + 2*4)/√(3^2+2^2+4^2) = 20/√29 * 3i-2j+4k
To find the vector projection of vector A onto vector B, you can use the formula:
Vector projection of A onto B = (A • B / |B|²) * B
1. Calculate the dot product of vectors A and B:
A • B = (6 * 3) + (3 * -2) + (2 * 4) = 18 - 6 + 8 = 20
2. Calculate the magnitude of vector B:
|B| = sqrt((3²) + (-2²) + (4²)) = sqrt(9 + 4 + 16) = sqrt(29)
3. Substitute the values into the formula:
Vector projection of A onto B = (20 / (sqrt(29))²) * (3i - 2j + 4k)
Simplify:
Vector projection of A onto B = (20 / 29) * (3i - 2j + 4k)
Therefore, the vector projection of vector A (6i + 3j + 2k) onto vector B (3i - 2j + 4k) is:
(20 / 29) * (3i - 2j + 4k)
To find the vector projection of a vector onto another vector, we can use the formula:
Projection of vector A onto vector B = (dot product of A and B) / ||B||^2 * vector B
Let's calculate step by step:
1. Calculate the dot product of the given vectors 6i+3j+2k and 3i-2j+4k:
dot product = (6 * 3) + (3 * -2) + (2 * 4)
= 18 - 6 + 8
= 20
2. Calculate the magnitude (or length) of the vector 3i-2j+4k:
magnitude = sqrt((3^2) + (-2^2) + (4^2))
= sqrt(9 + 4 + 16)
= sqrt(29)
3. Calculate the squared magnitude of vector B:
squared magnitude = ||B||^2 = (sqrt(29))^2
= 29
4. Calculate the vector projection by plugging the values into the formula:
Projection of vector A onto vector B = (dot product of A and B) / ||B||^2 * vector B
= (20 / 29) * (3i-2j+4k)
So, the vector projection of the vector 6i+3j+2k onto the vector 3i-2j+4k is (20/29) * (3i-2j+4k).