Question

The position vector of a particle is given by:
$\sf{\hat{r} = 0.6 \hat{i} \: + \: b \hat{j} \: + \: 0.8 \hat{k}}$
Find the value of "b"​

Answer 1

${\bold{\underline{\underline{Answer:}}}}$

${\bold{\therefore b=0}}$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\underline \bold{given : } \\ \implies \hat{r} = 0.6 \:\hat{i} + b \:\hat{j} + 0.8 \:\hat{k} \\ \\ \underline \bold{to \: find : } \\ \implies b = ?$

• According to given question :

$\implies \hat{r} = 0.6 \:\hat{i} + b \:\hat{j} + 0.8 \:\hat{j} \\ \\ \implies | \hat{r}| = |0.6 \:\hat{i} + b\: \hat{j} + 0.8 \:\hat{j}| \\ \\ \bold{ | \hat{r}| = 1} \\ \\ \implies {1}^{2} =({0.6})^{2} + {b}^{2} +( 0.8)^{2} \\ \\ \implies 0.36 + {b}^{2} + 0.64 = 1 \\ \\ \implies {b}^{2} + 1 = 1 \\ \\ \implies {b}^{2} = \cancel1 - \cancel1 \\ \\ \implies {b}^{2} = 0 \\ \\ \bold{\implies b = 0}$

Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

$\large{ \hat{r} = 0.6 \hat{i} + b \hat{j} + 0.8 \hat{k}}$

$\huge{\bold{\underline{Explanation:-}}}$

As here ${ \hat{r}}$ is given and it is a unit vector .

Taking Magnitude of unit vector gives a value of 1.

Glance on unit vector:-

A unit vector is a vector which has a magnitude of 1.

Also,

${\hat{i}, \hat{j} , \hat{k}}$ are unit vectors along x, y , and z axis.

Now,

$\large{ \hat{r} = 0.6 \hat{i} + b \hat{j} + 0.8 \hat{k}}$

Taking Magnitude,

$\large{ \implies| \hat{r} | = \sqrt{(0.6)^2 + (b)^2 + (0.8)^2}}$

Unit vector magnitude = 1.

$\large{ \implies 1 = \sqrt{0.36 + b^2 + 0.64}}$

$\large{ \implies 1 = \sqrt{1.00 + b^2}}$

Squaring on both sides,

$\large{ \implies(1)^2 = ( \sqrt{1.00 + b^2})}$

It becomes,

$\large{ \implies 1 = 1 + b^2}$

$\large{ \implies b^2 = 1 - 1}$

$\large{ \implies b^2 = 0}$

$\huge{\boxed{\boxed{b = 0}}}$

So,the value of b is "0".