A= 2i + 3j - k, B= 4i + 2j - 2k, Find a vector x parallel to A but has magnitude of B ?
Answer is √12/7 (2i+3j) - K
solve it , i have mcat admission test please solve it
Magnitude of a vector is its length:
if v = ai + bj + ck
|v| = sqrt(a^2 + b^2 + c^2)
So that's the length. It's a scalar quantity.
|b| = sqrt(16+4+4) = sqrt(24)
|v| = |b|
Parallel vectors: the dot product is 0.
so. A.v = 0
or 2a + 3b - 1c = 0
The thing is there isn't just one answer. I'm guessing you set c to be -1 or something to get a -1k in your answer.
After that you have two equations with a and b left. (one with a and b the other with a^2 and b^2. Which has 2 answers).
It feels like there should be 1 more piece of information in the question to get a unique answer though.
Thanks sir, I understand that information is lacking but I don't know why they have given such high level question for MCAT admission test :/ , anyway thanks
To find a vector parallel to vector A but with the magnitude of vector B, you can use the following steps:
Step 1: Calculate the magnitude of vector B.
The magnitude of vector B is given by the formula |B| = √(Bx² + By² + Bz²). From the given information, B = 4i + 2j - 2k. So, Bx = 4, By = 2, and Bz = -2. Plugging these values into the magnitude formula, we get:
|B| = √(4² + 2² + (-2)²)
|B| = √(16 + 4 + 4)
|B| = √24
|B| = 2√6
Step 2: Find the unit vector of A.
The unit vector of A is a vector with the same direction as A but with a magnitude of 1. To find the unit vector of A, divide each component of A by its magnitude |A|.
The magnitude of A is given by the formula |A| = √(Ax² + Ay² + Az²). From the given information, A = 2i + 3j - k. So, Ax = 2, Ay = 3, and Az = -1. Plugging these values into the magnitude formula, we get:
|A| = √(2² + 3² + (-1)²)
|A| = √(4 + 9 + 1)
|A| = √14
Dividing each component of A by its magnitude, we get:
Aunit = (2/√14)i + (3/√14)j + (-1/√14)k
Step 3: Calculate the scalar value to scale the unit vector of A to have a magnitude of B.
The scalar value required can be found by dividing |B| by |Aunit|, since the scalar magnitude scales the unit vector Aunit to have the magnitude of B.
Scalar = |B| / |Aunit|
Scalar = (2√6) / √14
Scalar = 2√(6/14)
Scalar = 2√(3/7)
Step 4: Multiply the unit vector of A by the scalar value.
Multiplying the unit vector of A by the scalar value, we get:
x = (2√(3/7))(2/√14)i + (2√(3/7))(3/√14)j + (2√(3/7))(-1/√14)k
x = (√12/7)i + (√36/7)j - (√3/7)k
x = (√12/7)(2i + 3j) - (√3/7)k
Therefore, the vector x parallel to A but with a magnitude of B is (√12/7)(2i + 3j) - (√3/7)k.