Question

Find the vector whose magnitude is 5 and which is in the direction of the vector 4i-3j+k.

Answer 1

Answer:

4i - 3j + 0k

Explanation:

Let the vector be 4i-3j+xk

Magnitude of a vector ai+bj+ck = sqrt (a^2 + b^2 + c^2)

Therefore the magnitude of the given vector = sqrt (4^2 + (-3)^2 + x^2)

                                                                     =  sqrt (16+9+x)

                                                                     =  sqrt (25+x)

                                                                     = 5

(We know that 5 is the square root of 25. So 'x' should be 0 to get it.)

So the vector should be 4i-3j+0k in to obtain the magnitude as 5.