Question

If vector A = 6i - 2j + 3k, determine
(a) A vector in the same direction as A with magnitude 2A
(b) A unit vector in the direction of A
(c) a vector opposite to A with magnitude of 4 m​

Answer 1

Answer:

(a) $2\vec A=12\hat i-4\hat j+6\hat k$

(b) $\displaystyle \vec{U_A}=12/7\hat i-4/7\hat j+6/7\hat k$

(c) $-4\vec{U_A}=-48/7\hat i+16/7\hat j-24/7\hat k$

Explanation:

Vectors

Given a vector

$\vec A=6\hat i-2\hat j+3\hat k$

We must determine the following:

a) A vector in the same direction as A with double magnitude 2A.

If the vector goes in the same direction but has a different magnitude, we only need to multiply each component by a common factor, in this case, by 2. Thus, the required vector is:

$2\vec A=12\hat i-4\hat j+6\hat k$

b) A unit vector in the same direction of A.

The unit vector needs to compute the magnitude of the vector:

$\mid A\mid=\sqrt{6^2+2^2+3^2}$

$\mid A\mid=\sqrt{36+4+9}=\sqrt{49}=7$

$\mid A\mid=7$

The unit vector is:

$\displaystyle \vec{U_A}=\frac{\vec A}{\mid \vec A\mid}$

$\displaystyle \vec{U_A}=\frac{12\hat i-4\hat j+6\hat k}{7}$

$\displaystyle \vec{U_A}=12/7\hat i-4/7\hat j+6/7\hat k$

c) A vector opposite to A with magnitude 4 m. We assume the original vector is also expressed in m.

The opposite vector to A is obtained simply by multiplying the unit vector by -1. To make its magnitude equal to 4, also multiply by 4. In all, we multiply the unit vector by -4:

$-4\vec{U_A}=-4(12/7\hat i-4/7\hat j+6/7\hat k)$

$-4\vec{U_A}=-48/7\hat i+16/7\hat j-24/7\hat k$