Question

A vector $\vec d$ has magnitude of $2.5m$ and points north. What are (a) the magnitude and (b) the direction of $4.0\vec d$ ? What are (c) the magnitude and (d) the direction of $ - 3.0\vec d$ ?

Answer 1

Hint: In order to solve this question we need to understand scalars and vectors. Scalar quantities are those quantities which have only magnitude defined and have no preferred direction, for example distance is a scalar quantity, and power, work are scalar quantities. Vector quantities are those quantities which have both magnitude and direction. Here we will solve for magnitude and direction of asked vectors with respect to the given vector which is in the North direction.

Complete step by step answer:
Initial magnitude of vector $\vec d$ is $\left| {\vec d} \right| = 2.5\,m$.
So for part (a), Let the new vector be $\vec D = 4.0\vec d$.
So magnitude of this vector is given by, $\left| {\vec D} \right| = \left| {4\vec d} \right|$
Solving further we get, $\left| {\vec D} \right| = 4\left| {\vec d} \right|$
Putting value we get, $\left| {\vec D} \right| = 4 \times 2.5m$
$\left| {\vec D} \right| = 10m$
So the magnitude of this vector is $10\,m$.

For (b) part, Direction of $\vec D$ is given by, $\hat D = \dfrac{{\vec D}}{{\left| {\vec D} \right|}}$
$\hat D = \dfrac{{4\vec d}}{{10}}$
$\Rightarrow \hat D = \dfrac{{2\vec d}}{5}$
Direction of this vector is given as the unit vector direction, $\hat D = \dfrac{{2\vec d}}{5}$ pointing towards North.

So for part (c), Let the new vector be ${\vec D_1} = - 3.0\vec d$.
So the magnitude of this vector is given by,
$\left| {{{\vec D}_1}} \right| = \left| { - 3.0\vec d} \right|$
Solving further we get,
$\left| {{{\vec D}_1}} \right| = 3.0\left| {\vec d} \right|$
Putting value we get,
$\left| {{{\vec D}_1}} \right| = 3 \times 2.5m$
$\Rightarrow \left| {{{\vec D}_1}} \right| = 7.5m$
So the magnitude of this vector is $7.5\,m$.

For (d) part, Direction of ${\vec D_1}$ is given by,
${\hat D_1} = \dfrac{{{{\vec D}_1}}}{{\left| {{{\vec D}_1}} \right|}}$
$\Rightarrow {\hat D_1} = \dfrac{{ - 3\vec d}}{{7.5}}$
$\therefore {\hat D_1} = \dfrac{{ - \vec d}}{{2.5}}$
Direction of this vector is given as unit vector direction, ${\hat D_1} = \dfrac{{ - \vec d}}{{2.5}}$ pointing towards South.

Note: It should be remembered that, here negative sign in direction indicates that given vector is in direction opposite to that of North direction which is South and positive sign indicates that vector is in same direction of given vector which is in North direction. And unit vector is always used to represent the direction of a vector and its magnitude is always one unit.